[algogeeks] Array problem

[payel roy]

There is a stream of numbers coming in and you have to find K largest
numbers out of the numbers received so far at any given time. Next problem
is that a constraint is added. memory is limited to m. m < k. How would you
achieve the goal still.

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Re: [algogeeks] Array Problem

[manish untwal]

@wladimir yes the problem seems to be that!!

On Tue, Dec 11, 2012 at 10:13 AM, Wladimir Tavares wrote:

> subset sum?
> Wladimir Araujo Tavares
> *Federal University of Ceará 
> Homepage  | 
> Maratona|
> *
>
>
>
>
>
> On Fri, Nov 16, 2012 at 2:46 AM, Pralay Biswas  > wrote:
>
>> Search for balance partitioning under Dynamic Programming. Doable in O(n)
>>
>>
>> On Thu, Nov 15, 2012 at 8:16 PM, bharat b 
>> wrote:
>>
>>> @ vishal : how can u divide an array into 2 groups whose difference is
>>> maximum in O(1). why max?
>>>
>>> solution : http://www.youtube.com/watch?v=GdnpQY2j064
>>>
>>>
>>>
>>>
>>> On Fri, Nov 16, 2012 at 9:22 AM, vishal chaudhary <
>>> vishal.cs.b...@gmail.com> wrote:
>>>
 Hi
 you can first sort the array which can be done in O(nlogn) complexity
 if the number of items in the array is n.
 Then using the indexing of arrays you can divide the array into two
 groups whose difference is going to be maximum and this can be done in O(1)
 complexity.
 So the complete algorithm is going to take O(nlogn) complexity.
 Kindly share an alternative algorithm if you find  one with lower
 complexity.

 Vishal


 On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra <
 reserve4placem...@gmail.com> wrote:

> Given an unsorted array, how to divide them into two equal arrays
> whose difference of sum is minimum.
>
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Indian Institute of Information Technology
Allahabad (2009-2013 batch)

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Re: [algogeeks] Array Problem

[Wladimir Tavares]

subset sum?
Wladimir Araujo Tavares
*Federal University of Ceará 
Homepage  |
Maratona|
*





On Fri, Nov 16, 2012 at 2:46 AM, Pralay Biswas
wrote:

> Search for balance partitioning under Dynamic Programming. Doable in O(n)
>
>
> On Thu, Nov 15, 2012 at 8:16 PM, bharat b wrote:
>
>> @ vishal : how can u divide an array into 2 groups whose difference is
>> maximum in O(1). why max?
>>
>> solution : http://www.youtube.com/watch?v=GdnpQY2j064
>>
>>
>>
>>
>> On Fri, Nov 16, 2012 at 9:22 AM, vishal chaudhary <
>> vishal.cs.b...@gmail.com> wrote:
>>
>>> Hi
>>> you can first sort the array which can be done in O(nlogn) complexity if
>>> the number of items in the array is n.
>>> Then using the indexing of arrays you can divide the array into two
>>> groups whose difference is going to be maximum and this can be done in O(1)
>>> complexity.
>>> So the complete algorithm is going to take O(nlogn) complexity.
>>> Kindly share an alternative algorithm if you find  one with lower
>>> complexity.
>>>
>>> Vishal
>>>
>>>
>>> On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra >> > wrote:
>>>
 Given an unsorted array, how to divide them into two equal arrays whose
 difference of sum is minimum.

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Re: [algogeeks] Array problem

[Dave]

@Ansum: Notice that the problem does not ask to give a method of making as 
many numbers as possible equal, but only what the maximum number is. Here 
is an algorithm for achieving an array with the equality numbers I 
specified:
 
1. If the sum of the numbers is a multiple of n, then avg = sum/n is the 
target number. If every number is not equal to avg, find a number a[i] in 
the array that is smaller than avg and a number a[j] that is larger than 
avg. Then set a[i] = a[i]+1 and a[j] = a[j]-1. Repeat until every number 
equals avg. Result is that n numbers are equal.
 
2. If the sum of the numbers is not a multiple of n, pick any number, say 
a[1], as the target value. If there is any number a[k] from k = 2 to n-1 
such that a[k] != a[1], increase or decrease a[k] by 1 to make it closer to 
a[1] and respectively decrease or increase a[n] by 1. Repeat until every 
number a[k] = a[1] for k = 2 to n-1. Then a[1] to a[n-1] are equal but a[n] 
cannot be equal because then the sum of the numbers would be a multiple of 
n. Result is that n-1 numbers are equal.
 
Since you don't have to produce the maximally equal array, the algorithm I 
gave is the solution. Here it is in code (with C's 0-based array indexing):
 
int maxEqual( int n, int a[])
{
int i, sum = 0;
for( i = 0 ; i < n ; ++i )
sum += a[i];
return if sum % n == 0 ? n : n - 1;
}
 
Dave

On Thursday, November 22, 2012 1:25:43 AM UTC-6, Ansum Baid wrote:

> @Dave: Can you give a little insight on your approach?
>
> On Wed, Nov 21, 2012 at 6:52 PM, Dave  >wrote:
>
>> @Ansum: Polycarpus should start by summing the numbers. If the sum is 
>> divisible by n, then n numbers can be made equal. If the sum is not 
>> divisible by n, then only n-1 numbers can be made equal.
>>  
>> Dave
>>
>> On Wednesday, November 21, 2012 12:18:54 PM UTC-6, Ansum Baid wrote:
>>
>>>  This question has been taken from codeforces.com. Any idea how to 
>>> solve this ?
>>>
>>> Polycarpus has an array, consisting of *n* integers *a*1, *a*2, ..., *a*
>>> *n*. Polycarpus likes it when numbers in an array match. That's why he 
>>> wants the array to have as many equal numbers as possible. For that 
>>> Polycarpus performs the following operation multiple times:
>>>
>>>
>>>- he chooses two elements of the array *a**i*, *a**j* (*i* ≠ *j*); 
>>>- he simultaneously increases number *a**i* by 1 and decreases 
>>>number *a**j* by 1, that is, executes *a**i* = *a**i* + 1 and *a**j*= 
>>>*a**j* - 1. 
>>>
>>> The given operation changes exactly two distinct array elements. 
>>> Polycarpus can apply the described operation an infinite number of times.
>>>
>>> Now he wants to know what maximum number of equal array elements he can 
>>> get if he performs an arbitrary number of such operation. Help Polycarpus.
>>> Input
>>>
>>> The first line contains integer *n* (1 ≤ *n* ≤ 105) — the array size. 
>>> The second line contains space-separated integers *a*1, *a*2, ..., *a**n
>>> * (|*a**i*| ≤ 104) — the original array.
>>> Output
>>>
>>> Print a single integer — the maximum number of equal array elements he 
>>> can get if he performs an arbitrary number of the given operation.
>>> Sample test(s)
>>>  input
>>>
>>> 2
>>> 2 1
>>>
>>> output
>>>
>>> 1
>>>
>>> input
>>>
>>> 3
>>> 1 4 1
>>>
>>> output
>>>
>>> 3
>>>
>>>
>>>  -- 
>>  
>>  
>>
>
>

-- 

Re: [algogeeks] Array problem

[Ansum Baid]

@Dave: Can you give a little insight on your approach?

On Wed, Nov 21, 2012 at 6:52 PM, Dave  wrote:

> @Ansum: Polycarpus should start by summing the numbers. If the sum is
> divisible by n, then n numbers can be made equal. If the sum is not
> divisible by n, then only n-1 numbers can be made equal.
>
> Dave
>
> On Wednesday, November 21, 2012 12:18:54 PM UTC-6, Ansum Baid wrote:
>
>>  This question has been taken from codeforces.com. Any idea how to solve
>> this ?
>>
>> Polycarpus has an array, consisting of *n* integers *a*1, *a*2, ..., *a**
>> n*. Polycarpus likes it when numbers in an array match. That's why he
>> wants the array to have as many equal numbers as possible. For that
>> Polycarpus performs the following operation multiple times:
>>
>>
>>- he chooses two elements of the array *a**i*, *a**j* (*i* ≠ *j*);
>>- he simultaneously increases number *a**i* by 1 and decreases number
>>*a**j* by 1, that is, executes *a**i* = *a**i* + 1 and *a**j* = *a**j*- 1
>>.
>>
>> The given operation changes exactly two distinct array elements.
>> Polycarpus can apply the described operation an infinite number of times.
>>
>> Now he wants to know what maximum number of equal array elements he can
>> get if he performs an arbitrary number of such operation. Help Polycarpus.
>> Input
>>
>> The first line contains integer *n* (1 ≤ *n* ≤ 105) — the array size.
>> The second line contains space-separated integers *a*1, *a*2, ..., *a**n*
>>  (|*a**i*| ≤ 104) — the original array.
>> Output
>>
>> Print a single integer — the maximum number of equal array elements he
>> can get if he performs an arbitrary number of the given operation.
>> Sample test(s)
>>  input
>>
>> 2
>> 2 1
>>
>> output
>>
>> 1
>>
>> input
>>
>> 3
>> 1 4 1
>>
>> output
>>
>> 3
>>
>>
>>  --
>
>
>

-- 

Re: [algogeeks] Array problem

[Dave]

@Ansum: Polycarpus should start by summing the numbers. If the sum is 
divisible by n, then n numbers can be made equal. If the sum is not 
divisible by n, then only n-1 numbers can be made equal.
 
Dave

On Wednesday, November 21, 2012 12:18:54 PM UTC-6, Ansum Baid wrote:

>  This question has been taken from codeforces.com. Any idea how to solve 
> this ?
>
> Polycarpus has an array, consisting of *n* integers *a*1, *a*2, ..., *a**n
> *. Polycarpus likes it when numbers in an array match. That's why he 
> wants the array to have as many equal numbers as possible. For that 
> Polycarpus performs the following operation multiple times:
>
>
>- he chooses two elements of the array *a**i*, *a**j* (*i* ≠ *j*); 
>- he simultaneously increases number *a**i* by 1 and decreases number *
>a**j* by 1, that is, executes *a**i* = *a**i* + 1 and *a**j* = *a**j*- 1
>. 
>
> The given operation changes exactly two distinct array elements. 
> Polycarpus can apply the described operation an infinite number of times.
>
> Now he wants to know what maximum number of equal array elements he can 
> get if he performs an arbitrary number of such operation. Help Polycarpus.
> Input
>
> The first line contains integer *n* (1 ≤ *n* ≤ 105) — the array size. The 
> second line contains space-separated integers *a*1, *a*2, ..., *a**n* (|*a
> **i*| ≤ 104) — the original array.
> Output
>
> Print a single integer — the maximum number of equal array elements he can 
> get if he performs an arbitrary number of the given operation.
> Sample test(s)
>  input
>
> 2
> 2 1
>
> output
>
> 1
>
> input
>
> 3
> 1 4 1
>
> output
>
> 3
>
>
>

-- 

[algogeeks] Array problem

[Ansum Baid]

This question has been taken from codeforces.com. Any idea how to solve
this ?

