Re: [algogeeks] Re: Amazon Interview Question

[Gary Drocella]

Counting sort does not run in O(1) space though.  Optimally it will run in 
O(K) space, where A is an array of integer numbers and K = max(A) - min(A)

On Saturday, February 9, 2013 9:52:01 PM UTC-5, Mohanabalan wrote:

 can use counting sort


 On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santosh...@gmail.comjavascript:
  wrote:

 If we can retrieve ith prime efficiently, we can do the following...
 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
 3.repeat it till end 



 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
  
 Given an array of integers where some numbers repeat once, some numbers 
 repeat twice and only one number repeats thrice, how do you find the number 
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution? 
 No hashmaps please!

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Re: [algogeeks] Re: Amazon Interview Question

[Gary Drocella]

A bit vector is basically just a sequence of bits such as a word or even an 
array of words.  Here is an example...
 
int x = 5;   // 32 bit word size on Intel IA-32 Architecture In C 
programming language.
 
A variable in C will be either located in a register in memory or in Main 
Memory.  You can even have bit vector data structures
reside on the hard disk. It just looks like this
 
10010101010010101...
 
If I'm interpreting the solution properly you will index into the bit 
vector based on the number found in the array A that is
 
for j = 1... A.size
currentNumber = A[j]  // The jth array element
bitVector[currentNumber] = ~ bitVector[currentNumber]   ; // Toggle the 
currentNumber bit
end for
 
Although this is a big memory save, you're still not using constant memory 
because the size of your memory is dependent on the size of a word let's 
say k
 
In worst case the space complexity will be 2*2^k = O(2^(K+1))

On Wednesday, April 3, 2013 9:58:23 AM UTC-8, ashekhar007 wrote:

 hi sourab please explain what bit vector1 and bit vector 2 really are can 
 you give an example please?

 On Saturday, February 16, 2013 11:20:59 PM UTC+5:30, sourabh wrote:

 you can solve this problem using bitvector/bitset.

 first scan :
 scan the array set the bit on odd occurrence and unset on even or 
 0 occurrence.

 second scan :
 shift all the odd occurring elements in beginning of array and even 
 towards end.

 third scan : till end of odd occurring elements.
 take another bit vector 
 on first occurence :if bit is set in bv1 then unset it and set it in bv2.
 on second occurence : if bv1 is not set and bv2 is set then these are the 
 array elements occuring 3rd time. 





 On Wed, Feb 13, 2013 at 9:27 PM, prakhar singh prakhars...@gmail.comwrote:

 Yes, thats a valid point Don.
 Thats what i meant when i wrote  //is that correct? in the comments on 
 the array line in code.


  int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?


 On Wed, Feb 13, 2013 at 9:09 PM, Don dond...@gmail.com wrote:

 The xor approach only works if there are no values which occur only
 once. But the problem statement indicates that some numbers occur
 once, some occur twice, and one occurs three times. So you will end up
 with prod equal to the xor of all of the values which occur once or
 three times. Put that in your input array and you'll find that you
 don't get the desired output.

 I don't know of a solution better than sorting and scanning the array.

 Don

 On Feb 12, 3:14 pm, prakhar singh prakharsngh.1...@gmail.com wrote:
  #includestdio.h
 
  int main()
  {
 int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
 int prod=a[0];int i;
 for(i=1;i(int)sizeof(a)/sizeof(a[0]);i++)
 {
prod ^= a[i];
 }
 printf(%d\n,prod);   //outputs 3, algorithm works as Sachin 
 described
  it;
 
  }
 
  On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
  sachinchital...@gmail.comwrote:
 
 
 
 
 
 
 
   use ex-or operation for all array elements..
   a^a=0
   a^a^a=a
 
   On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohanabala...@gmail.comwrote:
 
   can use counting sort
 
   On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santoshthot...@gmail.comwrote:
 
   If we can retrieve ith prime efficiently, we can do the 
 following...
   1.maintain a prod=1, start from 1st element, say a[0]=n find n th 
 prime
   2.check if (prod% (ith_prime * ith_prime )==0) then return i;
  else prod=prod*ith_prime;
   3.repeat it till end
 
   On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
 
   Given an array of integers where some numbers repeat once, some 
 numbers
   repeat twice and only one number repeats thrice, how do you find 
 the number
   that gets repeated 3 times?
 
   Does this problem have an O(n) time and O(1) space solution?
   No hashmaps please!
 
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   Contact# : +91 8086284349, 9892159511
   Oracle Corporation
 
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Re: [algogeeks] Re: Amazon Interview Question

[Parag Khanna]

use XOR



On Tue, Apr 30, 2013 at 6:12 AM, Gary Drocella gdroc...@gmail.com wrote:

 This will only work if each element in the array are relatively prime to
 one another, that is for any two elements x, y in array A the gcd(x,y) = 1,
 which is also just another way of saying no number divides another number
 in the array.  Once this rule is broken, then
 the algorithm will no longer work.  Here is a counter example

 A = { 4,3,4,2,4,2 } = {2^2, 3, 2^2, 2, 2^2, 2}

 You might be able to see now why this algorithm breaks down.  It is
 because the final product = 2^8*3^1 and of course 2^3 will easily divide
 this number, but would return the wrong solution.  It was of course a very
 good try!

 On Thursday, July 12, 2012 11:46:50 PM UTC-8, jatin wrote:


 1)Find product of the array and store it in say prod  o(n) and o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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Parag Khanna

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Re: [algogeeks] Re: Amazon interview question

[BackBencher]

:O the final required sum would be 4C0 * a5 - 4C1 * a4 + 4C2 * a3 - 4C3 * 
a2 + a1 how ? , did i missed something id yes can you paste the link or 
explain ?


Thanks
Shashank

On Wednesday, April 10, 2013 5:09:41 AM UTC+5:30, Shachindra A C wrote:

 Consider n = 5. Naming the array elements as a1,a2,a3,a4,a5 , the final 
 required sum would be 4C0 * a5 - 4C1 * a4 + 4C2 * a3 - 4C3 * a2 + a1.

 That is nothing but the pattern of a binomial expansion. Using this 
 method, the complexity can be reduced to O(n).

 Correct me if I'm wrong!

 On Tue, Apr 9, 2013 at 1:51 PM, Shashwat Kumar 
 shashwa...@gmail.comjavascript:
  wrote:

 Recursion and iteration don't differ in this algorithm. But avoid using 
 recursion if same can be done using iteration. In practical cases, system 
 does not allow very large depth in recursion. So for large values of n, 
 there can occur segmentation fault.


 On Tue, Apr 9, 2013 at 11:43 AM, rahul sharma 
 rahul2...@gmail.comjavascript:
  wrote:

 If you have any other solution ..please post that...i thnik recursion is 
 ok with base case...we need to scan again after first iteration...??


 On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma 
 rahul2...@gmail.comjavascript:
  wrote:

 i forgot to add base case..can add wen 2 elemnts are there then there 
 sum is stored and we reurn from there...i m in hurry,,,sry for that,,


 On Wed, Apr 10, 2013 at 12:11 AM, Don dond...@gmail.com 
 javascript:wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is 
 executed n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and 
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of 
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 
 then how???

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 -- 
 Shashwat Kumar
 Third year Undergraduate student
 Department of Computer Science and Engineering
 IIT Kharagpur




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 Regards,
 Shachindra A C
  

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Re: [algogeeks] Re: Amazon interview question

[pankaj joshi]

in this Problem if the array is
A[n] = {a0,a1,a(n-1),a(n)}

after the second iteration,
the value will be
{a0 -2*a2+a3, a2 -2*a3 + a4, a3-2*a4+a5,, a(n-2)-2a(n-1)+a(n)}

if we add all these terms then
all the middle elements will be canceled out so the remaining will be.

{a0-a2  - a(n-1)+a(n)}
which can be written as a general form
solution: - {a(n)-a(n-1)} - {a2-a0}

this will solve the problem in time complexity in O(1).

example:  A = {5, 3, 8, 9, 16}
solution : - {16-9}-{3-5}
{7} - {-2}
 9


On Wed, Apr 10, 2013 at 7:03 AM, Shashwat Anand m...@shashwat.me wrote:

  On 4/10/13 12:13 AM, rahul sharma wrote:

 If you have any other solution ..please post that...i thnik recursion is
 ok with base case...we need to scan again after first iteration...??

 First of all, the array size and number of iteration both won't be N or
 else the answer will always be 0.
 I take the following assumption, array have K elements and number of
 iteration is N.
 Now, if N = K, the return value will always be 0.
 For rest, we can decompose the array following the rule of adjacent
 element difference.

 Solution with O(NK) time complexity follows:

 int
 doit (vector int v, int N) {
 int k = (int) v.size () - 1;
 if (N  k) return 0;
 int c = 1;
 while (N--) {
 for (int i = k; i = c; i--)
 v [i] -= v [i - 1];
 for (int i = 0; i  c; i++)
 v [i] = 0;
 c++;
 }
 return accumulate (v.begin (), v.end (), 0);
 }

 int
 main ()

 int a [] = { 5, 3, 8, 9, 16 };
 vector int v (a, a + sizeof (a) / sizeof (a [0]));
 assert (doit (v, 0) == 41);
 assert (doit (v, 1) == 11);
 assert (doit (v, 2) == 9);
 assert (doit (v, 3) == -1);
 assert (doit (v, 4) == 21);
 assert (doit (v, 5) == 0);

 return 0;
 }

 However, I *strongly believe* the solution can be done in *linear time*.
 To code this with quadratic time complexity is a no-brainer.

 So, I took the array with K = 6 elements and decomposed.

 N = 0: [a1, a2, a3, a4, a5, a6]  = a1 + a2 + a3 + a4 + a4 + a6
 N = 1: [0, a2 - a1, a3 - a2, a4 - a3, a5 - a4, a6 - a5] = a6 - a1
 N = 2: [0, 0, a3 - 2a2 + a1, a4 - 2a3 + a2, a5 - 2a4 + a3, a6 - 2a5 + a4]
 = a6 - a5 - a2 + a1 = (a6 - a5) - (a2 - a1)
 N = 3: [0, 0, 0, a4 - 3a3 +3a2 - a1, a5 - 3a4 + 3a3 - a2, a6 - 3a5 + 3a4 -
 a3] = a6 - 2a5 +a4 - a3 + 2a2 - a1
 N = 4: [0, 0, 0, 0, a5 - 4a4 + 6a3 - 4a2 + a1, a6 - 4a5 + 6a4 - 4a3 + a2]
 = a6 - 3a5 + 2a4 + 2a3 - 3a2 + a1
 N = 5: [0, 0, 0, 0, 0, a6 - 5a5 + 10a4 - 10a3 + 5a2 - a1] = a6 - 5a5 +
 10a4 - 10a3 + 5a2 - a1
 N = 6: [0, 0, 0, 0, 0, 0] = 0

 The resulting solution does show some property of Binomial coefficient as
 pointed out by @Don in his hint (Pascal triangle).  I suppose this shall be
 the way to attack this problem.



 On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma rahul23111...@gmail.comwrote:

 i forgot to add base case..can add wen 2 elemnts are there then there sum
 is stored and we reurn from there...i m in hurry,,,sry for that,,


 On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is
 executed n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 then
 how???

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Re: [algogeeks] Re: Amazon interview question

[Shashwat Anand]

On 4/13/13 10:05 PM, pankaj joshi wrote:

in this Problem if the array is
A[n] = {a0,a1,a(n-1),a(n)}

after the second iteration,
the value will be
{a0 -2*a2+a3, a2 -2*a3 + a4, a3-2*a4+a5,, a(n-2)-2a(n-1)+a(n)}

if we add all these terms then
all the middle elements will be canceled out so the remaining will be.

{a0-a2  - a(n-1)+a(n)}
which can be written as a general form
solution: - {a(n)-a(n-1)} - {a2-a0}
this will solve the problem in time complexity in O(1).

example:  A = {5, 3, 8, 9, 16}
solution : - {16-9}-{3-5}
  {7} - {-2}
   9


No.

Your assertion is valid only for first two iteration.
Iteration 1: { a1 - a0, a2 - a1, ... an - an-1 } = an - a0
Iteration 2: { a2 - 2a1 + a0,  an - 2an-1 + an-2 } = (an - an-1) - 
(a1 - a0)
Iteration 3: an - 2an-1 +an-1 - a2 + 2a1 - a0, thereby nullifying the 
above observation.


At any state T, we need to find f (n, T) where coefficient of T is 0, to 
make it O(1) solution.

That does not seems to be the case here.

Say, for 4th iteration, what will be your answer ?  I don't see any 
closed form here.




On Wed, Apr 10, 2013 at 7:03 AM, Shashwat Anand m...@shashwat.me 
mailto:m...@shashwat.me wrote:


On 4/10/13 12:13 AM, rahul sharma wrote:

If you have any other solution ..please post that...i thnik
recursion is ok with base case...we need to scan again after
first iteration...??

First of all, the array size and number of iteration both won't be
N or else the answer will always be 0.
I take the following assumption, array have K elements and number
of iteration is N.
Now, if N = K, the return value will always be 0.
For rest, we can decompose the array following the rule of
adjacent element difference.

