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
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
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
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
: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
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 -
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
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,
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
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
int getsum(int *a, int n)
{
while(--n)
{
for(int i = 0; i n; ++i)
a[i] = a[i+1] - a[i];
}
return a[0];
}
I'm not really clear about how it is intended to work. It seems that
if you start with an array of N values, each iteration reduces the
number of values by 1, so
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
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
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
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
@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
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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Ans to *Boundary traversal of a tree. Write the code.
*1) you need to for preoder oder for left most tree with flag and post
order traversal for right most tree with flag.
2) the role of flag would be to decide wther to print or not like in case
of left subtree ,flag would be tree as u knw that
i have few questions regarding ur problems pratik :
1) A tree with only parent pointer, how to find LCA?
Doubt : do u mean only root of the tree is given , i am assuming root along
with two nodes address whose lca need to
find too is given , i am right ?
2) Find top k searched elements
these questions were asked for software dev. in amazon ? which round .
looks like straight easy questions...
On Sun, Mar 10, 2013 at 5:58 PM, Nishant Pandey
nishant.bits.me...@gmail.com wrote:
Ans to *Boundary traversal of a tree. Write the code.
*1) you need to for preoder oder for left
It looks as if you have just pasted some Amazon interview questions on this
forum.
These are pretty common questions.
Try to come up with your own answers.
Do some research on google and previous posts on this forum. You'll get
answers to all of them.
If you have some idea and want to discuss
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
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
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
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
@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
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
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
@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;
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,
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
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
@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
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.
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
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
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.
For the series like 2,4,3,9,4,16,5,25 ur algo runs in o(n*n/2) =o(n^2)
On Friday, 13 July 2012 13:16:50 UTC+5:30, 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
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
.
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
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
@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,
@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:
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
@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
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
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
: 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
+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
@Algo bard: No. You can do an O(n) time, O(n) space solution with a radix
sort, or you can do an O(n log n) time, O(1) space solution with a variety
of sorts.
Dave
On Thursday, July 12, 2012 12:25:02 AM UTC-5, algo bard wrote:
Given an array of integers where some numbers repeat once, some
If the assumption is that there is only one element which occurs
once , some elements repeat twice and only one element which repeats
thrice
then following is the code according to the assumption made
http://ideone.com/yseYy
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could u explain how would you use a trie for this??
On Thursday, June 14, 2012 1:01:00 PM UTC+5:30, Mohit Rathi wrote:
Hi,
*There are two arrays of length 100 each. Each of these has initially n
(n=100)
elements. First array contains names and the second array contains numbers
such that
Store each of the words in array in a trie and mark the end of the word by
its corresponding entry in the second array. Now if u are searching for a
word it'll take O(length of word) if there is a mismatch at any point you
know the word is not present in array1 and add it to the trie or else
You can use a trie with end of word marked by its corresponding entry in
array.
On Thursday, 14 June 2012 13:01:00 UTC+5:30, Mohit Rathi wrote:
Hi,
*There are two arrays of length 100 each. Each of these has initially n
(n=100)
elements. First array contains names and the second array
+1 for trie..
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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
Hassan geke should not be a valid string. The question states which have
the same substring following it so here e follows e. There is no
precondition that it has to follow immediate.
Utsav: can you clarify?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
nope geke is valid string..
here is the link from where question was taken
http://geeksforgeeks.org/forum/topic/amazon-interview-question-password-checker
On Wed, Jun 6, 2012 at 11:44 AM, Ashish Goel ashg...@gmail.com wrote:
Hassan geke should not be a valid string. The question states which
geke is valid. BTW if you changeif(i=len) toif(i0) my code
outputs geke is invalid.( what you desired)
if geke is invalid regarding to the question, then you can achieve the
answer in nLogn by sorting strings :s[0..n-1], s[1..n-1],s[n-1..n-1]
and comparing adjacent members.
Regards
@ashish:- geke is valid as repeated substrings should be immediate.
On Wed, Jun 6, 2012 at 1:44 PM, Hassan Monfared hmonfa...@gmail.com wrote:
geke is valid. BTW if you changeif(i=len) toif(i0) my code
outputs geke is invalid.( what you desired)
if geke is invalid regarding to the
Thanks, my approach
prepare a multimap of charcters and their positions by walking over the
string.
ashishis if is give string
the multimap will have
a-0
s-1,4,7
h-2,5
i-3,6
now for every character while walkingover the given string again, check
from its multimap, if it is repeated, if not
*tree substrings*
tree--|---mississippim .. mississippi
|
|---i--|---ssi--|---ssippi i .. ississippi
| | |
| | |---ppi issip,issipp,issippi
| |
| |---ppi
Hassan, can you explain your algo?
Best Regards
Ashish Goel
Think positive and find fuel in failure
+919985813081
+919966006652
On Mon, Jun 4, 2012 at 11:20 AM, Hassan Monfared hmonfa...@gmail.comwrote:
for
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The problem suggests that a character can't be more than once present and
thereby it can be done by just having s bitmap and if a char repeats, any
longer repeating substring will have those char repeated atleast twice,
hence O(n) solution.
Also, Hasaan: how is your algo O(n2) for for-while-for
Ashish:
the algorithm passes over string and check if there is any substring with
len=1 is repeated or not. if not, tries for substring with len 2,... and so
on.
max length of substring which can be repeated can be at most N/2.
Regards,
On Tue, Jun 5, 2012 at 10:48 AM, Ashish Goel
is geke is a invalid strng?