Polycarpus has an array, consisting of *n* integers *a*1, *a*2, ..., *a**n*.
Polycarpus likes it when numbers in an array match. That's why he wants the
array to have as many equal numbers as possible. For that Polycarpus
performs the following operation multiple times:


   - he chooses two elements of the array *a**i*, *a**j* (*i* ≠ *j*);
   - he simultaneously increases number *a**i* by 1 and decreases number *a*
   *j* by 1, that is, executes *a**i* = *a**i* + 1 and *a**j* = *a**j* - 1.

The given operation changes exactly two distinct array elements. Polycarpus
can apply the described operation an infinite number of times.

Now he wants to know what maximum number of equal array elements he can get
if he performs an arbitrary number of such operation. Help Polycarpus.
Input

The first line contains integer *n* (1 ≤ *n* ≤ 105) — the array size. The
second line contains space-separated integers *a*1, *a*2, ..., *a**n* (|*a**
i*| ≤ 104) — the original array.
Output

Print a single integer — the maximum number of equal array elements he can
get if he performs an arbitrary number of the given operation.
Sample test(s)
input

2
2 1

output

1

input

3
1 4 1

output

3

-- 

Re: [algogeeks] Array Problem

[bharat b]

@ vishal :let array be {5,2,1,1} ==> as per u'r algo =>{1,2},{1,5} are sets
difference is 3 .. but the sol is {5}{1,1,2} ==> diff = 1;

On Fri, Nov 16, 2012 at 10:12 AM, vishal chaudhary  wrote:

> Hi
> Sorry for that as i misinterpreted the question.
> for the difference to be minimum, i think(not completely sure) we can
> first sort the array
> and then we can start putting the elements at even index in the last part
> of the array and the odd ones in the starting in the new array
> you can do this in the same array itself i guess but you have to do some
> kind of shifting.
> by doing this for all the elements and dividing them into two groups.
> I hope this helps.
>
> Vishal
>
>
>
> On Fri, Nov 16, 2012 at 9:46 AM, bharat b wrote:
>
>> @ vishal : how can u divide an array into 2 groups whose difference is
>> maximum in O(1). why max?
>>
>> solution : http://www.youtube.com/watch?v=GdnpQY2j064
>>
>>
>>
>>
>> On Fri, Nov 16, 2012 at 9:22 AM, vishal chaudhary <
>> vishal.cs.b...@gmail.com> wrote:
>>
>>> Hi
>>> you can first sort the array which can be done in O(nlogn) complexity if
>>> the number of items in the array is n.
>>> Then using the indexing of arrays you can divide the array into two
>>> groups whose difference is going to be maximum and this can be done in O(1)
>>> complexity.
>>> So the complete algorithm is going to take O(nlogn) complexity.
>>> Kindly share an alternative algorithm if you find  one with lower
>>> complexity.
>>>
>>> Vishal
>>>
>>>
>>> On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra >> > wrote:
>>>
 Given an unsorted array, how to divide them into two equal arrays whose
 difference of sum is minimum.

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Re: [algogeeks] Array Problem

[vishal chaudhary]

Hi
Sorry for that as i misinterpreted the question.
for the difference to be minimum, i think(not completely sure) we can first
sort the array
and then we can start putting the elements at even index in the last part
of the array and the odd ones in the starting in the new array
you can do this in the same array itself i guess but you have to do some
kind of shifting.
by doing this for all the elements and dividing them into two groups.
I hope this helps.

Vishal


On Fri, Nov 16, 2012 at 9:46 AM, bharat b wrote:

> @ vishal : how can u divide an array into 2 groups whose difference is
> maximum in O(1). why max?
>
> solution : http://www.youtube.com/watch?v=GdnpQY2j064
>
>
>
>
> On Fri, Nov 16, 2012 at 9:22 AM, vishal chaudhary <
> vishal.cs.b...@gmail.com> wrote:
>
>> Hi
>> you can first sort the array which can be done in O(nlogn) complexity if
>> the number of items in the array is n.
>> Then using the indexing of arrays you can divide the array into two
>> groups whose difference is going to be maximum and this can be done in O(1)
>> complexity.
>> So the complete algorithm is going to take O(nlogn) complexity.
>> Kindly share an alternative algorithm if you find  one with lower
>> complexity.
>>
>> Vishal
>>
>>
>> On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra 
>> wrote:
>>
>>> Given an unsorted array, how to divide them into two equal arrays whose
>>> difference of sum is minimum.
>>>
>>> --
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Re: [algogeeks] Array Problem

[bharat b]

@ vishal : how can u divide an array into 2 groups whose difference is
maximum in O(1). why max?

solution : http://www.youtube.com/watch?v=GdnpQY2j064



On Fri, Nov 16, 2012 at 9:22 AM, vishal chaudhary
wrote:

> Hi
> you can first sort the array which can be done in O(nlogn) complexity if
> the number of items in the array is n.
> Then using the indexing of arrays you can divide the array into two groups
> whose difference is going to be maximum and this can be done in O(1)
> complexity.
> So the complete algorithm is going to take O(nlogn) complexity.
> Kindly share an alternative algorithm if you find  one with lower
> complexity.
>
> Vishal
>
>
> On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra 
> wrote:
>
>> Given an unsorted array, how to divide them into two equal arrays whose
>> difference of sum is minimum.
>>
>> --
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Re: [algogeeks] Array Problem

[vishal chaudhary]

Hi
you can first sort the array which can be done in O(nlogn) complexity if
the number of items in the array is n.
Then using the indexing of arrays you can divide the array into two groups
whose difference is going to be maximum and this can be done in O(1)
complexity.
So the complete algorithm is going to take O(nlogn) complexity.
Kindly share an alternative algorithm if you find  one with lower
complexity.

Vishal

On Wed, Nov 7, 2012 at 7:43 PM, Arun Kindra wrote:

> Given an unsorted array, how to divide them into two equal arrays whose
> difference of sum is minimum.
>
> --
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[algogeeks] Array Problem

[Arun Kindra]

Given an unsorted array, how to divide them into two equal arrays whose
difference of sum is minimum.

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Re: [algogeeks] array problem

[Tushar]

@srikanth
we can use segment trees to get sum of an interval
but there is another condition of sum of distinct numbers only. how can we 
take that into account in a segment tree?

On Thursday, 6 September 2012 17:35:59 UTC+5:30, srikanth reddy malipatel 
wrote:
>
> post the logic not the code!
> BTW this  problem can be done using segment trees.
>
>
> http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor
>  
>
> On Thu, Sep 6, 2012 at 4:51 PM, bharat b 
> > wrote:
>
>> Its better to write an O(n) solution for this problem as the # test cases 
>> are very high and #elements are also very huge..
>> use visited array for distinct numbers ... space--O(n).. time--O(n)
>>
>>
>>  On Sat, Aug 25, 2012 at 2:39 AM, michael miller 
>> 
>> > wrote:
>>
>>> Hi,
>>> You are given N numbers. You have to perform two kinds of operations:
>>> U x y - x-th number becomes equal to y.
>>> Q x y - calculate the sum of distinct numbers from x-th to y-th. It 
>>> means that the sum for the set {1, 2, 3, 2, 7} will be equal to 13 
>>> (1+2+3+7).
>>>
>>> 1<=N<=5 and 
>>> t is the number of test cases where   1<=t<=10
>>>
>>> all numbers fit in the 32 bit integer range...
>>>
>>> suggest some solution..
>>>
>>> here is my solution
>>> but it is giving wrong answer fo some unknown test case...plz suggest me 
>>> the test case where i am getting wrong answer
>>>
>>>
>>> #include
>>> #include
>>> int main()
>>> {
>>> int list[5],i,n,j,sum,k,l;char c;long t;
>>> scanf 
>>> ("%d",&n);
>>> for(i=0;i>> {
>>> scanf 
>>> ("%d",&list[i]);
>>>
>>>
>>> }
>>> scanf 
>>> ("%ld",&t);
>>>
>>>
>>> t=2*t;
>>> while(t)
>>> {
>>> sum=0;
>>> scanf 
>>> ("%c",&c);
>>>
>>>
>>> fflush(stdin);
>>> scanf 
>>> ("%d
>>> %d",&k,&l);   
>>>
>>>
>>> if(c=='Q')
>>> {
>>> for(i=k-1;i>> {
>>> for(j=i+1;j>> {
>>>if(list[i]==list[j])
>>>   break;
>>>else if((j==l-1) &&(list[i]!=list[j]))
>>>
>>>
>>>{
>>>   sum=sum+list[i];
>>>
>>>}
>>> }
>>>  }
>>>
>>>  printf 
>>> ("%d\n",sum+list[l-1]);
>>>
>>>
>>>  }
>>>  if(c=='U')
>>>  {
>>>  list[k-1]=l;
>>>
>>>  }
>>>  t--;
>>> }
>>> return 0;   
>>> }
>>>
>>>
>>>
>>>  -- 
>>> You received this message because you are subscribed to the Google 
>>> Groups "Algorithm Geeks" group.
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>>> .
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>>>
>>
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>>
>
>
>
> -- 
> Srikanth Reddy M
> (M.Sc Tech.) Information Systems
> BITS-PILANI
>
>  

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Re: [algogeeks] array problem

[Arman Kamal]

It will be even easier with BIT (Binary Indexed Tree), if you know how to
use it.