Solution with O(NK) time complexity follows:

int
doit (vector int v, int N) {
int k = (int) v.size () - 1;
if (N  k) return 0;
int c = 1;
while (N--) {
for (int i = k; i = c; i--)
v [i] -= v [i - 1];
for (int i = 0; i  c; i++)
v [i] = 0;
c++;
}
return accumulate (v.begin (), v.end (), 0);
}

int
main ()

int a [] = { 5, 3, 8, 9, 16 };
vector int v (a, a + sizeof (a) / sizeof (a [0]));
assert (doit (v, 0) == 41);
assert (doit (v, 1) == 11);
assert (doit (v, 2) == 9);
assert (doit (v, 3) == -1);
assert (doit (v, 4) == 21);
assert (doit (v, 5) == 0);

return 0;
}

However, I /strongly believe/ the solution can be done in *linear
time*.  To code this with quadratic time complexity is a no-brainer.

So, I took the array with K = 6 elements and decomposed.

N = 0: [a1, a2, a3, a4, a5, a6]  = a1 + a2 + a3 + a4 + a4 + a6
N = 1: [0, a2 - a1, a3 - a2, a4 - a3, a5 - a4, a6 - a5] = a6 - a1
N = 2: [0, 0, a3 - 2a2 + a1, a4 - 2a3 + a2, a5 - 2a4 + a3, a6 -
2a5 + a4] = a6 - a5 - a2 + a1 = (a6 - a5) - (a2 - a1)
N = 3: [0, 0, 0, a4 - 3a3 +3a2 - a1, a5 - 3a4 + 3a3 - a2, a6 - 3a5
+ 3a4 - a3] = a6 - 2a5 +a4 - a3 + 2a2 - a1
N = 4: [0, 0, 0, 0, a5 - 4a4 + 6a3 - 4a2 + a1, a6 - 4a5 + 6a4 -
4a3 + a2] = a6 - 3a5 + 2a4 + 2a3 - 3a2 + a1
N = 5: [0, 0, 0, 0, 0, a6 - 5a5 + 10a4 - 10a3 + 5a2 - a1] = a6 -
5a5 + 10a4 - 10a3 + 5a2 - a1
N = 6: [0, 0, 0, 0, 0, 0] = 0

The resulting solution does show some property of Binomial
coefficient as pointed out by @Don in his hint (Pascal triangle). 
I suppose this shall be the way to attack this problem.



On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma
rahul23111...@gmail.com mailto:rahul23111...@gmail.com wrote:

i forgot to add base case..can add wen 2 elemnts are there
then there sum is stored and we reurn from there...i m in
hurry,,,sry for that,,


On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com
mailto:dondod...@gmail.com wrote:

It is O(N^2) because the inner loop takes N steps to
execute and that
loop will be executed N times.

However, I would suggest not using recursion. There is no
reason to
not do it iteratively. Your recursive solution has no
base case so it
will recurse until your computer runs out of stack space,
at which
point it will crash.

Don

On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com
mailto:rahul23111...@gmail.com wrote:
  A = {5, 3, 8, 9, 16}
 After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
 After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
 Given an array, return sum after n iterations

 my sol/
 void abc(int arr[],n)
 {
 for(i=0;in;i++)
 arr[i]=arr[i+1]-arr[i];
 abc(arr,n-1);

 }

   

Re: [algogeeks] Re: Amazon interview question

[rahul sharma]

i forgot to add base case..can add wen 2 elemnts are there then there sum
is stored and we reurn from there...i m in hurry,,,sry for that,,


On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is executed
 n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 then
 how???

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Re: [algogeeks] Re: Amazon interview question

[rahul sharma]

If you have any other solution ..please post that...i thnik recursion is ok
with base case...we need to scan again after first iteration...??


On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma rahul23111...@gmail.comwrote:

 i forgot to add base case..can add wen 2 elemnts are there then there sum
 is stored and we reurn from there...i m in hurry,,,sry for that,,


 On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is
 executed n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 then
 how???

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Re: [algogeeks] Re: Amazon interview question

[Shashwat Kumar]

Recursion and iteration don't differ in this algorithm. But avoid using
recursion if same can be done using iteration. In practical cases, system
does not allow very large depth in recursion. So for large values of n,
there can occur segmentation fault.


On Tue, Apr 9, 2013 at 11:43 AM, rahul sharma rahul23111...@gmail.comwrote:

 If you have any other solution ..please post that...i thnik recursion is
 ok with base case...we need to scan again after first iteration...??


 On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma rahul23111...@gmail.comwrote:

 i forgot to add base case..can add wen 2 elemnts are there then there sum
 is stored and we reurn from there...i m in hurry,,,sry for that,,


 On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is
 executed n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 then
 how???

 --
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-- 
Shashwat Kumar
Third year Undergraduate student
Department of Computer Science and Engineering
IIT Kharagpur

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Re: [algogeeks] Re: Amazon interview question

[Shachindra A C]

Consider n = 5. Naming the array elements as a1,a2,a3,a4,a5 , the final
required sum would be 4C0 * a5 - 4C1 * a4 + 4C2 * a3 - 4C3 * a2 + a1.

That is nothing but the pattern of a binomial expansion. Using this method,
the complexity can be reduced to O(n).

Correct me if I'm wrong!

On Tue, Apr 9, 2013 at 1:51 PM, Shashwat Kumar shashwatkmr@gmail.comwrote:

 Recursion and iteration don't differ in this algorithm. But avoid using
 recursion if same can be done using iteration. In practical cases, system
 does not allow very large depth in recursion. So for large values of n,
 there can occur segmentation fault.


 On Tue, Apr 9, 2013 at 11:43 AM, rahul sharma rahul23111...@gmail.comwrote:

 If you have any other solution ..please post that...i thnik recursion is
 ok with base case...we need to scan again after first iteration...??


 On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma 
 rahul23111...@gmail.comwrote:

 i forgot to add base case..can add wen 2 elemnts are there then there
 sum is stored and we reurn from there...i m in hurry,,,sry for that,,


 On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com wrote:

 It is O(N^2) because the inner loop takes N steps to execute and that
 loop will be executed N times.

 However, I would suggest not using recursion. There is no reason to
 not do it iteratively. Your recursive solution has no base case so it
 will recurse until your computer runs out of stack space, at which
 point it will crash.

 Don

 On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com wrote:
   A = {5, 3, 8, 9, 16}
  After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
  After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
  Given an array, return sum after n iterations
 
  my sol/
  void abc(int arr[],n)
  {
  for(i=0;in;i++)
  arr[i]=arr[i+1]-arr[i];
  abc(arr,n-1);
 
  }
 
  I wana ask that the complexity is o(n) or o(n)2..as loop is
 executed n
  times..say n is 10...so fxn is called 10 timesi.e  10 n..and
 ignoring n
  it comes out to be...n..but if we implemeted with 2 loops then
  complexity is n2 ...and both sol are taking same no of
 iterations...please
  tell whether complexity is n or n2 for above codeif it is n2 then
 how???

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 --
 Shashwat Kumar
 Third year Undergraduate student
 Department of Computer Science and Engineering
 IIT Kharagpur




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Regards,
Shachindra A C

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Re: [algogeeks] Re: Amazon interview question

[Shashwat Anand]

On 4/10/13 12:13 AM, rahul sharma wrote:
If you have any other solution ..please post that...i thnik recursion 
is ok with base case...we need to scan again after first iteration...??
First of all, the array size and number of iteration both won't be N or 
else the answer will always be 0.
I take the following assumption, array have K elements and number of 
iteration is N.

Now, if N = K, the return value will always be 0.
For rest, we can decompose the array following the rule of adjacent 
element difference.


Solution with O(NK) time complexity follows:

int
doit (vector int v, int N) {
int k = (int) v.size () - 1;
if (N  k) return 0;
int c = 1;
while (N--) {
for (int i = k; i = c; i--)
v [i] -= v [i - 1];
for (int i = 0; i  c; i++)
v [i] = 0;
c++;
}
return accumulate (v.begin (), v.end (), 0);
}

int
main ()

int a [] = { 5, 3, 8, 9, 16 };
vector int v (a, a + sizeof (a) / sizeof (a [0]));
assert (doit (v, 0) == 41);
assert (doit (v, 1) == 11);
assert (doit (v, 2) == 9);
assert (doit (v, 3) == -1);
assert (doit (v, 4) == 21);
assert (doit (v, 5) == 0);

return 0;
}

However, I /strongly believe/ the solution can be done in *linear 
time*.  To code this with quadratic time complexity is a no-brainer.


So, I took the array with K = 6 elements and decomposed.

N = 0: [a1, a2, a3, a4, a5, a6]  = a1 + a2 + a3 + a4 + a4 + a6
N = 1: [0, a2 - a1, a3 - a2, a4 - a3, a5 - a4, a6 - a5] = a6 - a1
N = 2: [0, 0, a3 - 2a2 + a1, a4 - 2a3 + a2, a5 - 2a4 + a3, a6 - 2a5 + 
a4] = a6 - a5 - a2 + a1 = (a6 - a5) - (a2 - a1)
N = 3: [0, 0, 0, a4 - 3a3 +3a2 - a1, a5 - 3a4 + 3a3 - a2, a6 - 3a5 + 3a4 
- a3] = a6 - 2a5 +a4 - a3 + 2a2 - a1
N = 4: [0, 0, 0, 0, a5 - 4a4 + 6a3 - 4a2 + a1, a6 - 4a5 + 6a4 - 4a3 + 
a2] = a6 - 3a5 + 2a4 + 2a3 - 3a2 + a1
N = 5: [0, 0, 0, 0, 0, a6 - 5a5 + 10a4 - 10a3 + 5a2 - a1] = a6 - 5a5 + 
10a4 - 10a3 + 5a2 - a1

N = 6: [0, 0, 0, 0, 0, 0] = 0

The resulting solution does show some property of Binomial coefficient 
as pointed out by @Don in his hint (Pascal triangle).  I suppose this 
shall be the way to attack this problem.



On Wed, Apr 10, 2013 at 12:12 AM, rahul sharma 
rahul23111...@gmail.com mailto:rahul23111...@gmail.com wrote:


i forgot to add base case..can add wen 2 elemnts are there then
there sum is stored and we reurn from there...i m in hurry,,,sry
for that,,


On Wed, Apr 10, 2013 at 12:11 AM, Don dondod...@gmail.com
mailto:dondod...@gmail.com wrote:

It is O(N^2) because the inner loop takes N steps to execute
and that
loop will be executed N times.

However, I would suggest not using recursion. There is no
reason to
not do it iteratively. Your recursive solution has no base
case so it
will recurse until your computer runs out of stack space, at which
point it will crash.

Don

On Apr 9, 2:29 pm, rahul sharma rahul23111...@gmail.com
mailto:rahul23111...@gmail.com wrote:
  A = {5, 3, 8, 9, 16}
 After one iteration A = {3-5,8-3,9-8,16-9}={-2,5,1,7}
 After second iteration A = {5-(-2),1-5,7-1} sum =7+(-4)+6=9
 Given an array, return sum after n iterations

 my sol/
 void abc(int arr[],n)
 {
 for(i=0;in;i++)
 arr[i]=arr[i+1]-arr[i];
 abc(arr,n-1);

 }

 I wana ask that the complexity is o(n) or o(n)2..as loop
is executed n
 times..say n is 10...so fxn is called 10 timesi.e  10
n..and ignoring n
 it comes out to be...n..but if we implemeted with 2
loops then
 complexity is n2 ...and both sol are taking same no of
iterations...please
 tell whether complexity is n or n2 for above codeif it
is n2 then how???

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Re: [algogeeks] Re: Amazon Interview Question

[ashekhar007]

hi sourab please explain what bit vector1 and bit vector 2 really are can 
you give an example please?

On Saturday, February 16, 2013 11:20:59 PM UTC+5:30, sourabh wrote:

 you can solve this problem using bitvector/bitset.

 first scan :
 scan the array set the bit on odd occurrence and unset on even or 
 0 occurrence.

 second scan :
 shift all the odd occurring elements in beginning of array and even 
 towards end.

 third scan : till end of odd occurring elements.
 take another bit vector 
 on first occurence :if bit is set in bv1 then unset it and set it in bv2.
 on second occurence : if bv1 is not set and bv2 is set then these are the 
 array elements occuring 3rd time. 





 On Wed, Feb 13, 2013 at 9:27 PM, prakhar singh 
 prakhars...@gmail.comjavascript:
  wrote:

 Yes, thats a valid point Don.
 Thats what i meant when i wrote  //is that correct? in the comments on 
 the array line in code.


  int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?


 On Wed, Feb 13, 2013 at 9:09 PM, Don dond...@gmail.com javascript:wrote:

 The xor approach only works if there are no values which occur only
 once. But the problem statement indicates that some numbers occur
 once, some occur twice, and one occurs three times. So you will end up
 with prod equal to the xor of all of the values which occur once or
 three times. Put that in your input array and you'll find that you
 don't get the desired output.

 I don't know of a solution better than sorting and scanning the array.