On Tue, Jun 5, 2012 at 12:17 PM, Hassan Monfared hmonfa...@gmail.comwrote:
Ashish:
the algorithm passes over string and check if there is any substring with
len=1 is repeated or not. if not, tries for substring with len 2,... and so
on.
max length of substring
yes It's valid, cuz it doesn't have any repeated substring next together
On Tue, Jun 5, 2012 at 7:08 PM, Lomash Goyal lomesh.go...@gmail.com wrote:
is geke is a invalid strng?
On Tue, Jun 5, 2012 at 12:17 PM, Hassan Monfared hmonfa...@gmail.comwrote:
Ashish:
the algorithm passes over
@hasan :-ohhh sorry, i didn't see the outer loop
yeah, it works but it is O(n^2), required solution is of O(n).
On Mon, Jun 4, 2012 at 11:20 AM, Hassan Monfared hmonfa...@gmail.comwrote:
utsav: It works fine, I did little bug fixing on boundaries as here goes :
bool IsValid(string s)
{
i have not implemented it but i can you an idea how to approach it.
Go to Each suffix in suffix or suffix array(I would prefer suffix array as
it is easier) traverse the each suffix till you encounter the first letter
of the suffix you are traversing and check to see this suppose i is the
index
I think it can be done by modifying the h-array and by making some changes
in KMP-algorithm
On Mon, Jun 4, 2012 at 9:35 AM, Jeevitesh jeeviteshshekha...@gmail.comwrote:
i have not implemented it but i can you an idea how to approach it.
Go to Each suffix in suffix or suffix array(I would
So any string with two same characters is not valid??
for example :
GEEK has E followed by E.
So GEEK is also invalid?
On Jun 3, 1:49 pm, Hassan Monfared hmonfa...@gmail.com wrote:
bool IsValid(string s)
{
for(int len=0;lens.len/2;len++)
{
int start1=0,start2=len+1;
yes it's not valid
On Sun, Jun 3, 2012 at 5:36 PM, anugrah anugrah.agra...@gmail.com wrote:
So any string with two same characters is not valid??
for example :
GEEK has E followed by E.
So GEEK is also invalid?
On Jun 3, 1:49 pm, Hassan Monfared hmonfa...@gmail.com wrote:
bool
@hassan :- it will not work for many strings as you are checking from the
mid of strings. try out ababcdef,aabc.
@atul :- it should be done in O(n).
On Sun, Jun 3, 2012 at 11:54 PM, Hassan Monfared hmonfa...@gmail.comwrote:
yes it's not valid
On Sun, Jun 3, 2012 at 5:36 PM, anugrah
@jeevitesh :- yes i am also thinking of suffix tree,
but i am facing problem in implementing it. did you implement it ??
On Mon, Jun 4, 2012 at 9:11 AM, utsav sharma utsav.sharm...@gmail.comwrote:
@hassan :- it will not work for many strings as you are checking from the
mid of strings. try
utsav: It works fine, I did little bug fixing on boundaries as here goes :
bool IsValid(string s)
{
for(int len=1;len=s.size()/2;len++)
{
int start1=0,start2=len;
while(start2s.size())
{
int i=0;
for(;ilen start2+is.size() s[start1+i]==s[start2+i];i++);
if(i=len)
I think adjacency list can be implemented by vector of vector.
vector vectorint Nodes;
The size of vector Nodes is total no of nodes.
Every element of the vector will store all its adjacent edges.
Nodes[i] is a vector containing all the edges adjacent to node i.
So, we can copy the graph easily.
Here, can we use function for comparison ??
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Recursion can be used as a approach.
void disp( node * head, int i)
{
i++;
if(i n || head == NULL)
return;
disp(head-next,i);
if(head != NULL)
cout head-data endl;
}
first call this function with pointer to head and 0;
disp(first,0);
then
http://geeksforgeeks.org/archives/8014
On Wed, Aug 31, 2011 at 3:28 PM, Abhinav Vishwa avishw...@gmail.com wrote:
Recursion can be used as a approach.
void disp( node * head, int i)
{
i++;
if(i n || head == NULL)
return;
disp(head-next,i);
if(head != NULL)
We should return curr-next in the last statement of ur code
On Aug 18, 7:08 am, Dipankar Patro dip10c...@gmail.com wrote:
A slight change in above code:
make it
while(cur cur-next)
^^ other wise the code will crash at last element in a prefect list, with no
loop.
On 18 August 2011 07:36,
we can do that too.
But if I return cur then I can myself print cur, cur-next, cur-next-prev.
for verification.
On 18 August 2011 11:56, Vijay Kansal vijaykans...@gmail.com wrote:
We should return curr-next in the last statement of ur code
On Aug 18, 7:08 am, Dipankar Patro
Hi Piyuesh Same Question i Faced Some Times Ago, Have a look at algo/code
let me know if any test case missed ?
http://shashank7s.blogspot.com/2011/05/wap-to-detect-loop-in-doubly-linked.html
if we wants to remove loop from DLL, we can do in the same way we used to do
for singly linked
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
what will be the oldDiff initially. can u plz explain with a egsample
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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
@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
-
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
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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You received this message because you are
For an O(n) in place : http://arxiv.org/pdf/0805.1598
There are links to other algo's for the same problem with varying
degrees of difficulty. I think it would be a very high expectation
indeed, if they expected this solution from the interviewee.
On Feb 28, 9:27 pm, 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) {
/*
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}) =
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
@gaurav. They way I see it it's O(nlogn)?
you have n/4 for each level of the recursion tree and logn height. In
total its O(nlogn)
On Feb 28, 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,
@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
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
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) {
@arpit otherwise it wont b amzon quest..
space dude..space is constants
Thanks
Shashank
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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,
@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
I also encountered this question yesterday. This is my solution which i
tested for a few sample cases.
https://github.com/algoseeker/Interview/blob/master/Node.java
I maintained a pointer to the Node where there should be creation of a new
node. If the node created is left child, shift the
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