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Re: [algogeeks] array problem

[srikanth reddy malipatel]

post the logic not the code!
BTW this  problem can be done using segment trees.

http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor


On Thu, Sep 6, 2012 at 4:51 PM, bharat b wrote:

> Its better to write an O(n) solution for this problem as the # test cases
> are very high and #elements are also very huge..
> use visited array for distinct numbers ... space--O(n).. time--O(n)
>
>
> On Sat, Aug 25, 2012 at 2:39 AM, michael miller <
> wentworth.miller6...@gmail.com> wrote:
>
>> Hi,
>> You are given N numbers. You have to perform two kinds of operations:
>> U x y - x-th number becomes equal to y.
>> Q x y - calculate the sum of distinct numbers from x-th to y-th. It means
>> that the sum for the set {1, 2, 3, 2, 7} will be equal to 13 (1+2+3+7).
>>
>> 1<=N<=5 and
>> t is the number of test cases where   1<=t<=10
>>
>> all numbers fit in the 32 bit integer range...
>>
>> suggest some solution..
>>
>> here is my solution
>> but it is giving wrong answer fo some unknown test case...plz suggest me
>> the test case where i am getting wrong answer
>>
>>
>> #include
>> #include
>> int main()
>> {
>> int list[5],i,n,j,sum,k,l;char c;long t;
>> scanf 
>> ("%d",&n);
>> for(i=0;i> {
>> scanf 
>> ("%d",&list[i]);
>>
>> }
>> scanf 
>> ("%ld",&t);
>>
>> t=2*t;
>> while(t)
>> {
>> sum=0;
>> scanf 
>> ("%c",&c);
>>
>> fflush(stdin);
>> scanf 
>> ("%d 
>>%d",&k,&l);
>>
>> if(c=='Q')
>> {
>> for(i=k-1;i> {
>> for(j=i+1;j> {
>>if(list[i]==list[j])
>>   break;
>>else if((j==l-1) &&(list[i]!=list[j]))
>>
>>{
>>   sum=sum+list[i];
>>
>>}
>> }
>>  }
>>
>>  printf 
>> ("%d\n",sum+list[l-1]);
>>
>>  }
>>  if(c=='U')
>>  {
>>  list[k-1]=l;
>>
>>  }
>>  t--;
>> }
>> return 0;
>> }
>>
>>
>>
>>  --
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>> "Algorithm Geeks" group.
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>>
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-- 
Srikanth Reddy M
(M.Sc Tech.) Information Systems
BITS-PILANI

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Re: [algogeeks] array problem

[bharat b]

Its better to write an O(n) solution for this problem as the # test cases
are very high and #elements are also very huge..
use visited array for distinct numbers ... space--O(n).. time--O(n)

On Sat, Aug 25, 2012 at 2:39 AM, michael miller <
wentworth.miller6...@gmail.com> wrote:

> Hi,
> You are given N numbers. You have to perform two kinds of operations:
> U x y - x-th number becomes equal to y.
> Q x y - calculate the sum of distinct numbers from x-th to y-th. It means
> that the sum for the set {1, 2, 3, 2, 7} will be equal to 13 (1+2+3+7).
>
> 1<=N<=5 and
> t is the number of test cases where   1<=t<=10
>
> all numbers fit in the 32 bit integer range...
>
> suggest some solution..
>
> here is my solution
> but it is giving wrong answer fo some unknown test case...plz suggest me
> the test case where i am getting wrong answer
>
>
> #include
> #include
> int main()
> {
> int list[5],i,n,j,sum,k,l;char c;long t;
> scanf 
> ("%d",&n);
> for(i=0;i {
> scanf 
> ("%d",&list[i]);
> }
> scanf 
> ("%ld",&t);
> t=2*t;
> while(t)
> {
> sum=0;
> scanf 
> ("%c",&c);
> fflush(stdin);
> scanf 
> ("%d  
>   %d",&k,&l);
> if(c=='Q')
> {
> for(i=k-1;i {
> for(j=i+1;j {
>if(list[i]==list[j])
>   break;
>else if((j==l-1) &&(list[i]!=list[j]))
>{
>   sum=sum+list[i];
>
>}
> }
>  }
>
>  printf 
> ("%d\n",sum+list[l-1]);
>  }
>  if(c=='U')
>  {
>  list[k-1]=l;
>
>  }
>  t--;
> }
> return 0;
> }
>
>
>
>  --
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Re: [algogeeks] Array problem

[sachin sabbarwal]

now i get this!! i thought we have to calculate the sum upto (i-1)th index.
thanx for the clarifiacation.

On Wed, Mar 14, 2012 at 3:07 PM, Dheeraj Sharma  wrote:

> you have to calculate the sum of elements which are less than..that
> particular element...this is not the question of calculating cumulative sum
>
>
> On Wed, Mar 14, 2012 at 11:22 AM, sachin sabbarwal <
> algowithsac...@gmail.com> wrote:
>
>> @gaurav popli:  how about this one??
>>
>> findsummat(int arr[],int n)
>> {
>>int *sum ;
>>  sum =(int*)malloc(sizeof(int)*n);
>>
>> for(int i=0;i>   sum[i] = 0;
>>
>> for(int i=0;i>sum[i] = sum[i-1] + arr[i-1];
>> //now print the sum array
>>
>> }
>>
>> it works very well
>> plz tell me if anything is wrong with this solution.
>>
>>
>> On Tue, Mar 13, 2012 at 12:03 PM, atul anand wrote:
>>
>>> @piyush : sorry dude , didnt get your algo . actually you are using
>>> different array and i get confused which array to be considered when.
>>>
>>>
>>>
>>> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>>>
 1)First map the array numbers into the position in which they would be,
 if they are sorted,for example
 {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
 2)Now for each number ,find the cumulative frequency of index ( = the
 corresponding number in the map - 1).
 3)Output the cumulative frequency and increase the frequency  at the
 position (=the corresponding number in the map) by the number itself.
 Example
 {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
 Process the numbers now,
 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is
 0. so output 0
Increase the frequency at index 2 to the number 3.
 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
Increase the frequency at index 3 to the number 5.
 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
 position 1 by 1.
 Similarly for others.


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>>>
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>
>
>
> --
> *Dheeraj Sharma*
>
>
>
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Re: [algogeeks] Array problem

[Dheeraj Sharma]

you have to calculate the sum of elements which are less than..that
particular element...this is not the question of calculating cumulative sum

On Wed, Mar 14, 2012 at 11:22 AM, sachin sabbarwal  wrote:

> @gaurav popli:  how about this one??
>
> findsummat(int arr[],int n)
> {
>int *sum ;
>  sum =(int*)malloc(sizeof(int)*n);
>
> for(int i=0;i   sum[i] = 0;
>
> for(int i=0;isum[i] = sum[i-1] + arr[i-1];
> //now print the sum array
>
> }
>
> it works very well
> plz tell me if anything is wrong with this solution.
>
>
> On Tue, Mar 13, 2012 at 12:03 PM, atul anand wrote:
>
>> @piyush : sorry dude , didnt get your algo . actually you are using
>> different array and i get confused which array to be considered when.
>>
>>
>>
>> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>>
>>> 1)First map the array numbers into the position in which they would be,
>>> if they are sorted,for example
>>> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
>>> 2)Now for each number ,find the cumulative frequency of index ( = the
>>> corresponding number in the map - 1).
>>> 3)Output the cumulative frequency and increase the frequency  at the
>>> position (=the corresponding number in the map) by the number itself.
>>> Example
>>> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
>>> Process the numbers now,
>>> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0.
>>> so output 0
>>>Increase the frequency at index 2 to the number 3.
>>> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>>>Increase the frequency at index 3 to the number 5.
>>> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
>>> position 1 by 1.
>>> Similarly for others.
>>>
>>>
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>>
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Re: [algogeeks] Array problem

[sachin sabbarwal]

@gaurav popli:  how about this one??

findsummat(int arr[],int n)
{
   int *sum ;
 sum =(int*)malloc(sizeof(int)*n);

for(int i=0;iwrote:

> @piyush : sorry dude , didnt get your algo . actually you are using
> different array and i get confused which array to be considered when.
>
>
>
> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>
>> 1)First map the array numbers into the position in which they would be,
>> if they are sorted,for example
>> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
>> 2)Now for each number ,find the cumulative frequency of index ( = the
>> corresponding number in the map - 1).
>> 3)Output the cumulative frequency and increase the frequency  at the
>> position (=the corresponding number in the map) by the number itself.
>> Example
>> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
>> Process the numbers now,
>> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0.
>> so output 0
>>Increase the frequency at index 2 to the number 3.
>> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>>Increase the frequency at index 3 to the number 5.
>> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at position
>> 1 by 1.
>> Similarly for others.
>>
>>
>>  --
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Re: [algogeeks] Array problem

[atul anand]

@piyush : sorry dude , didnt get your algo . actually you are using
different array and i get confused which array to be considered when.


On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor  wrote:

> 1)First map the array numbers into the position in which they would be, if
> they are sorted,for example
> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
> 2)Now for each number ,find the cumulative frequency of index ( = the
> corresponding number in the map - 1).
> 3)Output the cumulative frequency and increase the frequency  at the
> position (=the corresponding number in the map) by the number itself.
> Example
> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
> Process the numbers now,
> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0.
> so output 0
>Increase the frequency at index 2 to the number 3.
> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>Increase the frequency at index 3 to the number 5.
> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at position
> 1 by 1.
> Similarly for others.
>
>
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Re: [algogeeks] Array problem

[santosh mahto]

I think we can use the counting sort method...

On Tue, Mar 13, 2012 at 1:51 AM, sunny agrawal wrote:

> @atul
> if its sum of numbers lesser than a[i] in left to i, then still i think it
> can be solved in O(nlgn) using Balanced Tree structures
> ie: if we use AVL tree, then we just need a  little care of how to update
> sum stored with rotations
> and required ans for ith index must be calculated just after the ith
> insertion
>
> and if its sum of  numbers lesser than i, then first sort(val, index pair)
> the array and find the prefix sum array would do, with some care for
> duplicates
>
> On Mon, Mar 12, 2012 at 11:09 PM, atul anand wrote:
>
>> @sunny : it was a reply to different question.
>> using AVL or RB for the given algo may screw it.
>>
>>
>> On Mon, Mar 12, 2012 at 11:00 PM, sunny agrawal 
>> wrote:
>>
>>> @atul
>>> TC can be bound to O(nlgn) using any balanced tree ie: AVL tree, RB tree
>>> as already mentioned in her last post
>>>
>>> On Mon, Mar 12, 2012 at 10:44 PM, atul anand wrote:
>>>
 @payal:
 TC will not alwayz be O(nlogn) bcozz we are not sure if the tree formed
 is balanced.
 so worst would be O(n^2)


 On Mon, Mar 12, 2012 at 9:16 PM, payal gupta wrote:

> @atul...
> if its the sum of the elements to the left of a[i] which are smaller
> the my approach works w/o any flaw
> here's the working code for ithttp://ideone.com/CH7VW
> if its the sum of all elements lesser than the element a[i] then this
> algo is surely wrong
> n we then have to proceed by the avl trees or some height balanced
> tree n the algo would be of TC-O(nlogn)
> btw nyc catch thnx...
>
> Regards,
> PAYAL GUPTA
>
>
>
> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>
>> 1)First map the array numbers into the position in which they would
>> be, if they are sorted,for example
>> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
>> 2)Now for each number ,find the cumulative frequency of index ( = the
>> corresponding number in the map - 1).
>> 3)Output the cumulative frequency and increase the frequency  at the
>> position (=the corresponding number in the map) by the number itself.
>> Example
>> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
>> Process the numbers now,
>> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is
>> 0. so output 0
>>Increase the frequency at index 2 to the number 3.
>> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output
>> 3.
>>Increase the frequency at index 3 to the number 5.
>> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
>> position 1 by 1.
>> Similarly for others.
>>
>>
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Re: [algogeeks] Array problem

[sunny agrawal]

@atul
if its sum of numbers lesser than a[i] in left to i, then still i think it
can be solved in O(nlgn) using Balanced Tree structures
ie: if we use AVL tree, then we just need a  little care of how to update
sum stored with rotations
and required ans for ith index must be calculated just after the ith
insertion

and if its sum of  numbers lesser than i, then first sort(val, index pair)
the array and find the prefix sum array would do, with some care for
duplicates

On Mon, Mar 12, 2012 at 11:09 PM, atul anand wrote:

> @sunny : it was a reply to different question.
> using AVL or RB for the given algo may screw it.
>
>
> On Mon, Mar 12, 2012 at 11:00 PM, sunny agrawal 
> wrote:
>
>> @atul
>> TC can be bound to O(nlgn) using any balanced tree ie: AVL tree, RB tree
>> as already mentioned in her last post
>>
>> On Mon, Mar 12, 2012 at 10:44 PM, atul anand wrote:
>>
>>> @payal:
>>> TC will not alwayz be O(nlogn) bcozz we are not sure if the tree formed
>>> is balanced.
>>> so worst would be O(n^2)
>>>
>>>
>>> On Mon, Mar 12, 2012 at 9:16 PM, payal gupta wrote:
>>>
 @atul...
 if its the sum of the elements to the left of a[i] which are smaller
 the my approach works w/o any flaw
 here's the working code for ithttp://ideone.com/CH7VW
 if its the sum of all elements lesser than the element a[i] then this
 algo is surely wrong
 n we then have to proceed by the avl trees or some height balanced tree
 n the algo would be of TC-O(nlogn)
 btw nyc catch thnx...