 Don

 On Feb 12, 3:14 pm, prakhar singh prakharsngh.1...@gmail.com wrote:
  #includestdio.h
 
  int main()
  {
 int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
 int prod=a[0];int i;
 for(i=1;i(int)sizeof(a)/sizeof(a[0]);i++)
 {
prod ^= a[i];
 }
 printf(%d\n,prod);   //outputs 3, algorithm works as Sachin 
 described
  it;
 
  }
 
  On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
  sachinchital...@gmail.comwrote:
 
 
 
 
 
 
 
   use ex-or operation for all array elements..
   a^a=0
   a^a^a=a
 
   On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohanabala...@gmail.comwrote:
 
   can use counting sort
 
   On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santoshthot...@gmail.comwrote:
 
   If we can retrieve ith prime efficiently, we can do the 
 following...
   1.maintain a prod=1, start from 1st element, say a[0]=n find n th 
 prime
   2.check if (prod% (ith_prime * ith_prime )==0) then return i;
  else prod=prod*ith_prime;
   3.repeat it till end
 
   On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
 
   Given an array of integers where some numbers repeat once, some 
 numbers
   repeat twice and only one number repeats thrice, how do you find 
 the number
   that gets repeated 3 times?
 
   Does this problem have an O(n) time and O(1) space solution?
   No hashmaps please!
 
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   Oracle Corporation
 
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 +91-8971547841 


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Re: [algogeeks] Re: Amazon Interview Question

[sourabh jain]

@sandeep he is talking about constant space.

On Tue, Mar 19, 2013 at 5:31 PM, sandeep kumar
sandeepkumar1...@gmail.comwrote:

 Hey what if while scanning through the array we create a BST n check
 simultaneously that :

 current node == current node's parent  current node == current node's
 left or right child


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Re: [algogeeks] Re: Amazon Interview Question

[sandeep kumar]

Hey what if while scanning through the array we create a BST n check
simultaneously that :

current node == current node's parent  current node == current node's
left or right child

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Re: [algogeeks] Re: Amazon Interview Question

[prakhar singh]

Yes, thats a valid point Don.
Thats what i meant when i wrote  //is that correct? in the comments on
the array line in code.

 int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?


On Wed, Feb 13, 2013 at 9:09 PM, Don dondod...@gmail.com wrote:

 The xor approach only works if there are no values which occur only
 once. But the problem statement indicates that some numbers occur
 once, some occur twice, and one occurs three times. So you will end up
 with prod equal to the xor of all of the values which occur once or
 three times. Put that in your input array and you'll find that you
 don't get the desired output.

 I don't know of a solution better than sorting and scanning the array.

 Don

 On Feb 12, 3:14 pm, prakhar singh prakharsngh.1...@gmail.com wrote:
  #includestdio.h
 
  int main()
  {
 int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
 int prod=a[0];int i;
 for(i=1;i(int)sizeof(a)/sizeof(a[0]);i++)
 {
prod ^= a[i];
 }
 printf(%d\n,prod);   //outputs 3, algorithm works as Sachin
 described
  it;
 
  }
 
  On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
  sachinchital...@gmail.comwrote:
 
 
 
 
 
 
 
   use ex-or operation for all array elements..
   a^a=0
   a^a^a=a
 
   On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohanabala...@gmail.comwrote:
 
   can use counting sort
 
   On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santoshthot...@gmail.comwrote:
 
   If we can retrieve ith prime efficiently, we can do the following...
   1.maintain a prod=1, start from 1st element, say a[0]=n find n th
 prime
   2.check if (prod% (ith_prime * ith_prime )==0) then return i;
  else prod=prod*ith_prime;
   3.repeat it till end
 
   On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
 
   Given an array of integers where some numbers repeat once, some
 numbers
   repeat twice and only one number repeats thrice, how do you find
 the number
   that gets repeated 3 times?
 
   Does this problem have an O(n) time and O(1) space solution?
   No hashmaps please!
 
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   Sachin Chitale
   Application Engineer SCJP, SCWCD
   Contact# : +91 8086284349, 9892159511
   Oracle Corporation
 
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Re: [algogeeks] Re: Amazon Interview Question

[sourabh jain]

you can solve this problem using bitvector/bitset.

first scan :
scan the array set the bit on odd occurrence and unset on even or
0 occurrence.

second scan :
shift all the odd occurring elements in beginning of array and even towards
end.

third scan : till end of odd occurring elements.
take another bit vector
on first occurence :if bit is set in bv1 then unset it and set it in bv2.
on second occurence : if bv1 is not set and bv2 is set then these are the
array elements occuring 3rd time.





On Wed, Feb 13, 2013 at 9:27 PM, prakhar singh
prakharsngh.1...@gmail.comwrote:

 Yes, thats a valid point Don.
 Thats what i meant when i wrote  //is that correct? in the comments on
 the array line in code.


  int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?


 On Wed, Feb 13, 2013 at 9:09 PM, Don dondod...@gmail.com wrote:

 The xor approach only works if there are no values which occur only
 once. But the problem statement indicates that some numbers occur
 once, some occur twice, and one occurs three times. So you will end up
 with prod equal to the xor of all of the values which occur once or
 three times. Put that in your input array and you'll find that you
 don't get the desired output.

 I don't know of a solution better than sorting and scanning the array.

 Don

 On Feb 12, 3:14 pm, prakhar singh prakharsngh.1...@gmail.com wrote:
  #includestdio.h
 
  int main()
  {
 int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
 int prod=a[0];int i;
 for(i=1;i(int)sizeof(a)/sizeof(a[0]);i++)
 {
prod ^= a[i];
 }
 printf(%d\n,prod);   //outputs 3, algorithm works as Sachin
 described
  it;
 
  }
 
  On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
  sachinchital...@gmail.comwrote:
 
 
 
 
 
 
 
   use ex-or operation for all array elements..
   a^a=0
   a^a^a=a
 
   On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohanabala...@gmail.comwrote:
 
   can use counting sort
 
   On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santoshthot...@gmail.comwrote:
 
   If we can retrieve ith prime efficiently, we can do the following...
   1.maintain a prod=1, start from 1st element, say a[0]=n find n th
 prime
   2.check if (prod% (ith_prime * ith_prime )==0) then return i;
  else prod=prod*ith_prime;
   3.repeat it till end
 
   On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
 
   Given an array of integers where some numbers repeat once, some
 numbers
   repeat twice and only one number repeats thrice, how do you find
 the number
   that gets repeated 3 times?
 
   Does this problem have an O(n) time and O(1) space solution?
   No hashmaps please!
 
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   Contact# : +91 8086284349, 9892159511
   Oracle Corporation
 
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Re: [algogeeks] Re: Amazon Interview Question

[BUBUN SHEKHAR]

Sachin Chitale,
Dude the xoring operation will give us xor of numbers with freq 1 and 3 
respectively. How do you filter out the number with the frequency 3??

On Tuesday, February 12, 2013 11:44:08 PM UTC+5:30, Sachin Chitale wrote:

 use ex-or operation for all array elements..
 a^a=0
 a^a^a=a


 On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohana...@gmail.comjavascript:
  wrote:

 can use counting sort


 On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santosh...@gmail.comjavascript:
  wrote:

 If we can retrieve ith prime efficiently, we can do the following...
 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
 3.repeat it till end



 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers 
 repeat twice and only one number repeats thrice, how do you find the 
 number 
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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 Contact# : +91 8086284349, 9892159511
 Oracle Corporation 


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Re: [algogeeks] Re: Amazon Interview Question

[sourabh jain]

Search for BitSet/BitVector in java .

On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
sachinchital...@gmail.comwrote:

 use ex-or operation for all array elements..
 a^a=0
 a^a^a=a


 On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohanabala...@gmail.comwrote:

 can use counting sort


 On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santoshthot...@gmail.comwrote:

 If we can retrieve ith prime efficiently, we can do the following...
 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
 3.repeat it till end



 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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 Regards,
 Sachin Chitale
 Application Engineer SCJP, SCWCD
 Contact# : +91 8086284349, 9892159511
 Oracle Corporation

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+91-8971547841

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Re: [algogeeks] Re: Amazon Interview Question

[arun singh chauhan]

@Sachin Chitale : Very good approach dude .
thumbs up +1

-- 
Arun Singh Chauhan
Engineer (RnD 2), Samsung Electronics 
Software Engineering Lab, Noida


On Tuesday, February 12, 2013 11:44:08 PM UTC+5:30, Sachin Chitale wrote:

 use ex-or operation for all array elements..
 a^a=0
 a^a^a=a


 On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B 
 mohana...@gmail.comjavascript:
  wrote:

 can use counting sort


 On Sun, Jul 15, 2012 at 6:37 PM, santosh thota 
 santosh...@gmail.comjavascript:
  wrote:

 If we can retrieve ith prime efficiently, we can do the following...
 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
 3.repeat it till end



 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers 
 repeat twice and only one number repeats thrice, how do you find the 
 number 
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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 Sachin Chitale
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 Contact# : +91 8086284349, 9892159511
 Oracle Corporation 


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Re: [algogeeks] Re: Amazon Interview Question

[Mohanabalan D B]

can use counting sort


On Sun, Jul 15, 2012 at 6:37 PM, santosh thota santoshthot...@gmail.comwrote:

 If we can retrieve ith prime efficiently, we can do the following...
 1.maintain a prod=1, start from 1st element, say a[0]=n find n th prime
 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
else prod=prod*ith_prime;
 3.repeat it till end



 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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Re: [algogeeks] Re: Amazon interview Question

[Ritesh Mishra]

@anurag : there is no need to SORT. as it will increase complexity O(n) to
O(n log n).
here is the correct code.. please look over it and notify me if i'm wrong .

T.C. = O( n )

// ex: 1 4 3 2 0 i'm explaining on behalf of it.
bool permute (int *arr , int N )
{
if ( N=1 ) return false;

for ( int i = N-2 ; i = 0 ; i-- )
{
if ( arr[i+1]  arr[i]) // now i points to 1
{
for ( int j = N-1 ; j  i ; j--)
{
if ( arr[i] arr[j]) // now j points to 2(just greater than
1 )
{
swap(arr[i],arr[j]) ;//this will generate 2 4 3 1 0
since 4310 are already sorted in reverse
 //thus no need to sort again in
inc order rather than reversing
i++ ; j = N-1;
while(ij)  swap(arr[i++],arr[j--]) ; // reversing the
seq 4..0 to 0..4
return true ;
}
}
}
}
return false ;
}


Regards,

Ritesh Kumar Mishra
Information Technology
Third Year Undergraduate
MNNIT Allahabad


On Mon, Dec 24, 2012 at 7:56 PM, marti amritsa...@gmail.com wrote:


 I REPEAT, THERE is no need to SORT;


 http://en.wikipedia.org/wiki/Permutation#Lexicographical%5Forder%5Fgeneration


 On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:

 For a given number, find the next greatest number which is just greater
 than previous one and made up of same digits.

  --




-- 

Re: [algogeeks] Re: Amazon interview Question

[Hariraman R]

Can anyone give me some idea if the given no is small like 12 then the next
one is 17


On Mon, Dec 24, 2012 at 7:56 PM, marti amritsa...@gmail.com wrote:


 I REPEAT, THERE is no need to SORT;


 http://en.wikipedia.org/wiki/Permutation#Lexicographical%5Forder%5Fgeneration


 On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:

 For a given number, find the next greatest number which is just greater
 than previous one and made up of same digits.

  --




-- 

Re: [algogeeks] Re: Amazon interview Question

[Anurag Gupta]

Hi Ritesh

Yeah, you are right. we do not need to sort. my bad
Thank you for explaining clearly


On Thu, Dec 27, 2012 at 4:29 AM, Ritesh Mishra rforr...@gmail.com wrote:

 @anurag : there is no need to SORT. as it will increase complexity O(n) to
 O(n log n).
 here is the correct code.. please look over it and notify me if i'm wrong .

 T.C. = O( n )

 // ex: 1 4 3 2 0 i'm explaining on behalf of it.
 bool permute (int *arr , int N )
 {
 if ( N=1 ) return false;

 for ( int i = N-2 ; i = 0 ; i-- )
 {
 if ( arr[i+1]  arr[i]) // now i points to 1
 {
 for ( int j = N-1 ; j  i ; j--)
 {
 if ( arr[i] arr[j]) // now j points to 2(just greater
 than 1 )
 {
 swap(arr[i],arr[j]) ;//this will generate 2 4 3 1 0
 since 4310 are already sorted in reverse
  //thus no need to sort again in
 inc order rather than reversing
 i++ ; j = N-1;
 while(ij)  swap(arr[i++],arr[j--]) ; // reversing the
 seq 4..0 to 0..4
 return true ;
 }
 }
 }
 }
 return false ;
 }


 Regards,

 Ritesh Kumar Mishra
 Information Technology
 Third Year Undergraduate
 MNNIT Allahabad


 On Mon, Dec 24, 2012 at 7:56 PM, marti amritsa...@gmail.com wrote:


 I REPEAT, THERE is no need to SORT;


 http://en.wikipedia.org/wiki/Permutation#Lexicographical%5Forder%5Fgeneration


 On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:

 For a given number, find the next greatest number which is just greater
 than previous one and made up of same digits.