 Regards,
 PAYAL GUPTA



 On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:

> 1)First map the array numbers into the position in which they would
> be, if they are sorted,for example
> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
> 2)Now for each number ,find the cumulative frequency of index ( = the
> corresponding number in the map - 1).
> 3)Output the cumulative frequency and increase the frequency  at the
> position (=the corresponding number in the map) by the number itself.
> Example
> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
> Process the numbers now,
> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is
> 0. so output 0
>Increase the frequency at index 2 to the number 3.
> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>Increase the frequency at index 3 to the number 5.
> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
> position 1 by 1.
> Similarly for others.
>
>
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>> --
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>> B.Tech. V year,CSI
>> Indian Institute Of Technology,Roorkee
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B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee

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Re: [algogeeks] Array problem

[atul anand]

@sunny : it was a reply to different question.
using AVL or RB for the given algo may screw it.

On Mon, Mar 12, 2012 at 11:00 PM, sunny agrawal wrote:

> @atul
> TC can be bound to O(nlgn) using any balanced tree ie: AVL tree, RB tree
> as already mentioned in her last post
>
> On Mon, Mar 12, 2012 at 10:44 PM, atul anand wrote:
>
>> @payal:
>> TC will not alwayz be O(nlogn) bcozz we are not sure if the tree formed
>> is balanced.
>> so worst would be O(n^2)
>>
>>
>> On Mon, Mar 12, 2012 at 9:16 PM, payal gupta  wrote:
>>
>>> @atul...
>>> if its the sum of the elements to the left of a[i] which are smaller the
>>> my approach works w/o any flaw
>>> here's the working code for ithttp://ideone.com/CH7VW
>>> if its the sum of all elements lesser than the element a[i] then this
>>> algo is surely wrong
>>> n we then have to proceed by the avl trees or some height balanced tree
>>> n the algo would be of TC-O(nlogn)
>>> btw nyc catch thnx...
>>>
>>> Regards,
>>> PAYAL GUPTA
>>>
>>>
>>>
>>> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>>>
 1)First map the array numbers into the position in which they would be,
 if they are sorted,for example
 {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
 2)Now for each number ,find the cumulative frequency of index ( = the
 corresponding number in the map - 1).
 3)Output the cumulative frequency and increase the frequency  at the
 position (=the corresponding number in the map) by the number itself.
 Example
 {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
 Process the numbers now,
 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is
 0. so output 0
Increase the frequency at index 2 to the number 3.
 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
Increase the frequency at index 3 to the number 5.
 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
 position 1 by 1.
 Similarly for others.


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Re: [algogeeks] Array problem

[sunny agrawal]

@atul
TC can be bound to O(nlgn) using any balanced tree ie: AVL tree, RB tree as
already mentioned in her last post

On Mon, Mar 12, 2012 at 10:44 PM, atul anand wrote:

> @payal:
> TC will not alwayz be O(nlogn) bcozz we are not sure if the tree formed is
> balanced.
> so worst would be O(n^2)
>
>
> On Mon, Mar 12, 2012 at 9:16 PM, payal gupta  wrote:
>
>> @atul...
>> if its the sum of the elements to the left of a[i] which are smaller the
>> my approach works w/o any flaw
>> here's the working code for ithttp://ideone.com/CH7VW
>> if its the sum of all elements lesser than the element a[i] then this
>> algo is surely wrong
>> n we then have to proceed by the avl trees or some height balanced tree n
>> the algo would be of TC-O(nlogn)
>> btw nyc catch thnx...
>>
>> Regards,
>> PAYAL GUPTA
>>
>>
>>
>> On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor wrote:
>>
>>> 1)First map the array numbers into the position in which they would be,
>>> if they are sorted,for example
>>> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
>>> 2)Now for each number ,find the cumulative frequency of index ( = the
>>> corresponding number in the map - 1).
>>> 3)Output the cumulative frequency and increase the frequency  at the
>>> position (=the corresponding number in the map) by the number itself.
>>> Example
>>> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
>>> Process the numbers now,
>>> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0.
>>> so output 0
>>>Increase the frequency at index 2 to the number 3.
>>> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>>>Increase the frequency at index 3 to the number 5.
>>> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at
>>> position 1 by 1.
>>> Similarly for others.
>>>
>>>
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-- 
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B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee

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Re: [algogeeks] Array problem

[payal gupta]

@atul...
if its the sum of the elements to the left of a[i] which are smaller the my
approach works w/o any flaw
here's the working code for ithttp://ideone.com/CH7VW
if its the sum of all elements lesser than the element a[i] then this algo
is surely wrong
n we then have to proceed by the avl trees or some height balanced tree n
the algo would be of TC-O(nlogn)
btw nyc catch thnx...

Regards,
PAYAL GUPTA


On Mon, Mar 12, 2012 at 5:19 PM, Piyush Kapoor  wrote:

> 1)First map the array numbers into the position in which they would be, if
> they are sorted,for example
> {30,50,10,60,77,88} ---> {2,3,1,4,5,6}
> 2)Now for each number ,find the cumulative frequency of index ( = the
> corresponding number in the map - 1).
> 3)Output the cumulative frequency and increase the frequency  at the
> position (=the corresponding number in the map) by the number itself.
> Example
> {3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
> Process the numbers now,
> 1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0.
> so output 0
>Increase the frequency at index 2 to the number 3.
> 2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
>Increase the frequency at index 3 to the number 5.
> 3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at position
> 1 by 1.
> Similarly for others.
>
>
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Re: [algogeeks] Array problem

[Piyush Kapoor]

1)First map the array numbers into the position in which they would be, if
they are sorted,for example
{30,50,10,60,77,88} ---> {2,3,1,4,5,6}
2)Now for each number ,find the cumulative frequency of index ( = the
corresponding number in the map - 1).
3)Output the cumulative frequency and increase the frequency  at the
position (=the corresponding number in the map) by the number itself.
Example
{3,5,1,6,7,8}  Map of which would be {2,3,1,4,5,6}
Process the numbers now,
1)3 comes ,find the cumulative frequency at index 1 ( = 2-1) which is 0. so
output 0
   Increase the frequency at index 2 to the number 3.
2)5 comes,find the CF at index 2( = 3-1) which is equal to 3 .output 3.
   Increase the frequency at index 3 to the number 5.
3)1 comes,CF at index 0 (=1-1) = 0 so output 0.increase the F at position 1
by 1.
Similarly for others.

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Re: [algogeeks] Array problem

[atul anand]

@piyush :  i dont knw what modification you have made to the BIT to make it
work for this problem .
please provide the code for better understanding or algo will do.

On Mon, Mar 12, 2012 at 3:56 PM, Piyush Kapoor  wrote:

> @atul anand : it will work,i can give u the code.
>
>
> On Mon, Mar 12, 2012 at 11:53 AM, sanjiv yadav wrote:
>
>> u r right.
>>
>>
>> On Mon, Mar 12, 2012 at 11:17 AM, atul anand wrote:
>>
>>> @sanjiv : wont work for this test case :-
>>>
>>> {1,5,3,6,2,7,8};
>>>
>>>
>>> On Mon, Mar 12, 2012 at 10:54 AM, sanjiv yadav 
>>> wrote:
>>>
 @atul anand- It will still work as follows---

 (3,0)
 /  \(5,0+3)
   (1,0) \(6,0+3+5)
 \(2,0+1)\(7,0+3+5+6)
   \(8,0+3+5+6+7)

 here, my logic is that if number is grater than its parent,then add the
 parent in the current sum,else keep it as such.

 check it and made correction in my logic if i m wrong.


 On Mon, Mar 12, 2012 at 10:33 AM, atul anand 
 wrote:

> @piyush : i dont think so BIT would work over here , we are not just
> reporting cumulative sum tilll index i.
>
> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor 
> wrote:
>
>> This can be done very easily with the help of a Binary Indexed
>> Tree,and it is very short to code as well.Simply process the numbers in
>> order,and for each number output the cumulative frequency of the index of
>> the number you are processing.
>>
>>
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 --
 Regards

 Sanjiv Yadav

 MobNo.-  8050142693

 Email Id-  sanjiv2009...@gmail.com

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>>
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>>
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>>
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> *2nd year,CSE
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Re: [algogeeks] Array problem

[Piyush Kapoor]

@atul anand : it will work,i can give u the code.

On Mon, Mar 12, 2012 at 11:53 AM, sanjiv yadav wrote:

> u r right.
>
>
> On Mon, Mar 12, 2012 at 11:17 AM, atul anand wrote:
>
>> @sanjiv : wont work for this test case :-
>>
>> {1,5,3,6,2,7,8};
>>
>>
>> On Mon, Mar 12, 2012 at 10:54 AM, sanjiv yadav 
>> wrote:
>>
>>> @atul anand- It will still work as follows---
>>>
>>> (3,0)
>>> /  \(5,0+3)
>>>   (1,0) \(6,0+3+5)
>>> \(2,0+1)\(7,0+3+5+6)
>>>   \(8,0+3+5+6+7)
>>>
>>> here, my logic is that if number is grater than its parent,then add the
>>> parent in the current sum,else keep it as such.
>>>
>>> check it and made correction in my logic if i m wrong.
>>>
>>>
>>> On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:
>>>
 @piyush : i dont think so BIT would work over here , we are not just
 reporting cumulative sum tilll index i.

 On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:

> This can be done very easily with the help of a Binary Indexed
> Tree,and it is very short to code as well.Simply process the numbers in
> order,and for each number output the cumulative frequency of the index of
> the number you are processing.
>
>
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>>>
>>>
>>>
>>> --
>>> Regards
>>>
>>> Sanjiv Yadav
>>>
>>> MobNo.-  8050142693
>>>
>>> Email Id-  sanjiv2009...@gmail.com
>>>
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*Regards,*
*Piyush Kapoor,*
*2nd year,CSE
IT-BHU*

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Re: [algogeeks] Array problem

[sanjiv yadav]

u r right.

On Mon, Mar 12, 2012 at 11:17 AM, atul anand wrote:

> @sanjiv : wont work for this test case :-
>
> {1,5,3,6,2,7,8};
>
>
> On Mon, Mar 12, 2012 at 10:54 AM, sanjiv yadav wrote:
>
>> @atul anand- It will still work as follows---
>>
>> (3,0)
>> /  \(5,0+3)
>>   (1,0) \(6,0+3+5)
>> \(2,0+1)\(7,0+3+5+6)
>>   \(8,0+3+5+6+7)
>>
>> here, my logic is that if number is grater than its parent,then add the
>> parent in the current sum,else keep it as such.
>>
>> check it and made correction in my logic if i m wrong.
>>
>>
>> On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:
>>
>>> @piyush : i dont think so BIT would work over here , we are not just
>>> reporting cumulative sum tilll index i.
>>>
>>> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:
>>>
 This can be done very easily with the help of a Binary Indexed Tree,and
 it is very short to code as well.Simply process the numbers in order,and
 for each number output the cumulative frequency of the index of the number
 you are processing.


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>>
>> Sanjiv Yadav
>>
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>>
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Re: [algogeeks] Array problem

[atul anand]

@sanjiv : wont work for this test case :-

{1,5,3,6,2,7,8};

On Mon, Mar 12, 2012 at 10:54 AM, sanjiv yadav wrote:

> @atul anand- It will still work as follows---
>
> (3,0)
> /  \(5,0+3)
>   (1,0) \(6,0+3+5)
> \(2,0+1)\(7,0+3+5+6)
>   \(8,0+3+5+6+7)
>
> here, my logic is that if number is grater than its parent,then add the
> parent in the current sum,else keep it as such.
>
> check it and made correction in my logic if i m wrong.
>
>
> On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:
>
>> @piyush : i dont think so BIT would work over here , we are not just
>> reporting cumulative sum tilll index i.
>>
>> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:
>>
>>> This can be done very easily with the help of a Binary Indexed Tree,and
>>> it is very short to code as well.Simply process the numbers in order,and
>>> for each number output the cumulative frequency of the index of the number
>>> you are processing.
>>>
>>>
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>>
>
>
>
> --
> Regards
>
> Sanjiv Yadav
>
> MobNo.-  8050142693
>
> Email Id-  sanjiv2009...@gmail.com
>
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Re: [algogeeks] Array problem

[sanjiv yadav]

@atul anand- It will still work as follows---

(3,0)
/  \(5,0+3)
  (1,0) \(6,0+3+5)
\(2,0+1)\(7,0+3+5+6)
  \(8,0+3+5+6+7)

here, my logic is that if number is grater than its parent,then add the
parent in the current sum,else keep it as such.

check it and made correction in my logic if i m wrong.

On Mon, Mar 12, 2012 at 10:33 AM, atul anand wrote:

> @piyush : i dont think so BIT would work over here , we are not just
> reporting cumulative sum tilll index i.
>
> On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor wrote:
>
>> This can be done very easily with the help of a Binary Indexed Tree,and
>> it is very short to code as well.Simply process the numbers in order,and
>> for each number output the cumulative frequency of the index of the number
>> you are processing.
>>
>>
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Sanjiv Yadav

MobNo.-  8050142693

Email Id-  sanjiv2009...@gmail.com

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Re: [algogeeks] Array problem

[atul anand]

@piyush : i dont think so BIT would work over here , we are not just
reporting cumulative sum tilll index i.

On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor  wrote:

> This can be done very easily with the help of a Binary Indexed Tree,and it
> is very short to code as well.Simply process the numbers in order,and for
> each number output the cumulative frequency of the index of the number you
> are processing.
>
>
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Re: [algogeeks] Array problem

[atul anand]

@payal : what will be be the structure of the augmented tree , i add 2 to
the given input. so input become.
ar[]={3,5,1,6,7,8,2};

On Sun, Mar 11, 2012 at 11:07 PM, payal gupta  wrote:

>  the algo vich i thought of is as follows-
>
> struct node{
> int data;
> struct node *left,*right;
> int lsum;   //for storing the leftsum
> int k;// for storing the storing the reqd sum which is leftsum + sum
> of the elements traversed till now on the path
> };
>
>
> Algo for insertion of nodes in bst->
> 1. add the new node with data as the value ,lsum set to 0 and k set to 0
> 2.continue adding further nodes and update the value of 'k' by the data to
> be added  whenever value is added to theleft subtree of the node
> 3. on adding the node to the right of the root set the value of leftsum to
> be summation of rootvalue+k of the nodes considered during traversal
>
>
>
>(3,0,0)
> \
> (5,3+0,0)
>
>(3,0,1)
> /\
> (1,0,0)(5,3+0,0)
>
>(3,0,1)
>/ \
> (1,0,0)  (5,3+0,0)
>\
>(6,1+5+3,0)
>
> (3,0,1)
> / \
>  (1,0,0)   (5,3+0,0)
>  \
>  (6,1+5+3,0)
>\
>(7,1+5+3+6,0)
>
> (3,0,1)
> / \
> (1,0,0)  (5,3+0,0)
> \
> (6,1+5+3,0)
>   \
>   (7,1+3+5+6,0)
> \
> (8,3+1+5+6+7,0)
>
>
> where the node-(data,leftsum,k)
>
>
>
>
>
>
> On Sun, Mar 11, 2012 at 7:06 PM, sanjiv yadav wrote:
>
>> u r right payal but
>> can u expln o(n) time complexity..
>>
>>
>> On Sun, Mar 11, 2012 at 6:10 PM, payal gupta  wrote:
>>
>>> By Augmented BST-
>>> TC-O(n)
>>>
>>>
>>> On Sun, Mar 11, 2012 at 3:08 PM, Gaurav Popli 
>>> wrote:
>>>
 given an array of size n...
 create an array of size n such that ai where ai is the element in the
 new array at index location i is equal to sum of all elements in
 original array which are smaller than element at posn i...

 e.g
 ar[]={3,5,1,6,7,8};

 ar1[]={0,3,0,9,15,22};

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>>
>>
>>
>> --
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>>
>> Sanjiv Yadav
>>
>> MobNo.-  8050142693
>>
>> Email Id-  sanjiv2009...@gmail.com
>>
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Re: [algogeeks] Array problem

[Dheeraj Sharma]

we can use self balancing BST for this problem to yield the complexity
O(nlogn) ..where every node will contain the sum of the node values on it
left sub tree .. you can check this post..its pretty similar (Method 2)

http://www.geeksforgeeks.org/archives/17235

On Mon, Mar 12, 2012 at 12:58 AM, Piyush Kapoor  wrote:

> This can be done very easily with the help of a Binary Indexed Tree,and it
> is very short to code as well.Simply process the numbers in order,and for
> each number output the cumulative frequency of the index of the number you
> are processing.
>
>
>  --
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-- 
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Re: [algogeeks] Array problem

[Piyush Kapoor]

This can be done very easily with the help of a Binary Indexed Tree,and it
is very short to code as well.Simply process the numbers in order,and for
each number output the cumulative frequency of the index of the number you
are processing.

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Re: [algogeeks] Array problem

[payal gupta]

 the algo vich i thought of is as follows-

struct node{
int data;
struct node *left,*right;
int lsum;   //for storing the leftsum
int k;// for storing the storing the reqd sum which is leftsum + sum of
the elements traversed till now on the path
};


Algo for insertion of nodes in bst->
1. add the new node with data as the value ,lsum set to 0 and k set to 0
2.continue adding further nodes and update the value of 'k' by the data to
be added  whenever value is added to theleft subtree of the node
3. on adding the node to the right of the root set the value of leftsum to
be summation of rootvalue+k of the nodes considered during traversal



   (3,0,0)
\
(5,3+0,0)

   (3,0,1)
/\
(1,0,0)(5,3+0,0)

   (3,0,1)
   / \
(1,0,0)  (5,3+0,0)
   \
   (6,1+5+3,0)

(3,0,1)
/ \
 (1,0,0)   (5,3+0,0)
 \
 (6,1+5+3,0)
   \
   (7,1+5+3+6,0)

(3,0,1)
/ \
(1,0,0)  (5,3+0,0)
\
(6,1+5+3,0)
  \
  (7,1+3+5+6,0)
\
(8,3+1+5+6+7,0)


where the node-(data,leftsum,k)






On Sun, Mar 11, 2012 at 7:06 PM, sanjiv yadav wrote:

> u r right payal but
> can u expln o(n) time complexity..
>
>
> On Sun, Mar 11, 2012 at 6:10 PM, payal gupta  wrote:
>
>> By Augmented BST-
>> TC-O(n)
>>
>>
>> On Sun, Mar 11, 2012 at 3:08 PM, Gaurav Popli wrote:
>>
>>> given an array of size n...
>>> create an array of size n such that ai where ai is the element in the
>>> new array at index location i is equal to sum of all elements in
>>> original array which are smaller than element at posn i...
>>>
>>> e.g
>>> ar[]={3,5,1,6,7,8};
>>>
>>> ar1[]={0,3,0,9,15,22};
>>>
>>> --
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>>>
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>
>
>
> --
> Regards
>
> Sanjiv Yadav
>
> MobNo.-  8050142693
>
> Email Id-  sanjiv2009...@gmail.com
>
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Re: [algogeeks] Array problem

[sanjiv yadav]

u r right payal but
can u expln o(n) time complexity..

On Sun, Mar 11, 2012 at 6:10 PM, payal gupta  wrote:

> By Augmented BST-
> TC-O(n)
>
>
> On Sun, Mar 11, 2012 at 3:08 PM, Gaurav Popli wrote:
>
>> given an array of size n...
>> create an array of size n such that ai where ai is the element in the
>> new array at index location i is equal to sum of all elements in
>> original array which are smaller than element at posn i...
>>
>> e.g
>> ar[]={3,5,1,6,7,8};
>>
>> ar1[]={0,3,0,9,15,22};
>>
>> --
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Sanjiv Yadav

MobNo.-  8050142693

Email Id-  sanjiv2009...@gmail.com

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Re: [algogeeks] Array problem

[payal gupta]

By Augmented BST-
TC-O(n)

On Sun, Mar 11, 2012 at 3:08 PM, Gaurav Popli wrote:

> given an array of size n...
> create an array of size n such that ai where ai is the element in the
> new array at index location i is equal to sum of all elements in
> original array which are smaller than element at posn i...
>
> e.g
> ar[]={3,5,1,6,7,8};
>
> ar1[]={0,3,0,9,15,22};
>
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[algogeeks] Array problem

[Gaurav Popli]

given an array of size n...
create an array of size n such that ai where ai is the element in the
new array at index location i is equal to sum of all elements in
original array which are smaller than element at posn i...

e.g
ar[]={3,5,1,6,7,8};

ar1[]={0,3,0,9,15,22};

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Re: [algogeeks] array problem

[Amol Sharma]

wellthat will be a different question.in my question i have never
said that value of the element lies between 0 and k moreover...i don't
want the count...i just want the element which is repeated b times...