  --




  --






-- 
Regards
Anurag Gupta
IV Year Computer Engineering
Delhi Technological University
(Formerly Delhi College of Engineering)

-- 

Re: [algogeeks] Re: Amazon interview Question

[Anurag Gupta]

@marti

your answer is completely wrong (check for 234987156221 ans is 2349871*61225
* whereas your answer would be 2349871*65225*)
and we do need to sort


On Mon, Dec 17, 2012 at 9:10 AM, marti amritsa...@gmail.com wrote:

 Yeah thanks Sandeep, theres an error in example...it should be
 5436827.However there is no need to sort.


 On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:

 For a given number, find the next greatest number which is just greater
 than previous one and made up of same digits.

  --






-- 
Regards
Anurag Gupta
IV Year Computer Engineering
Delhi Technological University
(Formerly Delhi College of Engineering)

-- 

Re: [algogeeks] Re: Amazon interview Question

[Shubham Sandeep]

hello all... anwer to previous example would be 5436827 instead of
5436872please correct it :)

On Sun, Dec 16, 2012 at 11:48 PM, marti amritsa...@gmail.com wrote:

 Here is what you do

 EG: 5436782

 ans is 5436872

 how did we arrive?

 FInd least index i, such that a[i-1] = a[i] starting from rigthmost
 5436782
   (8)
 Now , Find least index j such that a[j-1] = a[i-1]:

 5436782
(7)

 swap them

 = 5436872

 Now swap all values between i and j.

 Pseudocode in Python:

 *def get_next(a):*
 *N = len(a)*
 *i = len(a)-1*
 *
 *
 *while a[i-1] = a[i]:*
 *i -= 1*
 *
 *
 *j = N*
 *
 *
 *while a[j-1] = a[i-1]:*
 *j -= 1*
 *
 *
 *a[i-1], a[j-1] = a[j-1], a[i-1]*
 *
 *
 *i += 1*
 *j = N*
 *
 *
 *while i  j:*
 *a[i-1], a[j-1] = a[j-1], a[i-1]*
 *i+=1*
 *j-=1*
 *
 *
 *return a*


 Source:
 http://nicolas-lara.blogspot.in/2009/01/permutations.html


 On Friday, December 14, 2012 11:56:16 AM UTC+5:30, tapan rathi wrote:

 For a given number, find the next greatest number which is just greater
 than previous one and made up of same digits.

  --






-- 
Regards,
SHUBHAM SANDEEP
IT 3rd yr.
NIT ALD.

-- 

Re: [algogeeks] Re: Amazon interview Question

[ramchandra sankhala]

Let the number is stored in an array a[n] with MSB at index 0...
1. i = n-1;
   reduce  i  till a[i]=a[i-1]  i  0.

If here i =0 means give number is largest possible number made out of digits
otherwise i is pointing to a digit such that a[i]a[i+1]

2. find smallest digit from a[i+1 to n-1] just larger than a[i].
   lets assume it come out as a digit of index j.
3. swap a[i] and a[j].
Finally sort a[i+1 to n-1] in increasing order...

suppose a=5436782
so
after step 1, i=4
after step 2, j=5
after step 4, a=5436872
finally after sorting  a=5436827


--
Ram Chandra Sankhala | 3rd Year  | Computer Science  Engineering | IIT Roorkee

-- 

Re: [algogeeks] Re: Amazon Interview Question

[jatin]

ya that's right :( any other solution..
On Saturday, 14 July 2012 03:24:41 UTC+5:30, Dave wrote:

 @Jatin: Take the array {1, 2, 4, 8, ..., x, 2, 4, 8, ..., x, x}, where x 
 is the largest power of two that will fit in your integer data type. Here 1 
 occurs once, 2, 4, 8, ... each occur twice, and x occurs three times. 
 Ignoring that you can't calculate the product of the numbers due to 
 overflow, the cube of each number in the array will be a factor of the 
 product, so you will have to check every number to see if it occurs three 
 times. Those checks will be O(n), so the whole algorithm will be O(n^2). 
 Correct me if I am wrong.
  
 As another example, just include a zero in the array and see if your 
 algorithm still works as written.
  
 Dave

 On Friday, July 13, 2012 9:59:31 AM UTC-5, jatin sethi wrote:

  as long as we are using goto with proper comments to ensure that it 
 won't decrease the readability we can use them and ther's no harm in it! 
 Secondly the worst case for my algo is o(n) .?..correct me if i am wrong
 On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not 
 advisable. 
 Plus this will exceed O(n) in the worst case, as we may keep visiting 
 the goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step 
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt 
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.comwrote:

 @jatin:
 Your first method may be proved wrong. 
  
 Here is a counter test case:
  
 Suppose the array is:
  
 27 729 19683 2 3 3 27 3 81 729
  
 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs 
 twice, 27 occurs twice, and 3 occurs thrice.
  
 You are supposed to return 3 
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=product(say)
  
 Upon traversing the second time, it will return 27, as 
 product%(27*27*27) is equal to zero!
  
 regards.
  

  
 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)
  
 but this solution requires o(logn) space though. 

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:
  
  
 1)Find product of the array and store it in say prod  o(n) and 
 o(1) 
 2)now traverse the array and check if  
  
 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while
  
 //now check is this is the element that is occuring three times 
 o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some 
 numbers repeat twice and only one number repeats thrice, how do you 
 find 
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution? 
 No hashmaps please!

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Re: [algogeeks] Re: Amazon Interview Question

[adarsh kumar]

@jatin:
Your first method may be proved wrong.

Here is a counter test case:

Suppose the array is:

27 729 19683 2 3 3 27 3 81 729

Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice, 27
occurs twice, and 3 occurs thrice.

You are supposed to return 3
But as per your method, the product will be computed as
729*19683*2*3*3*27*3*81*729=product(say)

Upon traversing the second time, it will return 27, as product%(27*27*27)
is equal to zero!

regards.



On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

 --
 You received this message because you are subscribed to the Google Groups
 Algorithm Geeks group.
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Re: [algogeeks] Re: Amazon Interview Question

[vindhya chhabra]

@adarsh
But i think jatin has asked to check for the number( we achieved in step 1)
occuring thrice or not..and in this check 27 will rule out.but i doubt the
algo given by Jatin runs in O(n) time. please comment.

On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong.

 Here is a counter test case:

 Suppose the array is:

 27 729 19683 2 3 3 27 3 81 729

 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice,
 27 occurs twice, and 3 occurs thrice.

 You are supposed to return 3
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=product(say)

 Upon traversing the second time, it will return 27, as product%(27*27*27)
 is equal to zero!

 regards.



 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

 --
 You received this message because you are subscribed to the Google Groups
 Algorithm Geeks group.
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 https://groups.google.com/d/msg/algogeeks/-/lmzhJ-GSdigJ.

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-- 
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RE: [algogeeks] Re: Amazon Interview Question

[adarsh kumar]

Ya I didn't see that part, sorry. And in general, using goto is not
advisable.
Plus this will exceed O(n) in the worst case, as we may keep visiting the
goto again and again. Not sure of its exact time complexity.
--
From: vindhya chhabra
Sent: 13-07-2012 17:46
To: algogeeks@googlegroups.com
Subject: Re: [algogeeks] Re: Amazon Interview Question

@adarsh
But i think jatin has asked to check for the number( we achieved in step 1)
occuring thrice or not..and in this check 27 will rule out.but i doubt the
algo given by Jatin runs in O(n) time. please comment.

On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong.

 Here is a counter test case:

 Suppose the array is:

 27 729 19683 2 3 3 27 3 81 729

 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice,
 27 occurs twice, and 3 occurs thrice.

 You are supposed to return 3
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=product(say)

 Upon traversing the second time, it will return 27, as product%(27*27*27)
 is equal to zero!

 regards.



 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some numbers
 repeat twice and only one number repeats thrice, how do you find the number
 that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

 --
 You received this message because you are subscribed to the Google Groups
 Algorithm Geeks group.
 To view this discussion on the web visit
 https://groups.google.com/d/msg/algogeeks/-/lmzhJ-GSdigJ.

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 To unsubscribe from this group, send email to
 algogeeks+unsubscr...@googlegroups.com.
 For more options, visit this group at
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-- 
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Re: [algogeeks] Re: Amazon Interview Question

[jatin]

 as long as we are using goto with proper comments to ensure that it won't 
decrease the readability we can use them and ther's no harm in it! 
Secondly the worst case for my algo is o(n) .?..correct me if i am wrong
On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not 
 advisable. 
 Plus this will exceed O(n) in the worst case, as we may keep visiting the 
 goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step 
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt 
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong. 
  
 Here is a counter test case:
  
 Suppose the array is:
  
 27 729 19683 2 3 3 27 3 81 729
  
 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice, 
 27 occurs twice, and 3 occurs thrice.
  
 You are supposed to return 3 
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=product(say)
  
 Upon traversing the second time, it will return 27, as product%(27*27*27) 
 is equal to zero!
  
 regards.
  

  
 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)
  
 but this solution requires o(logn) space though. 

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:
  
  
 1)Find product of the array and store it in say prod  o(n) and o(1) 
 2)now traverse the array and check if  
  
 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while
  
 //now check is this is the element that is occuring three times o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some 
 numbers repeat twice and only one number repeats thrice, how do you find 
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution? 
 No hashmaps please!

 -- 
 You received this message because you are subscribed to the Google 
 Groups Algorithm Geeks group.
 To view this discussion on the web visit 
 https://groups.google.com/d/msg/algogeeks/-/lmzhJ-GSdigJ. 

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 -- 
 Vindhya Chhabra



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Re: [algogeeks] Re: Amazon Interview Question

[Dave]

@Jatin: Take the array {1, 2, 4, 8, ..., x, 2, 4, 8, ..., x, x}, where x is 
the largest power of two that will fit in your integer data type. Here 1 
occurs once, 2, 4, 8, ... each occur twice, and x occurs three times. 
Ignoring that you can't calculate the product of the numbers due to 
overflow, the cube of each number in the array will be a factor of the 
product, so you will have to check every number to see if it occurs three 
times. Those checks will be O(n), so the whole algorithm will be O(n^2). 
Correct me if I am wrong.
 
As another example, just include a zero in the array and see if your 
algorithm still works as written.
 
Dave

On Friday, July 13, 2012 9:59:31 AM UTC-5, jatin sethi wrote:

  as long as we are using goto with proper comments to ensure that it won't 
 decrease the readability we can use them and ther's no harm in it! 
 Secondly the worst case for my algo is o(n) .?..correct me if i am wrong
 On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not 
 advisable. 
 Plus this will exceed O(n) in the worst case, as we may keep visiting the 
 goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step 
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt 
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong. 
  
 Here is a counter test case:
  
 Suppose the array is:
  
 27 729 19683 2 3 3 27 3 81 729
  
 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice, 
 27 occurs twice, and 3 occurs thrice.
  
 You are supposed to return 3 
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=product(say)
  
 Upon traversing the second time, it will return 27, as 
 product%(27*27*27) is equal to zero!
  
 regards.
  

  
 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)
  
 but this solution requires o(logn) space though. 

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:
  
  
 1)Find product of the array and store it in say prod  o(n) and 
 o(1) 
 2)now traverse the array and check if  
  
 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while
  
 //now check is this is the element that is occuring three times 
 o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some 
 numbers repeat twice and only one number repeats thrice, how do you find 
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution? 
 No hashmaps please!

 -- 
 You received this message because you are subscribed to the Google 
 Groups Algorithm Geeks group.
 To view this discussion on the web visit 
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Re: [algogeeks] Re: Amazon Interview Question

[algo bard]

No it's not O(n). Coz you're scanning the array and in case you do not find
a solution, you're jumping back to the start of the loop. That's equivalent
to a nested loop. And I think in the worst case...it'll turn out to be an
O(n^2) solution. Please correct me if I'm wrong.

On Fri, Jul 13, 2012 at 8:29 PM, jatin jatinseth...@gmail.com wrote:

  as long as we are using goto with proper comments to ensure that it won't
 decrease the readability we can use them and ther's no harm in it!
 Secondly the worst case for my algo is o(n) .?..correct me if i am wrong

 On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not
 advisable.
 Plus this will exceed O(n) in the worst case, as we may keep visiting the
 goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong.

 Here is a counter test case:

 Suppose the array is:

 27 729 19683 2 3 3 27 3 81 729

 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice,
 27 occurs twice, and 3 occurs thrice.

 You are supposed to return 3
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=**product(say)

 Upon traversing the second time, it will return 27, as
 product%(27*27*27) is equal to zero!

 regards.



 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and
 o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times
 o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some
 numbers repeat twice and only one number repeats thrice, how do you find
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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Re: [algogeeks] Re: Amazon Interview Question

[Shruti Gupta]

@jatin: even i think it will b more than O(n).. infact it will be
O(n-square) in the worst case as if each hit is spurious(until the last
element of course) we will have to traverse the array for each spurious
hit.. thus giving the worst case time of O(n-square)


On Fri, Jul 13, 2012 at 8:29 PM, jatin jatinseth...@gmail.com wrote:

  as long as we are using goto with proper comments to ensure that it won't
 decrease the readability we can use them and ther's no harm in it!
 Secondly the worst case for my algo is o(n) .?..correct me if i am wrong

 On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not
 advisable.
 Plus this will exceed O(n) in the worst case, as we may keep visiting the
 goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong.