hope u got the difference.
--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
 







On Thu, Feb 16, 2012 at 11:30 PM, atul anand wrote:

> if i am not wrong , solution is given in this thread and with less
> complexity :-
>
>
> http://groups.google.com/group/algogeeks/browse_thread/thread/44dd396b22595142/6632ae276b99d4ad?hl=en&lnk=gst&q=Array+Problem+%2B+tushar#6632ae276b99d4ad
>
>
>
> On Thu, Feb 16, 2012 at 5:54 PM, Devansh Gupta 
> wrote:
>
>> On 2/16/12, Amol Sharma  wrote:
>> > k,n,b are known and 0> > --
>> >
>> >
>> > Amol Sharma
>> > Third Year Student
>> > Computer Science and Engineering
>> > MNNIT Allahabad
>> >  
>> > <
>> http://in.linkedin.com/pub/amol-sharma/21/79b/507><
>> http://www.simplyamol.blogspot.com/>
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Thu, Feb 16, 2012 at 7:56 AM, saurabh singh 
>> wrote:
>> >
>> >> k,n and b are known to us?
>> >> Saurabh Singh
>> >> B.Tech (Computer Science)
>> >> MNNIT
>> >> blog:geekinessthecoolway.blogspot.com
>> >>
>> >>
>> >>
>> >> On Wed, Feb 15, 2012 at 11:37 PM, Amol Sharma
>> >> wrote:
>> >>
>> >>> Given an array of size n*k+b.In this array n elements are repeated k
>> >>> times and 1 elements are repeated b times.find the Elements which is
>> >>> repeated b time.( O(n*k+b) expected )
>> >>> --
>> >>>
>> >>>
>> >>> Amol Sharma
>> >>> Third Year Student
>> >>> Computer Science and Engineering
>> >>> MNNIT Allahabad
>> >>> 
>> >>> <
>> http://in.linkedin.com/pub/amol-sharma/21/79b/507><
>> http://www.simplyamol.blogspot.com/>
>> >>>
>> >>>
>> >>>
>> >>>
>> >>>  --
>> >>> You received this message because you are subscribed to the Google
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>> *MNNIT, Allahabad*
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Re: [algogeeks] array problem

[atul anand]

if i am not wrong , solution is given in this thread and with less
complexity :-

http://groups.google.com/group/algogeeks/browse_thread/thread/44dd396b22595142/6632ae276b99d4ad?hl=en&lnk=gst&q=Array+Problem+%2B+tushar#6632ae276b99d4ad


On Thu, Feb 16, 2012 at 5:54 PM, Devansh Gupta wrote:

> On 2/16/12, Amol Sharma  wrote:
> > k,n,b are known and 0 > --
> >
> >
> > Amol Sharma
> > Third Year Student
> > Computer Science and Engineering
> > MNNIT Allahabad
> >  
> > <
> http://in.linkedin.com/pub/amol-sharma/21/79b/507><
> http://www.simplyamol.blogspot.com/>
> >
> >
> >
> >
> >
> >
> > On Thu, Feb 16, 2012 at 7:56 AM, saurabh singh 
> wrote:
> >
> >> k,n and b are known to us?
> >> Saurabh Singh
> >> B.Tech (Computer Science)
> >> MNNIT
> >> blog:geekinessthecoolway.blogspot.com
> >>
> >>
> >>
> >> On Wed, Feb 15, 2012 at 11:37 PM, Amol Sharma
> >> wrote:
> >>
> >>> Given an array of size n*k+b.In this array n elements are repeated k
> >>> times and 1 elements are repeated b times.find the Elements which is
> >>> repeated b time.( O(n*k+b) expected )
> >>> --
> >>>
> >>>
> >>> Amol Sharma
> >>> Third Year Student
> >>> Computer Science and Engineering
> >>> MNNIT Allahabad
> >>> 
> >>> <
> http://in.linkedin.com/pub/amol-sharma/21/79b/507><
> http://www.simplyamol.blogspot.com/>
> >>>
> >>>
> >>>
> >>>
> >>>  --
> >>> You received this message because you are subscribed to the Google
> Groups
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> >>>
> >>
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> --
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> *Devansh Gupta*
> *B.Tech Third Year*
> *MNNIT, Allahabad*
>
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Re: [algogeeks] array problem

[Devansh Gupta]

On 2/16/12, Amol Sharma  wrote:
> k,n,b are known and 0 --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  
> 
>
>
>
>
>
>
> On Thu, Feb 16, 2012 at 7:56 AM, saurabh singh  wrote:
>
>> k,n and b are known to us?
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT
>> blog:geekinessthecoolway.blogspot.com
>>
>>
>>
>> On Wed, Feb 15, 2012 at 11:37 PM, Amol Sharma
>> wrote:
>>
>>> Given an array of size n*k+b.In this array n elements are repeated k
>>> times and 1 elements are repeated b times.find the Elements which is
>>> repeated b time.( O(n*k+b) expected )
>>> --
>>>
>>>
>>> Amol Sharma
>>> Third Year Student
>>> Computer Science and Engineering
>>> MNNIT Allahabad
>>> 
>>> 
>>>
>>>
>>>
>>>
>>>  --
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>>
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>


-- 
Thanks and Regards
*Devansh Gupta*
*B.Tech Third Year*
*MNNIT, Allahabad*

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Re: [algogeeks] array problem

[saurabh singh]

k,n and b are known to us?
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com



On Wed, Feb 15, 2012 at 11:37 PM, Amol Sharma wrote:

> Given an array of size n*k+b.In this array n elements are repeated k times
> and 1 elements are repeated b times.find the Elements which is repeated b
> time.( O(n*k+b) expected )
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  
> 
>
>
>
>
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Re: [algogeeks] Array Problem

[atul anand]

yeah...only if given array is sorted ..we can do it in O(n) and O(1) fetch
with all given restriction

On Wed, Feb 15, 2012 at 10:14 AM, Kartik Sachan wrote:

> @atul if array is not sorted we have to take extra array for
> this...otherwise this cannn't be done in O(n) time.
>
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Re: [algogeeks] Array Problem

[Kartik Sachan]

@atul if array is not sorted we have to take extra array for
this...otherwise this cannn't be done in O(n) time.

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Re: [algogeeks] Array Problem

[atul anand]

@kartik : question says inplace . so using hashing would be violation...
i dont think so it can be done if array is unsorted and with given
restriction

On Wed, Feb 15, 2012 at 10:05 AM, Kartik Sachan wrote:

> if n is less than 10^6 hasing works fine ..and we count in O(1) time only
>
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Re: [algogeeks] Array Problem

[Kartik Sachan]

if n is less than 10^6 hasing works fine ..and we count in O(1) time only

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[algogeeks] Array Problem

[TUSHAR]

Given an array of size N having numbers in the range 0 to k where
k<=N, preprocess the array inplace and in linear time in such a way
that after preprocessing you should be able to return count of the
input element in O(1).

Please give some idea !!
Thanks..

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[algogeeks] Array Problem??

[Vikram Singh]

Given an array say A=(4,3,1,2). An array B is formed out of this in
such a way that B[i] = no. of elements in A, occuring on rhs of A[i],
which are less then A[i].
eg.for the A given, B is (3,2,0,0).
Here A of length n only contains elements from 1 to n that too
distinct..
Now the problem is:
1). You are given with any such A. Find out corresponding B?

2). You are given with any such B. Find out corresponding A?

Please provide solution in O(n),if possible??

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Re: [algogeeks] ARRAY PROBLEM

[Wladimir Tavares]

First,
you need change 0 to -1
Wladimir Araujo Tavares
*Federal University of Ceará 
Homepage  |
Maratona|
*




On Thu, Sep 29, 2011 at 7:35 AM, Wladimir Tavares wrote:

> Algorithm Kadane
>
> http://www.algorithmist.com/index.php/Kadane%27s_Algorithm
>
> http://www.cs.ucf.edu/~reinhard/classes/cop3503/lectures/AlgAnalysis04.pdf
>
>
> http://struts2spring.wordpress.com/2009/11/02/finding-the-maximum-contiguous-subsequence-in-a-one-dimensional-array/
>
>
> Wladimir Araujo Tavares
> *Federal University of Ceará 
> Homepage  | 
> Maratona|
> *
>
>
>
>
>
> On Thu, Sep 29, 2011 at 5:28 AM, apoorv gupta wrote:
>
>> sachan!!amol ke rum par jaakar pooch le.
>>
>>
>> On Thu, Sep 29, 2011 at 1:10 PM, kartik sachan 
>> wrote:
>>
>>> ok amol i will do it..but i am unable to convience myself
>>> that this algo will give the desire result
>>>
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>>
>>
>>
>> --
>> *
>> 
>> Thanks And Sincere Regards
>> Apoorv Gupta
>> Btech 3rd Year
>> Computer Science And Engineering
>> MNNIT Allahabad
>> *
>>
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>
>

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Re: [algogeeks] ARRAY PROBLEM

[Wladimir Tavares]

Algorithm Kadane

http://www.algorithmist.com/index.php/Kadane%27s_Algorithm

http://www.cs.ucf.edu/~reinhard/classes/cop3503/lectures/AlgAnalysis04.pdf

http://struts2spring.wordpress.com/2009/11/02/finding-the-maximum-contiguous-subsequence-in-a-one-dimensional-array/


Wladimir Araujo Tavares
*Federal University of Ceará 
Homepage  |
Maratona|
*




On Thu, Sep 29, 2011 at 5:28 AM, apoorv gupta wrote:

> sachan!!amol ke rum par jaakar pooch le.
>
>
> On Thu, Sep 29, 2011 at 1:10 PM, kartik sachan wrote:
>
>> ok amol i will do it..but i am unable to convience myself that
>> this algo will give the desire result
>>
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>>
>
>
>
> --
> *
> 
> Thanks And Sincere Regards
> Apoorv Gupta
> Btech 3rd Year
> Computer Science And Engineering
> MNNIT Allahabad
> *
>
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Re: [algogeeks] ARRAY PROBLEM

[apoorv gupta]

sachan!!amol ke rum par jaakar pooch le.

On Thu, Sep 29, 2011 at 1:10 PM, kartik sachan wrote:

> ok amol i will do it..but i am unable to convience myself that
> this algo will give the desire result
>
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-- 
*

Thanks And Sincere Regards
Apoorv Gupta
Btech 3rd Year
Computer Science And Engineering
MNNIT Allahabad
*

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Re: [algogeeks] ARRAY PROBLEM

[kartik sachan]

ok amol i will do it..but i am unable to convience myself that
this algo will give the desire result

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Re: [algogeeks] ARRAY PROBLEM

[Amol Sharma]

@kartik...try to implement the algo using pen and paper take 2-3 extra
variables for storing index also along with the variable max.the same
algo will also give you the required subarray indexes along with its
length...
--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad







On Thu, Sep 29, 2011 at 12:58 PM, DIVIJ WADHAWAN  wrote:

> Impossible to solve in O(n) I suppose
>
>
> On Thu, Sep 29, 2011 at 12:34 PM, kartik sachan 
> wrote:
>
>> @AMOL i want array index i.e i to j that will be max subarry which has
>> equal no of zero and one's
>>
>>
>> but i think ur soln is not providing this
>>
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Re: [algogeeks] ARRAY PROBLEM

[DIVIJ WADHAWAN]

Impossible to solve in O(n) I suppose

On Thu, Sep 29, 2011 at 12:34 PM, kartik sachan wrote:

> @AMOL i want array index i.e i to j that will be max subarry which has
> equal no of zero and one's
>
>
> but i think ur soln is not providing this
>
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Re: [algogeeks] ARRAY PROBLEM

[kartik sachan]

@AMOL i want array index i.e i to j that will be max subarry which has equal
no of zero and one's


but i think ur soln is not providing this

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Re: [algogeeks] ARRAY PROBLEM

[Shravan Kumar]

O/P should be 00111010  and sub array is exclusive of start index, inclusive
of end index.

Nice solution

On Thu, Sep 29, 2011 at 11:58 AM, UTKARSH SRIVASTAV  wrote:

> @Amol +1
>
> On Thu, Sep 29, 2011 at 11:52 AM, Amol Sharma wrote:
>
>> given array-
>> 1  0  0  1  1  1  0  1  0  1
>> for our convenience lets replace 0 by -1
>> array becomes
>> 1 -1 -1  1  1  1 -1  1 -1  1
>>
>> take another array count which which represents the sum till that index
>> sum array becomes
>> 1  0 -1  0  1  2  1  2  1  2 //count array
>>
>> now make an important observation that if a number repeats in the count
>> array then between that two indexes equal number of zeros and ones have been
>> added...
>>
>> now take 2 arrays of size 2n+1 and index them from -n to n.lets call
>> them array a_front and a_backthe idea behind indexing is that the
>> possible value of sum varies from -n(all 0's) to +n(all 1's)
>>
>>
>> for *a_front*
>>   traverse the count array from front and for each number update index of
>> it's first occurrence in the a_front array
>>in this case...a_front[1]=0,a_front[0]=1,a_front[-1]=2,a_front[2]=6...only
>> these four entries will be updated in the a_front array
>>
>> for *a_back*
>>   traverse the count array from back and for each number update index of
>> it's last occurrence in the a_back array
>>  in this case...a_back[2]=9,a_back[1]=8,a_back[0]=3,a_back[-1]=2..only
>> these four entries will be updated in the a_back array
>>
>> now traverse both a_front and a_back simultaneously
>> max(a_back[i]-a_front[i]) will indicate the maximum subarray with equal
>> zeros and ones.
>>
>> Hope this helps !!
>>
>>
>>
>> --
>>
>>
>> Amol Sharma
>> Third Year Student
>> Computer Science and Engineering
>> MNNIT Allahabad
>>  
>> 
>>
>>
>>
>>
>>
>> On Thu, Sep 29, 2011 at 10:39 AM, kartik sachan 
>> wrote:
>>
>>>Given a binary array ( array containing only 0s and 1s ). You have to
>>> print the sub-array with
>>> maximum number of equal 1s and 0s.
>>> INPUT   OUTPUT
>>> 1001110101  0011101
>>>
>>>
>>> complex-O(n)
>>>
>>> --
>>>
>>> *WITH REGARDS,*
>>> *
>>> *
>>> *KARTIK SACHAN*
>>>
>>> *B.TECH 3rd YEAR*
>>> *COMPUTER SCIENCE AND ENGINEERING*
>>> *NIT ALLAHABAD*
>>>
>>>  --
>>> You received this message because you are subscribed to the Google Groups
>>> "Algorithm Geeks" group.
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>>>
>>
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>
>
>
> --
> *UTKARSH SRIVASTAV
> CSE-3
> B-Tech 3rd Year
> @MNNIT ALLAHABAD*
>
>
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Re: [algogeeks] ARRAY PROBLEM

[UTKARSH SRIVASTAV]

@Amol +1

On Thu, Sep 29, 2011 at 11:52 AM, Amol Sharma wrote:

> given array-
> 1  0  0  1  1  1  0  1  0  1
> for our convenience lets replace 0 by -1
> array becomes
> 1 -1 -1  1  1  1 -1  1 -1  1
>
> take another array count which which represents the sum till that index
> sum array becomes
> 1  0 -1  0  1  2  1  2  1  2 //count array
>
> now make an important observation that if a number repeats in the count
> array then between that two indexes equal number of zeros and ones have been
> added...
>
> now take 2 arrays of size 2n+1 and index them from -n to n.lets call
> them array a_front and a_backthe idea behind indexing is that the
> possible value of sum varies from -n(all 0's) to +n(all 1's)
>
>
> for *a_front*
>   traverse the count array from front and for each number update index of
> it's first occurrence in the a_front array
>in this case...a_front[1]=0,a_front[0]=1,a_front[-1]=2,a_front[2]=6...only
> these four entries will be updated in the a_front array
>
> for *a_back*
>   traverse the count array from back and for each number update index of
> it's last occurrence in the a_back array
>  in this case...a_back[2]=9,a_back[1]=8,a_back[0]=3,a_back[-1]=2..only
> these four entries will be updated in the a_back array
>
> now traverse both a_front and a_back simultaneously
> max(a_back[i]-a_front[i]) will indicate the maximum subarray with equal
> zeros and ones.
>
> Hope this helps !!
>
>
>
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  
> 
>
>
>
>
>
> On Thu, Sep 29, 2011 at 10:39 AM, kartik sachan 
> wrote:
>
>>Given a binary array ( array containing only 0s and 1s ). You have to
>> print the sub-array with
>> maximum number of equal 1s and 0s.
>> INPUT   OUTPUT
>> 1001110101  0011101
>>
>>
>> complex-O(n)
>>
>> --
>>
>> *WITH REGARDS,*
>> *
>> *
>> *KARTIK SACHAN*
>>
>> *B.TECH 3rd YEAR*
>> *COMPUTER SCIENCE AND ENGINEERING*
>> *NIT ALLAHABAD*
>>
>>  --
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>>
>
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-- 
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CSE-3
B-Tech 3rd Year
@MNNIT ALLAHABAD*

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Re: [algogeeks] ARRAY PROBLEM

[Amol Sharma]

complexity O(n)
extra space O(n)
--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
 






On Thu, Sep 29, 2011 at 11:52 AM, Amol Sharma wrote:

> given array-
> 1  0  0  1  1  1  0  1  0  1
> for our convenience lets replace 0 by -1
> array becomes
> 1 -1 -1  1  1  1 -1  1 -1  1
>
> take another array count which which represents the sum till that index
> sum array becomes
> 1  0 -1  0  1  2  1  2  1  2 //count array
>
> now make an important observation that if a number repeats in the count
> array then between that two indexes equal number of zeros and ones have been
> added...
>
> now take 2 arrays of size 2n+1 and index them from -n to n.lets call
> them array a_front and a_backthe idea behind indexing is that the
> possible value of sum varies from -n(all 0's) to +n(all 1's)
>
>
> for *a_front*
>   traverse the count array from front and for each number update index of
> it's first occurrence in the a_front array
>in this case...a_front[1]=0,a_front[0]=1,a_front[-1]=2,a_front[2]=6...only
> these four entries will be updated in the a_front array
>
> for *a_back*
>   traverse the count array from back and for each number update index of
> it's last occurrence in the a_back array
>  in this case...a_back[2]=9,a_back[1]=8,a_back[0]=3,a_back[-1]=2..only
> these four entries will be updated in the a_back array
>
> now traverse both a_front and a_back simultaneously
> max(a_back[i]-a_front[i]) will indicate the maximum subarray with equal
> zeros and ones.
>
> Hope this helps !!
>
>
>
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  
> 
>
>
>
>
>
> On Thu, Sep 29, 2011 at 10:39 AM, kartik sachan 
> wrote:
>
>>Given a binary array ( array containing only 0s and 1s ). You have to
>> print the sub-array with
>> maximum number of equal 1s and 0s.
>> INPUT   OUTPUT
>> 1001110101  0011101
>>
>>
>> complex-O(n)
>>
>> --
>>
>> *WITH REGARDS,*
>> *
>> *
>> *KARTIK SACHAN*
>>
>> *B.TECH 3rd YEAR*
>> *COMPUTER SCIENCE AND ENGINEERING*
>> *NIT ALLAHABAD*
>>
>>  --
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>> algogeeks+unsubscr...@googlegroups.com.
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>> http://groups.google.com/group/algogeeks?hl=en.
>>
>
>

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Re: [algogeeks] ARRAY PROBLEM

[Amol Sharma]

given array-
1  0  0  1  1  1  0  1  0  1
for our convenience lets replace 0 by -1
array becomes
1 -1 -1  1  1  1 -1  1 -1  1

take another array count which which represents the sum till that index
sum array becomes
1  0 -1  0  1  2  1  2  1  2 //count array

now make an important observation that if a number repeats in the count
array then between that two indexes equal number of zeros and ones have been
added...

now take 2 arrays of size 2n+1 and index them from -n to n.lets call
them array a_front and a_backthe idea behind indexing is that the
possible value of sum varies from -n(all 0's) to +n(all 1's)


for *a_front*
  traverse the count array from front and for each number update index of
it's first occurrence in the a_front array
   in this case...a_front[1]=0,a_front[0]=1,a_front[-1]=2,a_front[2]=6...only
these four entries will be updated in the a_front array

for *a_back*
  traverse the count array from back and for each number update index of
it's last occurrence in the a_back array
 in this case...a_back[2]=9,a_back[1]=8,a_back[0]=3,a_back[-1]=2..only these
four entries will be updated in the a_back array

now traverse both a_front and a_back simultaneously
max(a_back[i]-a_front[i]) will indicate the maximum subarray with equal
zeros and ones.

Hope this helps !!