 Here is a counter test case:

 Suppose the array is:

 27 729 19683 2 3 3 27 3 81 729

 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice,
 27 occurs twice, and 3 occurs thrice.

 You are supposed to return 3
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=**product(say)

 Upon traversing the second time, it will return 27, as
 product%(27*27*27) is equal to zero!

 regards.



 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and
 o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times
 o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some
 numbers repeat twice and only one number repeats thrice, how do you find
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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RE: [algogeeks] Re: Amazon Interview Question

[adarsh kumar]

+1.

--
From: Shruti Gupta
Sent: 14-07-2012 08:08
To: algogeeks@googlegroups.com
Subject: Re: [algogeeks] Re: Amazon Interview Question

@jatin: even i think it will b more than O(n).. infact it will be
O(n-square) in the worst case as if each hit is spurious(until the last
element of course) we will have to traverse the array for each spurious
hit.. thus giving the worst case time of O(n-square)


On Fri, Jul 13, 2012 at 8:29 PM, jatin jatinseth...@gmail.com wrote:

  as long as we are using goto with proper comments to ensure that it won't
 decrease the readability we can use them and ther's no harm in it!
 Secondly the worst case for my algo is o(n) .?..correct me if i am wrong

 On Friday, 13 July 2012 18:12:41 UTC+5:30, adarsh kumar wrote:

 Ya I didn't see that part, sorry. And in general, using goto is not
 advisable.
 Plus this will exceed O(n) in the worst case, as we may keep visiting the
 goto again and again. Not sure of its exact time complexity.
 --
 From: vindhya chhabra
 Sent: 13-07-2012 17:46
 To: algogeeks@googlegroups.com
 Subject: Re: [algogeeks] Re: Amazon Interview Question

 @adarsh
 But i think jatin has asked to check for the number( we achieved in step
 1) occuring thrice or not..and in this check 27 will rule out.but i doubt
 the algo given by Jatin runs in O(n) time. please comment.

 On Fri, Jul 13, 2012 at 5:17 PM, adarsh kumar algog...@gmail.com wrote:

 @jatin:
 Your first method may be proved wrong.

 Here is a counter test case:

 Suppose the array is:

 27 729 19683 2 3 3 27 3 81 729

 Here, 81 occurs once, 19683 occurs once, 2 occurs once,729 occurs twice,
 27 occurs twice, and 3 occurs thrice.

 You are supposed to return 3
 But as per your method, the product will be computed as
 729*19683*2*3*3*27*3*81*729=**product(say)

 Upon traversing the second time, it will return 27, as
 product%(27*27*27) is equal to zero!

 regards.



 On Fri, Jul 13, 2012 at 1:29 PM, @jatin jatinseth...@gmail.com wrote:

 Or we can make a BST from array list in o(n)
 then traverse it inorder-o(logn)

 but this solution requires o(logn) space though.

 On Friday, 13 July 2012 13:16:50 UTC+5:30, jatin sethi wrote:


 1)Find product of the array and store it in say prod  o(n) and
 o(1)
 2)now traverse the array and check if

 static int i;
 tag:
 while(in)
 if( prod %(ar[i]*arr[i]*arr[i] ) ==0)
 break;
 //this may be the required element
 //e-o-while

 //now check is this is the element that is occuring three times
 o(n)
 if(number is not the required one then)
 goto tag;

 On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:

 Given an array of integers where some numbers repeat once, some
 numbers repeat twice and only one number repeats thrice, how do you find
 the number that gets repeated 3 times?

 Does this problem have an O(n) time and O(1) space solution?
 No hashmaps please!

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Re: [algogeeks] Re: Amazon Interview Question

[Amol Sharma]

+1 for trie..
--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
 http://gplus.to/amolsharma99
http://twitter.com/amolsharma99http://in.linkedin.com/pub/amol-sharma/21/79b/507http://www.simplyamol.blogspot.com/






On Fri, Jun 15, 2012 at 5:21 PM, enchantress elaenjoy...@gmail.com wrote:

 se a trie with end of word marked by its corresponding entry in array.

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Re: [algogeeks] Re: Amazon Interview Question

[Wladimir Tavares]

Hi Guys,

The @pacific solution is the best?

Wladimir Araujo Tavares
*Federal University of Ceará

*




On Tue, Jun 28, 2011 at 7:26 AM, sunny agrawal sunny816.i...@gmail.comwrote:

 you can initialize it to (Max-Min+1)
 where Max = max of all elements
 Min = min of all elements

 Or simple initialise it to a large integer like 0x7fff for 32 bit
 numbers.



 On Tue, Jun 28, 2011 at 3:32 PM, vikas mehta...@gmail.com wrote:

 what will be the oldDiff initially. can u plz explain with a egsample

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 Indian Institute Of Technology,Roorkee


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Re: [algogeeks] Re: Amazon Interview Question

[vikas]

what will be the oldDiff initially. can u plz explain with a egsample

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Re: [algogeeks] Re: Amazon Interview Question

[sunny agrawal]

you can initialize it to (Max-Min+1)
where Max = max of all elements
Min = min of all elements

Or simple initialise it to a large integer like 0x7fff for 32 bit
numbers.


On Tue, Jun 28, 2011 at 3:32 PM, vikas mehta...@gmail.com wrote:

 what will be the oldDiff initially. can u plz explain with a egsample

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-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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Re: [algogeeks] Re: Amazon Interview Question

[anshu mishra]

@pacific you are perfectly right but the order is not o(kn) its is
O(k*n*log(n)) because to get the least number u have to use priority queue
nd at every iteration (from 1 to k*n) u have to push and pop one single
element.
-- 
Anshuman Mishra
IIIT Allahabad
-

anshumishra6...@gmail.com
rit2007...@iiita.ac.in

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Re: [algogeeks] Re: Amazon Interview Question

[pacific :-)]

My approach :

 Have a pointer to the start (smallest of the array) of each of the N
arrays.
 Until all pointers reach end of respective arrays :
take the smallest value from all of the pointers
and compute the difference between the smallest and the current pointers
of each of the arrays
let newdiff be the smallest among all the diffs
if newdiff  olddiff :
   olddiff = newdiff
   only advance the smallest number pointer

Correctness :

I believe that if the smallest difference occurs between a and b ,(a is
smallest) you will come across that comparison and find the least.

Complexity : 0(kn) , it should be the best because you atleast need to read
all of the input.

Please correct me if i'm wrong.


On Mon, May 9, 2011 at 8:55 PM, bittu shashank7andr...@gmail.com wrote:

 see here  let me know if anything wrong..??


 http://shashank7s.blogspot.com/2011/05/wap-to-find-minimum-difference-between.html



 Thanks  Regrads
 Shashank   the Best Way to Escape From The problem is to Solve it
 Computer Science  Engg.
 BIT Mesra

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Re: [algogeeks] Re: Amazon Interview question

[Vinay Pandey]

I forgot to mention
Time complexity: O(n), Space complexity: O(1)
Assuming you accept my solution :-)

On Mon, Feb 28, 2011 at 9:27 PM, Vinay Pandey babbupan...@gmail.com wrote:

 Hi,

 Here is my solution, let me know:


 /* a helper function */
 void swap(int* arr, int index1, int index2) {
 /* this is only to swap to integers */
  arr[index1] = arr[index1]^arr[index2];
 arr[index2] = arr[index1]^arr[index2];
  arr[index1] = arr[index1]^arr[index2];
 }

 /* Actual switching */
 void switch(int* arr, int length) {
 if(length%2!=0  length2) {
  coutCannot switch in this case;
 return;
 }
  int index = length-1, first=0, second=0;
 bool flag = 0;
 while(index=2) {
  if(flag == 0) {
 swap(arr, (index+1)/2, index);
 }
  if(flag == 1) {
 swap(arr, index/2 + 1, index);
 swap(arr, (index+1)/2, index);
  }
 index-=2;
 flag=(flag==0)?1:0;
  }
 }


 On Mon, Feb 28, 2011 at 4:32 PM, Dave dave_and_da...@juno.com wrote:

 @Arpit: The problem is that for certain values of n, you get more than
 one cycle, and the cycles are disjoint. You have to find all of the
 disjoint cycles and move the elements in each one.

 Dave

 On Feb 28, 12:25 pm, Arpit Sood soodfi...@gmail.com wrote:
  well space complexity should be mentioned in the question then, anyway,
  start with the second element put it at its correct location(say x),
 then
  take x put it at its correct location, this was when you do this n-1
 time,
  you will get the correct answer because it forms a cycle.
 
 
 
  On Mon, Feb 28, 2011 at 11:33 PM, bittu shashank7andr...@gmail.com
 wrote:
   @arpit otherwise it wont b amzon quest..
   space dude..space is constants
 
   Thanks
   Shashank
 
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  - Show quoted text -

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Re: [algogeeks] Re: Amazon Interview question

[gaurav gupta]

Here is one recursive solution I propose :

consider example a1,a2,a3,a4,b1,b2,b3,b4

1. if n is even
swap second Half of first array with first half of second array
so it would be a1,a2,b1,b2,a3,a4,b3,b4
and it  can be solved recursively
so rearrange({a1,a2,a3,a4,b1,b2,b3,b4}) = rearrange({a1,a2,b1,b2}),
rearrange({a3,a4,b3,b4});
2. if n is odd (a1,a2,a3,a4,a5,b1,b2,b3,b4,b5)
swap second Half+1 of first array with first half+1 of second array
so it would be a1,a2,b1,b2,*b3,a3*,a4,a5,b4,b5
swap bolded elements( nth and n+1 th) a1,a2,b1,b2,a3,b3,a4,a5,b4,b5
and it  can be solved recursively by dividing in two sub parts
rearrange({a1,a2,a3,a4,a5,b1,b2,b3,b4,b5}) = swap(b3,a3),
rearrange({a1,a2,b1,b2}), rearrange({a4,a5,b4,b5});

limiting case would be
if n==2 //since the pair is in required order
return;

Time complexity analysis ( n is size of total array a+b):

T(n) = 2*T(n/2) + n/4   ( n/4 is swapping time of second Half of first array
with first half of second array)

T(n) = 2^k T(1) + n/4( 1+1/2+1/4+...+1/(2^k))
= 2^k+n/4( 1-1/(2^k))*2k = log(n)
=n+ (n-1)/2
=O(n)


What say?

On Mon, Feb 28, 2011 at 11:55 PM, Arpit Sood soodfi...@gmail.com wrote:

 well space complexity should be mentioned in the question then, anyway,
 start with the second element put it at its correct location(say x), then
 take x put it at its correct location, this was when you do this n-1 time,
 you will get the correct answer because it forms a cycle.


 On Mon, Feb 28, 2011 at 11:33 PM, bittu shashank7andr...@gmail.comwrote:

 @arpit otherwise it wont b amzon quest..
 space dude..space is constants


 Thanks
 Shashank

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Re: [algogeeks] Re: Amazon Interview question

[Vinay Pandey]

Hi,

Here is my solution, let me know:


/* a helper function */
void swap(int* arr, int index1, int index2) {
/* this is only to swap to integers */
arr[index1] = arr[index1]^arr[index2];
arr[index2] = arr[index1]^arr[index2];
arr[index1] = arr[index1]^arr[index2];
}

/* Actual switching */
void switch(int* arr, int length) {
if(length%2!=0  length2) {
coutCannot switch in this case;
return;
}
int index = length-1, first=0, second=0;
bool flag = 0;
while(index=2) {
if(flag == 0) {
swap(arr, (index+1)/2, index);
}
if(flag == 1) {
swap(arr, index/2 + 1, index);
swap(arr, (index+1)/2, index);
}
index-=2;
flag=(flag==0)?1:0;
}
}


On Mon, Feb 28, 2011 at 4:32 PM, Dave dave_and_da...@juno.com wrote:

 @Arpit: The problem is that for certain values of n, you get more than
 one cycle, and the cycles are disjoint. You have to find all of the
 disjoint cycles and move the elements in each one.

 Dave

 On Feb 28, 12:25 pm, Arpit Sood soodfi...@gmail.com wrote:
  well space complexity should be mentioned in the question then, anyway,
  start with the second element put it at its correct location(say x), then
  take x put it at its correct location, this was when you do this n-1
 time,
  you will get the correct answer because it forms a cycle.
 
 
 
  On Mon, Feb 28, 2011 at 11:33 PM, bittu shashank7andr...@gmail.com
 wrote:
   @arpit otherwise it wont b amzon quest..
   space dude..space is constants
 
   Thanks
   Shashank
 
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Re: [algogeeks] Re: Amazon Interview question

[Arpit Sood]

Are there any constraints in the problem, because it seems straight forward.

if number of elements are 2n indexed from 0 to 2n-1

for i=0 to n-1:
   new_array[i*2]=old_array[i];
   new_array[i*2+1]=old_array[i+n];

On Mon, Feb 28, 2011 at 7:41 PM, bittu shashank7andr...@gmail.com wrote:

 @jalaj U needs to clarify becoz what i can say that dat is overwritten
 in ur explanation so we loosing the original  data where we are saving
 when we swapping the elements ur explanation seems to be right but
 little confusing

 @ujjwal  i haven't tested ur code but i think its O(n^2)  then why not
 try for this 20  Line Simple  code of O(n^2)

 Finally we wants O(n) better solution fro this which exist for
 this ???

 #includestring.h

 void swap(char *c,char *p)
 {
  char tmp;
  tmp=*c;
  *c=*p;
  *p=tmp;

 }

 int main (int argc, char const* argv[])
 {
char str[] = 1234abcd;
int i,j;
int len = strlen(str)/2;
//swap str[len-1] and str[len] and so on
for ( i = 0; i  len-1; i += 1) {
for ( j = len-1-i; j = len+i; j += 2)
{
 printf(i=%d j=%d c1=%c \t c2=%c
 \n,i,j,str[j],str[j+1]);
swap(str[j],str[j+1]);
}
}
printf(%s \n, str);
return 0;
 }

 Time Complexcity O(n^2)  O(n) Needed..


 Thanks  Regards
 Shashank  The Best Way to escape From The Problem is Solve It

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Re: [algogeeks] Re: Amazon Interview question

[murthy.krishn...@gmail.com]

how abt dis guys ??

#include stdio.h
#include string.h
#define MAX 100

int main()
{
int n;
int i;
int j;
int it;
char input[MAX];
char tmp;

scanf(%s,input);

n = strlen(input);

i = j = n/2;

for(it=1; itn-2; it++) {
if(it%2 == 1) {
tmp = input[j];
input[j] = input[it];
input[it] = tmp;
j++;
} else if( it  i ) {
tmp = input[i];
input[i] = input[it];
input[it] = tmp;
} else {
tmp = input[it];
input[it] = input[it+1];
input[it+1] = tmp;
}
}

printf(\n%s\n,input);

return 0;
}

On Mon, Feb 28, 2011 at 8:29 PM, Arpit Sood soodfi...@gmail.com wrote:

 Are there any constraints in the problem, because it seems straight
 forward.

 if number of elements are 2n indexed from 0 to 2n-1

 for i=0 to n-1:
new_array[i*2]=old_array[i];
new_array[i*2+1]=old_array[i+n];

 On Mon, Feb 28, 2011 at 7:41 PM, bittu shashank7andr...@gmail.com wrote:

 @jalaj U needs to clarify becoz what i can say that dat is overwritten
 in ur explanation so we loosing the original  data where we are saving
 when we swapping the elements ur explanation seems to be right but
 little confusing

 @ujjwal  i haven't tested ur code but i think its O(n^2)  then why not
 try for this 20  Line Simple  code of O(n^2)

 Finally we wants O(n) better solution fro this which exist for
 this ???

 #includestring.h

 void swap(char *c,char *p)
 {
  char tmp;
  tmp=*c;
  *c=*p;
  *p=tmp;

 }

 int main (int argc, char const* argv[])
 {
char str[] = 1234abcd;
int i,j;
int len = strlen(str)/2;
//swap str[len-1] and str[len] and so on
for ( i = 0; i  len-1; i += 1) {
for ( j = len-1-i; j = len+i; j += 2)
{
 printf(i=%d j=%d c1=%c \t c2=%c
 \n,i,j,str[j],str[j+1]);
swap(str[j],str[j+1]);
}
}
printf(%s \n, str);
return 0;
 }

 Time Complexcity O(n^2)  O(n) Needed..


 Thanks  Regards
 Shashank  The Best Way to escape From The Problem is Solve It

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Re: [algogeeks] Re: Amazon Interview question

[Arpit Sood]

well space complexity should be mentioned in the question then, anyway,
start with the second element put it at its correct location(say x), then
take x put it at its correct location, this was when you do this n-1 time,
you will get the correct answer because it forms a cycle.

On Mon, Feb 28, 2011 at 11:33 PM, bittu shashank7andr...@gmail.com wrote:

 @arpit otherwise it wont b amzon quest..
 space dude..space is constants


 Thanks
 Shashank

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Re: [algogeeks] Re: amazon interview question

[Soumya Prasad Ukil]

O(m+n).
Search from rightmost top corner. Count the no of zero from right and go to
left, i.e. traverse through columns from right of the first row. When you
find a column having 0, go down to lower row. Check the correspondent column
is 1 or not. If it is, follow the above step or else go down to next lower
row.

On 24 December 2010 16:02, MAC macatad...@gmail.com wrote:

 updated :)

 Given a boolean matrix with all the true elements on left side in the row
 so that each row can be broken into two parts left part containing only true
 elements and right part containing only false elements. Find the row with
 max number of true elements in it.


 00011
 0
 1

 true is 0 false is 1 .. so we will say 2nd row has maximum 1  and 3rd row
 has mximum 0

 --
 thanks
 --mac




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Re: [algogeeks] Re: amazon interview question

[Senthilnathan Maadasamy]

@SoumyaNice Solution.

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Re: [algogeeks] Re: Amazon Interview Question

[saravana kumar]

It can be done easily by counting sort
On Wed, Dec 15, 2010 at 5:36 AM, Soumya Prasad Ukil
ukil.sou...@gmail.comwrote:

 Have a look : http://geeksforgeeks.org/?p=1488


 On 15 December 2010 05:19, Saurabh Koar saurabhkoar...@gmail.com wrote:

 @ Bittu:

 Lets analyze your code with iterations:

 the array contains  1 3 3 1 2 3 5 2 3
 count contains   0 2 2 4 0 1 0 0 0

 now 1st iteration:
 i=8,7,6 the inner loop will nt be executed;

 i=5; so j=0 only; as count[5]=1;
 Now arr[pos](i.e array[0])=5

 i=4; the inner loop fails

 i=3; so j=0 to 3 as count[3]=4
 Now array[1]=array[2]=array[3]=array[4]=3

 i=2; so j=0 to 1 as count[2]=2
 so array[5]=array[6]=2

 and likewise array[7]=array[8]=1

 so the final array 5 3 3 3 3 2 2 1 1

 It doesnt match with the desired output.

 Correct me if I m missing something.



 On 12/15/10, bittu shashank7andr...@gmail.com wrote:
  @ankur,saurabh,soumya..
 
  ya ankur i forget to remove range from dare also no need to find
  range  for this..\
 
  put size-1 intead of range so that malloc will alocate the memory to
  all  elements in array ..no hope its fine...
 
 
  what i did is that
  i took counter array  thta cvontains the  no of times particular
  elements  comes
  then i have sort d array in reverse  order...  prints accoding to
  count array..thats it... and the last line  of actual question says..
 
  ..in case of a tie  print the number which occurred first ..i
  don't think d way i hav solved dis..
 
  still please let me no..if m wrong..
 
  Regards
  Shashank Mani Narayan  Don't b evil U can earn while u Learn
  Computer Science  enginerring
  BIrla Institute of Technology,Mesra
  Cell 9166674831
 
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Re: [algogeeks] Re: Amazon Interview Question

[Soumya Prasad Ukil]

Have a look : http://geeksforgeeks.org/?p=1488

On 15 December 2010 05:19, Saurabh Koar saurabhkoar...@gmail.com wrote:

 @ Bittu:

 Lets analyze your code with iterations:

 the array contains  1 3 3 1 2 3 5 2 3
 count contains   0 2 2 4 0 1 0 0 0

 now 1st iteration:
 i=8,7,6 the inner loop will nt be executed;

 i=5; so j=0 only; as count[5]=1;
 Now arr[pos](i.e array[0])=5

 i=4; the inner loop fails

 i=3; so j=0 to 3 as count[3]=4
 Now array[1]=array[2]=array[3]=array[4]=3

 i=2; so j=0 to 1 as count[2]=2
 so array[5]=array[6]=2

 and likewise array[7]=array[8]=1

 so the final array 5 3 3 3 3 2 2 1 1

 It doesnt match with the desired output.

 Correct me if I m missing something.



 On 12/15/10, bittu shashank7andr...@gmail.com wrote:
  @ankur,saurabh,soumya..
 
  ya ankur i forget to remove range from dare also no need to find
  range  for this..\
 
  put size-1 intead of range so that malloc will alocate the memory to
  all  elements in array ..no hope its fine...
 
 
  what i did is that
  i took counter array  thta cvontains the  no of times particular
  elements  comes
  then i have sort d array in reverse  order...  prints accoding to
  count array..thats it... and the last line  of actual question says..
 
  ..in case of a tie  print the number which occurred first ..i
  don't think d way i hav solved dis..
 
  still please let me no..if m wrong..
 
  Regards
  Shashank Mani Narayan  Don't b evil U can earn while u Learn
  Computer Science  enginerring
  BIrla Institute of Technology,Mesra
  Cell 9166674831
 
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Re: [algogeeks] Re: Amazon Interview Question

[Ankur Khurana]

@sajj: if the range of number is not given then ?

On Tue, Dec 14, 2010 at 11:41 PM, sajj sajjv...@gmail.com wrote:
 you can use hash table for this dude.

 On Dec 14, 8:22 pm, Prims topcode...@gmail.com wrote:
 given some positive numbers
 output the numbers in decreasing order of frequency..in case of a tie
 print the number which occurred first

 for eg: 1,3,3,1,2,3,5,2,3
 the output should be 11225

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Re: [algogeeks] Re: Amazon Interview Question

[Ankur Khurana]

what you can do is create a vector pairint,int and sort it on the
baisis of secondary key where it will be it's frequency. if tha range
is not given. If the range is in permissible limits, then hash table
is the answer. I suppose.

On Tue, Dec 14, 2010 at 11:45 PM, Ankur Khurana
ankur.kkhur...@gmail.com wrote:
 @sajj: if the range of number is not given then ?

 On Tue, Dec 14, 2010 at 11:41 PM, sajj sajjv...@gmail.com wrote:
 you can use hash table for this dude.

 On Dec 14, 8:22 pm, Prims topcode...@gmail.com wrote:
 given some positive numbers
 output the numbers in decreasing order of frequency..in case of a tie
 print the number which occurred first

 for eg: 1,3,3,1,2,3,5,2,3
 the output should be 11225

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Re: [algogeeks] Re: Amazon Interview Question

[Soumya Prasad Ukil]

bittu, in stead of writing your code, put your logic here. It is not the
place to show how capable one is to write a program.

On 14 December 2010 11:00, bittu shashank7andr...@gmail.com wrote:

 #include stdlib.h
 #includestdio.h
 #includeconio.h


 int main()
 {
int array[]={1,3,3,1,2,3,5,2,3};
int size=sizeof(array)/sizeof(int);
int min,max;
max=min=array[0];
int i=0;

  for(i = 1; i  size; i++)
  {
if (array[i]  min)
  min = array[i];
else if (array[i]  max)
  max = array[i];
  }

  int range = max - min + 1;
  int *count =(int *) malloc(range * sizeof(int));

for(i = 0; i  size; i++)
count[i] = 0;

for(i = 0; i  size; i++)
count[array[i]]++;



int pos=0;

for(i=size-1;i=0;i--)
{
  for(int j=0;jcount[i];j++)
  {
array[pos]=i;
pos++;
  }
}
//free(count);

for(int i=0;isize;i++)
printf(%d \n ,array[i]);

getch();
 }

 in case of a tie
 print the number which occurred first  ...i think dis situation never
 occurs...as ..array is sorted..

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Re: [algogeeks] Re: Amazon Interview Question

[Ankur Murarka]

I think ankur khanna's solution is appropriate. couldn't get what bittu was
trying to do completely.. could you just explain it once please!

On Wed, Dec 15, 2010 at 10:35 AM, Soumya Prasad Ukil
ukil.sou...@gmail.comwrote:

 bittu, in stead of writing your code, put your logic here. It is not the
 place to show how capable one is to write a program.

 On 14 December 2010 11:00, bittu shashank7andr...@gmail.com wrote:

 #include stdlib.h
 #includestdio.h
 #includeconio.h


 int main()
 {
int array[]={1,3,3,1,2,3,5,2,3};
int size=sizeof(array)/sizeof(int);
int min,max;
max=min=array[0];
int i=0;

  for(i = 1; i  size; i++)
  {
if (array[i]  min)
  min = array[i];
else if (array[i]  max)
  max = array[i];
  }

  int range = max - min + 1;
  int *count =(int *) malloc(range * sizeof(int));

for(i = 0; i  size; i++)
count[i] = 0;

for(i = 0; i  size; i++)
count[array[i]]++;



int pos=0;

for(i=size-1;i=0;i--)
{
  for(int j=0;jcount[i];j++)
  {
array[pos]=i;
pos++;
  }
}
//free(count);

for(int i=0;isize;i++)
printf(%d \n ,array[i]);

getch();
 }

 in case of a tie
 print the number which occurred first  ...i think dis situation never
 occurs...as ..array is sorted..

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Re: [algogeeks] Re: Amazon Interview Question

[Nikhil Agarwal]

@gene: can you do dry run a test case:

a[0]-a[n-1]
0 1 2 31 34 135

and if u c

On Tue, Dec 7, 2010 at 8:55 AM, Gene gene.ress...@gmail.com wrote:

 I should have mentioned that this problem only makes sense if the
 array values must be unique.

 On Dec 6, 8:20 pm, Gene gene.ress...@gmail.com wrote:
  You guys are working too hard.  Think about the sequence of numbers
  [ A[i] - i | i = 0,1,2,...n-1 ].  You are trying to probe this
  sequence to find the number of zeros.  If you think about it for a
  while, you will see that this sequence is non-decreasing.   It must be
  a segment of zero or more negative numbers followed by a segment of
  zero or more zeros followed by a segment of zero or more positive
  numbers.  Therefore you can easily use two binary searches to find the
  index of the leftmost and rightmost zeros.   This identifies all the
  elements where A[i]=i in O(2 log n) = O(log n) time.  Of course if the
  searches fail, you have no elements at all where A[i]=i.
 
  On Dec 5, 9:46 am, Nikhil Agarwal nikhil.bhoja...@gmail.com wrote:
 
 
 
   @Divesh I have updated my algo and Array A[1,2,3.n] is sorted with
   unique elements.CALL FIND_EQUAL(A,1,n)
 
   int FIND_EQUAL(A,start,end)
   1.Go to the middle element
   2. If A[mid]mid)
   3.  if(start==mid)
   4  return mid-1;
   5return FIND_EQUAL(A,1,mid-1);
   6  if(A[mid]=mid)
   7if(mid==end)
   8  return mid;
   9return FIND_EQUAL(A,mid+1,end);
 
   On Sun, Dec 5, 2010 at 7:36 PM, coolfrog$
   dixit.coolfrog.div...@gmail.comwrote:
 
@Nikhil
run you algo ..
on test case
index - 1 2 3 4 5
value - 1 2 3 4 5
 
ouput is mid+1= 3+1=4
but it should be 5...
correct me if i am wrong... and u have assumed  all are positive,
 hence
base index should be 1
 
On Sun, Dec 5, 2010 at 4:41 PM, Nikhil Agarwal 
 nikhil.bhoja...@gmail.comwrote:
 
If All the elements are unique and sorted in ascending order then we
 can
design an algorithm for O(lgn) and all nos are positive.
 
Run FIND_EQUAL(A,0,N-1) [A0,A1 and A2 are the array elements]
 
FIND_EQUAL(A,start,end)
1.Go to the middle element
2.If A[mid]==mid)
return mid+1;
if(A[mid]mid)
   then FIND_EQUAL(A,start,mid-1)
else
FIND_EQUAL(A,mid+1,end)
 
Runs in O(lgn)
 
On Sun, Dec 5, 2010 at 12:20 PM, jai gupta sayhelloto...@gmail.com
 wrote:
 
Any algorithm must in worst case lead to O(n) if the elements are
 not
unique. Take the case:
1 2 3 4 5 6
as all the elements satisfy the condition of of key==index . we
 must go
someway to each element.
Hence O(n).
 
This may be somehow made less if the element are given to be unique
 by
using heap.
 
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Re: [algogeeks] Re: Amazon Interview Question

[Nikhil Agarwal]

@Divesh I have updated my algo and Array A[1,2,3.n] is sorted with
unique elements.CALL FIND_EQUAL(A,1,n)

int FIND_EQUAL(A,start,end)
1.Go to the middle element
2. If A[mid]mid)
3.  if(start==mid)
4  return mid-1;
5return FIND_EQUAL(A,1,mid-1);
6  if(A[mid]=mid)
7if(mid==end)
8  return mid;
9return FIND_EQUAL(A,mid+1,end);




On Sun, Dec 5, 2010 at 7:36 PM, coolfrog$
dixit.coolfrog.div...@gmail.comwrote:

 @Nikhil
 run you algo ..
 on test case
 index - 1 2 3 4 5
 value - 1 2 3 4 5

 ouput is mid+1= 3+1=4
 but it should be 5...
 correct me if i am wrong... and u have assumed  all are positive, hence
 base index should be 1


 On Sun, Dec 5, 2010 at 4:41 PM, Nikhil Agarwal 
 nikhil.bhoja...@gmail.comwrote:

 If All the elements are unique and sorted in ascending order then we can
 design an algorithm for O(lgn) and all nos are positive.

 Run FIND_EQUAL(A,0,N-1) [A0,A1 and A2 are the array elements]

 FIND_EQUAL(A,start,end)
 1.Go to the middle element
 2.If A[mid]==mid)
 return mid+1;
 if(A[mid]mid)
then FIND_EQUAL(A,start,mid-1)
 else
 FIND_EQUAL(A,mid+1,end)

 Runs in O(lgn)





 On Sun, Dec 5, 2010 at 12:20 PM, jai gupta sayhelloto...@gmail.comwrote:

 Any algorithm must in worst case lead to O(n) if the elements are not
 unique. Take the case:
 1 2 3 4 5 6
 as all the elements satisfy the condition of of key==index . we must go
 someway to each element.
 Hence O(n).

 This may be somehow made less if the element are given to be unique by
 using heap.

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 Computer Science  Engineering,
 National Institute Of Technology, Durgapur,India
 http://tech-nikk.blogspot.com
 http://beta.freshersworld.com/communities/nitd


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Re: [algogeeks] Re: Amazon Interview Question

[Nikhil Agarwal]

If All the elements are unique and sorted in ascending order then we can
design an algorithm for O(lgn) and all nos are positive.

Run FIND_EQUAL(A,0,N-1) [A0,A1 and A2 are the array elements]

FIND_EQUAL(A,start,end)
1.Go to the middle element
2.If A[mid]==mid)
return mid+1;
if(A[mid]mid)
   then FIND_EQUAL(A,start,mid-1)
else
FIND_EQUAL(A,mid+1,end)

Runs in O(lgn)





On Sun, Dec 5, 2010 at 12:20 PM, jai gupta sayhelloto...@gmail.com wrote:

 Any algorithm must in worst case lead to O(n) if the elements are not
 unique. Take the case:
 1 2 3 4 5 6
 as all the elements satisfy the condition of of key==index . we must go
 someway to each element.
 Hence O(n).

 This may be somehow made less if the element are given to be unique by
 using heap.

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Re: [algogeeks] Re: Amazon Interview Question

[Ashim Kapoor]

yaar I can use  simple O(n) sweep to find out who all are equal, I think it
can't be less than this

On Sat, Dec 4, 2010 at 8:05 PM, Abioy Sun abioy@gmail.com wrote:

 2010/12/4 ankit sablok ankit4...@gmail.com:
  as all the elements are sorted in the array make a min heap of the
  array elements and as min heap is a tree of keys querying a min heap
  or a binary search tree requires operations with time equal to the
  height of the tree which is log(n) hence time for querying a min heap
 I think you might be use a log(n) time to find out a element whose
 value is equal to some certain index, so the complexity could be
 n*log(n)?

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Re: [algogeeks] Re: Amazon Interview Question

[coolfrog$]

@Nikhil
run you algo ..
on test case
index - 1 2 3 4 5
value - 1 2 3 4 5

ouput is mid+1= 3+1=4
but it should be 5...
correct me if i am wrong... and u have assumed  all are positive, hence base
index should be 1

On Sun, Dec 5, 2010 at 4:41 PM, Nikhil Agarwal nikhil.bhoja...@gmail.comwrote:

 If All the elements are unique and sorted in ascending order then we can
 design an algorithm for O(lgn) and all nos are positive.

 Run FIND_EQUAL(A,0,N-1) [A0,A1 and A2 are the array elements]

 FIND_EQUAL(A,start,end)
 1.Go to the middle element
 2.If A[mid]==mid)
 return mid+1;
 if(A[mid]mid)
then FIND_EQUAL(A,start,mid-1)
 else
 FIND_EQUAL(A,mid+1,end)

 Runs in O(lgn)





 On Sun, Dec 5, 2010 at 12:20 PM, jai gupta sayhelloto...@gmail.comwrote:

 Any algorithm must in worst case lead to O(n) if the elements are not
 unique. Take the case:
 1 2 3 4 5 6
 as all the elements satisfy the condition of of key==index . we must go
 someway to each element.
 Hence O(n).

 This may be somehow made less if the element are given to be unique by
 using heap.

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 Computer Science  Engineering,
 National Institute Of Technology, Durgapur,India
 http://tech-nikk.blogspot.com
 http://beta.freshersworld.com/communities/nitd


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Re: [algogeeks] Re: Amazon Interview Question

[Nikhil Agarwal]

@ashim please refer my post.I posted an o(lgn) algo.i.e. I am copying again
below with an update

If All the elements are unique and sorted in ascending order then we can
design an algorithm for O(lgn) and all nos are positive.

Run FIND_EQUAL(A,0,N-1) [A0,A1 and A2 are the array elements]

FIND_EQUAL(A,start,end)
1.Go to the middle element
2.If A[mid]==mid||mid==0)
{
if(mid==0A[0]!=0)
return 0;
return mid+1;
}
if(A[mid]mid)
   then FIND_EQUAL(A,start,mid-1)
else
FIND_EQUAL(A,mid+1,end)

Runs in O(lgn)



On Sat, Dec 4, 2010 at 8:08 PM, Ashim Kapoor ashimkap...@gmail.com wrote:

 yaar I can use  simple O(n) sweep to find out who all are equal, I think it
 can't be less than this


 On Sat, Dec 4, 2010 at 8:05 PM, Abioy Sun abioy@gmail.com wrote:

 2010/12/4 ankit sablok ankit4...@gmail.com:
  as all the elements are sorted in the array make a min heap of the
  array elements and as min heap is a tree of keys querying a min heap
  or a binary search tree requires operations with time equal to the
  height of the tree which is log(n) hence time for querying a min heap
 I think you might be use a log(n) time to find out a element whose
 value is equal to some certain index, so the complexity could be
 n*log(n)?

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Re: [algogeeks] Re: Amazon Interview Question

[Abioy Sun]

2010/12/5 juver++ avpostni...@gmail.com:
 Wrong.
 Counterexample: 1 2 2 2 2 6
suppose you mean 1 3 3 3 3 6?

1) divide-conquer

2) search binary, in each sub array [s, t],
mid = (s+t)/2
if t  array[mid]:
　　// no need to check the part after mid, all will fail
else if s  array[mid]:
 // no need to check the part before mid, all will fail

// if we have the elements unique
if mid  array[mid]:
 // no need to check the part after mid, all will fail
else if mid  array[mid]:
 // no need to check the part before mid, all will fail

3) when the elements are unique, in [s, t],
if (array[s] = s  array[t] = t)
 // all in [s, t] is ok

but I think its complexity will be O(n) in the worse case if elements
could be equal.

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[vikash jain]

i think this prob cannot be solved by DP then...
Because anytime a new value coming into the non decreasing sub-array
obtained by the
DP equation can be disrupted(like in the above example).
I think a backtracking approach cud prove useful.
(Recursion)
anytime we get a new element we can do two things :
Either delete it or decrement the previous values equal to it.
We do both by calling the  same function twice for each element.
Plz comment on this...

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[vikash jain]

@gene ..if u just give an example herethat  will make things more
clear...

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[gaurav singhal]

@ Gene:

Sorry  I misunderstood the problem.
I thought the other operation is of increment rather than decrement... My
bad

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[rahul patil]

Just read the comments.  You will get logic.
1 read global variables
2 start with main
3 read rec (a recursive fn)

The main logic is that whether to keep the no in the final no (by
decrementing it)
or to completely remove it.


#include stdio.h
#define SIZE 4
int result[SIZE];/*Array for storing current state of solution*/
int final_cost = 10; /*set minimum cost to some large value*/
int curr_ans[SIZE];  /*sorted array with the minimum cost*/

/*copies result into curr_ans*/
void save_arr(int *result)
{
  int i;
  for (i=0 ;iSIZE ;i++) {
curr_ans[i] = result[i];
  }
}

/*print the final solution*/
void print_ans()
{
  int i;
  for (i=0 ;iSIZE ;i++) {
printf ( %d,curr_ans[i]);
  }
}


void rec (int *arr,int index,int min,int cost)
{
  if (index = 0) {
/*
 *   keep the arr[index] in the solution
 *   i.e copy arr[index] in curr_ans[index]
 *   will increase cost by
 *   1  arr[index] - min   if arr[index]  min
 *   (where min is the min no among arr[index] to arr[SIZE-1])
 *   2  0   if arr[index]  min and will set min = arr[index]
 *   (where min is the min no among arr[index] to arr[SIZE-1])
 *
 *   After this call a fn rec (recursively) with index-1
 *   new cost calculated and new min
 */
if (arr[index] = min) {
result[index] = arr[index];
rec (arr,index-1,arr[index],cost);
} else {
result[index] = min;
cost += (arr[index] - min);
rec (arr,index-1,min,cost);
cost -= (arr[index] - min);
}
  /*
 *   remove the arr[index] from the solution
 *   copy 0 in curr_ans[index] increases the
 *   cost by arr[index] as we are removig it
 */
result[index] = 0;
cost +=  arr[index];
rec (arr,index-1,min,cost);
  } else if (index == -1) {
if (cost  final_cost) {
final_cost = cost;
save_arr(result);
}
return;
  }

}

int main(int argc,char *argv[])
{
  int i;
  int arr[SIZE] = {10,3,11,12};
  rec(arr,SIZE-1,arr[SIZE]+1,0);
  /*
   * start with the last index i.e traverse array from end to start
   * as we call rec with index-1 each time
   */
  printf (\n minimum cost = %d \n sorted array =,final_cost);
  print_ans();
  return 0;
}


On Sun, Aug 29, 2010 at 2:18 PM, jagadish jagadish1...@gmail.com wrote:


 @Rahul Patil:
 I ran your  code on some basic test cases and i found it to be
 correct!

 Can you please explain the logic you used to arrive at the solution?

 Thanks :-)

 On Aug 29, 12:25 pm, rahul patil rahul.deshmukhpa...@gmail.com
 wrote:
  check out this solution.I think this works correct
  will explain logic if u find it correct.
 
  #include stdio.h
  #define SIZE 4
  int result[SIZE];
  int final_cost = 10;
  int curr_ans[SIZE];
 
  void save_arr(int *result)
  {
int i;
for (i=0 ;iSIZE ;i++) {
  curr_ans[i] = result[i];
}
 
  }
 
  void print_ans()
  {
int i;
for (i=0 ;iSIZE ;i++) {
  printf ( %d,curr_ans[i]);
}}
 
  void rec (int *arr,int index,int min,int cost)
  {
if (index = 0) {
  // keep the arr[index]
  if (arr[index] = min) {
  result[index] = arr[index];
  rec (arr,index-1,arr[index],cost);
  } else {
  result[index] = min;
  cost += (arr[index] - min);
  rec (arr,index-1,min,cost);
  cost -= (arr[index] - min);
  }
 
  // remove the arr[index]
  result[index] = 0;
  cost +=  arr[index];
  rec (arr,index-1,min,cost);
} else if (index == -1) {
  if (cost  final_cost) {
  final_cost = cost;
  save_arr(result);
  }
  return;
}
 
  }
 
  int main(int argc,char *argv[])
  {
int i;
int arr[SIZE] = {10,3,11,12};
rec(arr,SIZE-1,arr[SIZE]+1,0);
printf (\n minimum cost = %d \n sorted array =,final_cost);
print_ans();
return 0;
 
  }
 
  On Aug 29, 1:17 am, Neeraj Gupta neeraj.gu...@nsitonline.in wrote:
 
   On Sun, Aug 29, 2010 at 1:35 AM, Gene gene.ress...@gmail.com wrote:
My algorithm proposal wasn't correct, but I can't see how to get 8.
You need to decrement 14, 15, 16, 13 to 11.  This costs 14.
 
So do I.
 
   May be he has not noticed that incrementing a number is not an option.
 ;)
 
 On Aug 28, 5:41 am, gaurav singhal singhal08gau...@gmail.com
 wrote:
 @ Gene :
 
 Output for
 
 int a[] = { 14, 15, 16, 13, 11, 18 };
 
 is coming out to be 14
 
 whereas minimum cost will be : 8
 
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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[gaurav singhal]

@ Gene :


Output for

int a[] = { 14, 15, 16, 13, 11, 18 };

is coming out to be 14

whereas minimum cost will be : 8

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[Neeraj Gupta]

On Sun, Aug 29, 2010 at 1:35 AM, Gene gene.ress...@gmail.com wrote:

 My algorithm proposal wasn't correct, but I can't see how to get 8.
 You need to decrement 14, 15, 16, 13 to 11.  This costs 14.

 So do I.
May be he has not noticed that incrementing a number is not an option. ;)


  On Aug 28, 5:41 am, gaurav singhal singhal08gau...@gmail.com wrote:
  @ Gene :
 
  Output for
 
  int a[] = { 14, 15, 16, 13, 11, 18 };
 
  is coming out to be 14
 
  whereas minimum cost will be : 8

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[GOBIND KUMAR]

This code is giving incorrect answer in some inputs.If possible someone
correct it.



#includestdio.h
#includeconio.h
static int SIZE=5;
void print(int *array);
int del(int *array,int i);
int decrement(int *array,int i);
void print(int *array)
 {
 printf(\n);
 int i=0;
 for(;iSIZE;i++)
 printf(%d\t,array[i]);printf(\n);
 }
int del(int *array,int i)
{
 int x=array[i];
 for(;i=SIZE-2;i++)
array[i-1]=array[i+1];
 SIZE-=1;
 return x;
 printf(\nSize,SIZE);
}
int decrement(int *array,int i)
{int c=0;
 while(array[i]array[i+1]){
   c++;
   array[i]=array[i]-1;
   return c;
}
}
int main()
{
int i,j,array[SIZE],count=0,flag=0;
for(i=0;iSIZE;i++)
  scanf(%d,array[i]);
while(!flag){
flag=1;
j=1;
i=0;
do{
 if(array[i]array[j]){
   if((array[i]-array[j])array[j])
 count=del(array,j);
   else
 count=decrement(array,i);flag=0;
 }
 j++;i++;
}while(j=SIZE);
}
print(array);
printf(\nTotal cost=%d,count);
getch();
   return 0;
}

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Re: [algogeeks] Re: Amazon interview Question (Array yet again!)

[Aravind]

greedy won't work on 8,2,2,2,2,2,2...

On Sat, Aug 28, 2010 at 7:39 AM, Gene gene.ress...@gmail.com wrote:

 This is a nice problem.  It looks like a trivial matroid, so a greedy
 algorithm will work fine.  The obvious greedy algorithm is to work
 left-to-right and incorporate elements into the sorted order one-by-
 one.  In each case, you have 2 choices.  The first is to decrement
 elements to the left by the amount needed to restore non-decreasing
 order.  The second is to delete the new element.  The cost of each is
 easy to calculate.  Pick the choice with least cost and continue.
 This algorithm is O(n^2).  There may be a faster way to do it, but I
 can't see one.

 #include stdio.h

 int make_nondecreasing(int *a, int n)
 {
  int i, j, dec, dec_cost, total_cost;

  total_cost = 0;
  for (i = 0; i  n - 1; i++) {

// If we find a decrease...
if (a[i]  a[i + 1]) {

  // Find cost of decrementing all to the left.
  dec_cost = dec = a[i] - a[i + 1];
  for (j = i - 1; j = 0; j--) {

// Find decrement that would be needed.
dec += a[j] - a[j + 1];

// If no decement, we're done.
if (dec = 0)
  break;

// Count cost of decrement.
dec_cost += dec;
  }

  // Compare decrement cost with deletion cost.
  if (dec_cost  a[i + 1]) {

// Decrement is cheaper.  Do it.
for (j = i; j = 0; j--) {
  if (a[j]  a[i + 1])
a[j] = a[i + 1];
}
total_cost += dec_cost;
  }
  else {

// Deletion is cheaper.  Do it.
total_cost += a[i + 1];
for (j = i + 1; j  n - 1; j++)
  a[j] = a[j + 1];
--n;
  }
}
  }
  return total_cost;
 }

 int main(void)
 {
  int a[] = { 14, 15, 16, 13, 11, 18 };
  //int a[] = { 4, 3, 5, 6};
  //int a[] = { 10, 3, 11, 12 };
  int cost = make_nondecreasing(a, sizeof a / sizeof a[0]);
  printf(cost=%d\n, cost);
  return 0;
 }

 On Aug 27, 12:15 pm, jagadish jagadish1...@gmail.com wrote:
  You are given an array of positive integers. Convert it to a sorted
  array with minimum cost. Only valid operation are
  1) Decrement - cost = 1
  2) Delete an element completely from the array - cost = value of
  element
 
  For example:
  4,3,5,6, - cost 1
  10,3,11,12 - cost 3

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Re: [algogeeks] Re: AMAZON Interview Question

[ankur aggarwal]

@harit..
logic plz.. not the code..

On Wed, Jul 14, 2010 at 9:50 PM, harit agarwal agarwalha...@gmail.comwrote:

 this is O(n) solution and using O(n) space...

 #includeiostream
 #includevector
 #includestack
 using namespace std;
 void leader_count(vectorint v,int *ar)
 {
 stackint s;
 int n=v.size();
 int i=n-1;
  while(i=0)
 {
 if(s.empty())
  {
 ar[--n]=0;
 s.push(i);
  i--;
 }
 else
  {
 if(v[i] = v[s.top()])
 {
  ar[--n]=s.top();
 s.push(i);
 i--;
  }
 else
 {
  s.pop();
 }
 }
  }
 for(int i=v.size()-1;i=0;i--)
 if(ar[i]!=0)
  ar[i]=ar[ar[i]]+1;

 }
 main()
 {
 int i;
  vectorint v;
 while(cini)
 v.push_back(i);
  int *ar=new int[v.size()];
 leader_count(v,ar);
 for(int i=0;iv.size();i++)
 coutar[i]  ;
 }

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Re: [algogeeks] Re: AMAZON Interview Question

[Anand]

One approach will be while creating a BST and also store position of the
element in the original array. While constructing ar_low check for two
condition 1. element should be less than the given element and also position
of element should greater than the given element.

On Tue, Jul 13, 2010 at 4:01 AM, Vipul Agrawal vipul.itbh...@gmail.comwrote:

 // sort ar[]
 and store in temp[]   -- o(nlogn)

 for(i=0 to n-1)
 {
 //search position of ar[i]  in temp[]  binary search --o(logn)
 ar_low[i] = pos-1;
delete temp[pos];
 }


 in binary search increase pos until next element is same .



 On Tue, Jul 13, 2010 at 4:09 PM, Amir hossein Shahriari 
 amir.hossein.shahri...@gmail.com wrote:

 make a balanced bst which also has the size of subtree at each node
 start from ar[n-1] and insert each element and see what is it's rank in
 BST
 for finding the rank when inserting each time you pick the right subtree
 add size of left subtree to rank
 O(nlogn)

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Re: [algogeeks] Re: AMAZON Interview Question

[Chakravarthi Muppalla]

I guess it is a dynamic programming problem.

On Wed, Jul 14, 2010 at 11:34 AM, Anand anandut2...@gmail.com wrote:

 One approach will be while creating a BST and also store position of the
 element in the original array. While constructing ar_low check for two
 condition 1. element should be less than the given element and also position
 of element should greater than the given element.


 On Tue, Jul 13, 2010 at 4:01 AM, Vipul Agrawal vipul.itbh...@gmail.comwrote:

 // sort ar[]
 and store in temp[]   -- o(nlogn)

 for(i=0 to n-1)
 {
 //search position of ar[i]  in temp[]  binary search --o(logn)
 ar_low[i] = pos-1;
delete temp[pos];
 }


 in binary search increase pos until next element is same .



 On Tue, Jul 13, 2010 at 4:09 PM, Amir hossein Shahriari 
 amir.hossein.shahri...@gmail.com wrote:

 make a balanced bst which also has the size of subtree at each node
 start from ar[n-1] and insert each element and see what is it's rank in
 BST
 for finding the rank when inserting each time you pick the right subtree
 add size of left subtree to rank
 O(nlogn)

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Re: [algogeeks] Re: AMAZON Interview Question

[harit agarwal]

this is O(n) solution and using O(n) space...

#includeiostream
#includevector
#includestack
using namespace std;
void leader_count(vectorint v,int *ar)
{
stackint s;
int n=v.size();
int i=n-1;
while(i=0)
{
if(s.empty())
{
ar[--n]=0;
s.push(i);
i--;
}
else
{
if(v[i] = v[s.top()])
{
ar[--n]=s.top();
s.push(i);
i--;
}
else
{
s.pop();
}
}
}
for(int i=v.size()-1;i=0;i--)
if(ar[i]!=0)
ar[i]=ar[ar[i]]+1;

}
main()
{
int i;
vectorint v;
while(cini)
v.push_back(i);
int *ar=new int[v.size()];
leader_count(v,ar);
for(int i=0;iv.size();i++)
coutar[i]  ;
}

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Re: [algogeeks] Re: AMAZON Interview Question

[srikanth sg]

actually the problem is to make BST construction in O(nlogn)...
we need rotations  which changes the structure so you wont be able to
distinguish between elements right to a particular elements, elements left
to it . ( in original array )..

On Tue, Jul 13, 2010 at 12:52 AM, Tech Id tech.login@gmail.com wrote:

 Initialise all elements of ar_low with 0

 for (int i=0; in-1; i++) {
for (int j=i+1; jn-1; j++) {
if (ar[j] = ar[i])
ar_low[i]++ ;
}
 }

 O(n^2)


 For O(nlogn), create a BST - O(nlogn)
 Traverse the tree, counting children on the left side of each node and
 putting it ar_low - O(n)

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Re: [algogeeks] Re: AMAZON Interview Question

[Amir hossein Shahriari]

make a balanced bst which also has the size of subtree at each node
start from ar[n-1] and insert each element and see what is it's rank in BST
for finding the rank when inserting each time you pick the right subtree add
size of left subtree to rank
O(nlogn)

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Re: [algogeeks] Re: AMAZON Interview Question

[Vipul Agrawal]

// sort ar[]
and store in temp[]   -- o(nlogn)

for(i=0 to n-1)
{
//search position of ar[i]  in temp[]  binary search --o(logn)
ar_low[i] = pos-1;
   delete temp[pos];
}


in binary search increase pos until next element is same .


On Tue, Jul 13, 2010 at 4:09 PM, Amir hossein Shahriari 
amir.hossein.shahri...@gmail.com wrote:

 make a balanced bst which also has the size of subtree at each node
 start from ar[n-1] and insert each element and see what is it's rank in BST
 for finding the rank when inserting each time you pick the right subtree
 add size of left subtree to rank
 O(nlogn)

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