--


Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad







On Thu, Sep 29, 2011 at 10:39 AM, kartik sachan wrote:

>Given a binary array ( array containing only 0s and 1s ). You have to
> print the sub-array with
> maximum number of equal 1s and 0s.
> INPUT   OUTPUT
> 1001110101  0011101
>
>
> complex-O(n)
>
> --
>
> *WITH REGARDS,*
> *
> *
> *KARTIK SACHAN*
>
> *B.TECH 3rd YEAR*
> *COMPUTER SCIENCE AND ENGINEERING*
> *NIT ALLAHABAD*
>
>  --
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[algogeeks] ARRAY PROBLEM

[kartik sachan]

   Given a binary array ( array containing only 0s and 1s ). You have to
print the sub-array with
maximum number of equal 1s and 0s.
INPUT   OUTPUT
1001110101  0011101


complex-O(n)

-- 

*WITH REGARDS,*
*
*
*KARTIK SACHAN*

*B.TECH 3rd YEAR*
*COMPUTER SCIENCE AND ENGINEERING*
*NIT ALLAHABAD*

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Re: [algogeeks] array problem

[Puneet Chawla]

+1 Sanjay

On Sun, Aug 21, 2011 at 2:59 PM, Dheeraj Sharma  wrote:

> yeah bt..when we talk abt the complexity..we consider abt the worst case
>
> On 8/21/11, himanshu kansal  wrote:
> > problem: There is an array containing integers.
> > for every bit in the integer,you have to print a 1 if no of 1s
> > corresponding to that bit is more than no of 0s corresponding to that
> > bit (counting that bit in all the integers) otherwise print a 0(if no
> > of 0s corresponding to that bit are more).
> >
> > this you have to do for all bits in the integers.
> >
> > assumption:integers are of 32bits.
> > no of integers in array are odd...(i.e. there is no case like no. of
> > 1s=no. of 0s)
> >
> > i have  done this by counting the no of 1s and 0s for all bits.
> >
> > but can anyone suggest any other efficient approach (somewhat using
> > bitwise operators).
> >
> > --
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> >
> >
>
>
> --
> *Dheeraj Sharma*
> Comp Engg.
> NIT Kurukshetra
> +91 8950264227
>
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Re: [algogeeks] array problem

[Dheeraj Sharma]

yeah bt..when we talk abt the complexity..we consider abt the worst case

On 8/21/11, himanshu kansal  wrote:
> problem: There is an array containing integers.
> for every bit in the integer,you have to print a 1 if no of 1s
> corresponding to that bit is more than no of 0s corresponding to that
> bit (counting that bit in all the integers) otherwise print a 0(if no
> of 0s corresponding to that bit are more).
>
> this you have to do for all bits in the integers.
>
> assumption:integers are of 32bits.
> no of integers in array are odd...(i.e. there is no case like no. of
> 1s=no. of 0s)
>
> i have  done this by counting the no of 1s and 0s for all bits.
>
> but can anyone suggest any other efficient approach (somewhat using
> bitwise operators).
>
> --
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>


-- 
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+91 8950264227

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Re: [algogeeks] array problem

[sagar pareek]

This problem can be reduced if we are taking whole 32 bits...
Mean left most all 0's bits are also including
then if number is less than 65535 (2^16-1) then make it 0
as 16 bits are at least zero in this case

On Sun, Aug 21, 2011 at 2:19 PM, Sanjay Rajpal  wrote:

> let n be the no.of integers in the array :
>
> int i=1,a;
> int zero,one;
> for(int a=1;a<=32;a++)
> {
> zero=0;
> one=0;
> for(int j=0;j {
> if(a[j] & i)
> {
> one++;
> }
> else
> {
> zero++;
> }
> }
> if(one > zero)
> {
> printf("1s are more \n");
> }
> else
> {
> printf("0s are more \n");
> }
> i=i<<1;
> }
>
> Correct me if m wrong.
>
> Sanju
> :)
>
>
>
> On Sun, Aug 21, 2011 at 1:28 AM, Dheeraj Sharma <
> dheerajsharma1...@gmail.com> wrote:
>
>> yeah i took it in the another way..i ll post it v soon
>>
>> On 8/21/11, himanshu kansal  wrote:
>> > problem: There is an array containing integers.
>> > for every bit in the integer,you have to print a 1 if no of 1s
>> > corresponding to that bit is more than no of 0s corresponding to that
>> > bit (counting that bit in all the integers) otherwise print a 0(if no
>> > of 0s corresponding to that bit are more).
>> >
>> > this you have to do for all bits in the integers.
>> >
>> > assumption:integers are of 32bits.
>> > no of integers in array are odd...(i.e. there is no case like no. of
>> > 1s=no. of 0s)
>> >
>> > i have  done this by counting the no of 1s and 0s for all bits.
>> >
>> > but can anyone suggest any other efficient approach (somewhat using
>> > bitwise operators).
>> >
>> > --
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>> Groups
>> > "Algorithm Geeks" group.
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>> >
>> >
>>
>>
>> --
>> *Dheeraj Sharma*
>> Comp Engg.
>> NIT Kurukshetra
>> +91 8950264227
>>
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Re: [algogeeks] array problem

[Sanjay Rajpal]

let n be the no.of integers in the array :

int i=1,a;
int zero,one;
for(int a=1;a<=32;a++)
{
zero=0;
one=0;
for(int j=0;j zero)
{
printf("1s are more \n");
}
else
{
printf("0s are more \n");
}
i=i<<1;
}

Correct me if m wrong.

Sanju
:)



On Sun, Aug 21, 2011 at 1:28 AM, Dheeraj Sharma  wrote:

> yeah i took it in the another way..i ll post it v soon
>
> On 8/21/11, himanshu kansal  wrote:
> > problem: There is an array containing integers.
> > for every bit in the integer,you have to print a 1 if no of 1s
> > corresponding to that bit is more than no of 0s corresponding to that
> > bit (counting that bit in all the integers) otherwise print a 0(if no
> > of 0s corresponding to that bit are more).
> >
> > this you have to do for all bits in the integers.
> >
> > assumption:integers are of 32bits.
> > no of integers in array are odd...(i.e. there is no case like no. of
> > 1s=no. of 0s)
> >
> > i have  done this by counting the no of 1s and 0s for all bits.
> >
> > but can anyone suggest any other efficient approach (somewhat using
> > bitwise operators).
> >
> > --
> > You received this message because you are subscribed to the Google Groups
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> >
> >
>
>
> --
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> +91 8950264227
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Re: [algogeeks] array problem

[Dheeraj Sharma]

yeah i took it in the another way..i ll post it v soon

On 8/21/11, himanshu kansal  wrote:
> problem: There is an array containing integers.
> for every bit in the integer,you have to print a 1 if no of 1s
> corresponding to that bit is more than no of 0s corresponding to that
> bit (counting that bit in all the integers) otherwise print a 0(if no
> of 0s corresponding to that bit are more).
>
> this you have to do for all bits in the integers.
>
> assumption:integers are of 32bits.
> no of integers in array are odd...(i.e. there is no case like no. of
> 1s=no. of 0s)
>
> i have  done this by counting the no of 1s and 0s for all bits.
>
> but can anyone suggest any other efficient approach (somewhat using
> bitwise operators).
>
> --
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Re: [algogeeks] array problem

[Puneet Chawla]

Similary as we  are counting set bits count 0's nd cmpare nd set 1 if
coutn(1)>count(0) for each integer in array

On Sun, Aug 21, 2011 at 1:44 PM, Sanjay Rajpal  wrote:

> @Dheeraj : I think u should review the problem again.
> What u have posted is a way to find no. of set bits in a number.
> But problem is check a bit at a position in all number for no. of 1s and
> 0s, not in a single number.
> This has to be done for all 32 bits.
>
> Sanju
> :)
>
>
>
> On Sun, Aug 21, 2011 at 1:11 AM, Dheeraj Sharma <
> dheerajsharma1...@gmail.com> wrote:
>
>> while(a)
>> (
>> a&=(a-1)
>> count++
>> )
>> counts number of 1s in number 'a'..
>> Loop can be breaken if count exceeds 16..
>>
>> On 8/21/11, himanshu kansal  wrote:
>> > problem: There is an array containing integers.
>> > for every bit in the integer,you have to print a 1 if no of 1s
>> > corresponding to that bit is more than no of 0s corresponding to that
>> > bit (counting that bit in all the integers) otherwise print a 0(if no
>> > of 0s corresponding to that bit are more).
>> >
>> > this you have to do for all bits in the integers.
>> >
>> > assumption:integers are of 32bits.
>> > no of integers in array are odd...(i.e. there is no case like no. of
>> > 1s=no. of 0s)
>> >
>> > i have  done this by counting the no of 1s and 0s for all bits.
>> >
>> > but can anyone suggest any other efficient approach (somewhat using
>> > bitwise operators).
>> >
>> > --
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>> Groups
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>> > For more options, visit this group at
>> > http://groups.google.com/group/algogeeks?hl=en.
>> >
>> >
>>
>>
>> --
>> *Dheeraj Sharma*
>> Comp Engg.
>> NIT Kurukshetra
>> +91 8950264227
>>
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Re: [algogeeks] array problem

[Sanjay Rajpal]

@Dheeraj : I think u should review the problem again.
What u have posted is a way to find no. of set bits in a number.
But problem is check a bit at a position in all number for no. of 1s and 0s,
not in a single number.
This has to be done for all 32 bits.

Sanju
:)



On Sun, Aug 21, 2011 at 1:11 AM, Dheeraj Sharma  wrote:

> while(a)
> (
> a&=(a-1)
> count++
> )
> counts number of 1s in number 'a'..
> Loop can be breaken if count exceeds 16..
>
> On 8/21/11, himanshu kansal  wrote:
> > problem: There is an array containing integers.
> > for every bit in the integer,you have to print a 1 if no of 1s
> > corresponding to that bit is more than no of 0s corresponding to that
> > bit (counting that bit in all the integers) otherwise print a 0(if no
> > of 0s corresponding to that bit are more).
> >
> > this you have to do for all bits in the integers.
> >
> > assumption:integers are of 32bits.
> > no of integers in array are odd...(i.e. there is no case like no. of
> > 1s=no. of 0s)
> >
> > i have  done this by counting the no of 1s and 0s for all bits.
> >
> > but can anyone suggest any other efficient approach (somewhat using
> > bitwise operators).
> >
> > --
> > You received this message because you are subscribed to the Google Groups
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> >
> >
>
>
> --
> *Dheeraj Sharma*
> Comp Engg.
> NIT Kurukshetra
> +91 8950264227
>
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Re: [algogeeks] array problem

[Dheeraj Sharma]

while(a)
(
a&=(a-1)
count++
)
counts number of 1s in number 'a'..
Loop can be breaken if count exceeds 16..

On 8/21/11, himanshu kansal  wrote:
> problem: There is an array containing integers.
> for every bit in the integer,you have to print a 1 if no of 1s
> corresponding to that bit is more than no of 0s corresponding to that
> bit (counting that bit in all the integers) otherwise print a 0(if no
> of 0s corresponding to that bit are more).
>
> this you have to do for all bits in the integers.
>
> assumption:integers are of 32bits.
> no of integers in array are odd...(i.e. there is no case like no. of
> 1s=no. of 0s)
>
> i have  done this by counting the no of 1s and 0s for all bits.
>
> but can anyone suggest any other efficient approach (somewhat using
> bitwise operators).
>
> --
> You received this message because you are subscribed to the Google Groups
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>
>


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Re: [algogeeks] array problem

[Sanjay Rajpal]

for this problem, we will have to check all the bits i.e. 32*n where n is
the number of integers in the array.
n bits in one go. So i think your approach will be fine.

Using bitwise operators will not make any difference.

if someone knows better solution ,plz post it.

Sanju
:)



On Sun, Aug 21, 2011 at 12:47 AM, himanshu kansal <
himanshukansal...@gmail.com> wrote:

> problem: There is an array containing integers.
> for every bit in the integer,you have to print a 1 if no of 1s
> corresponding to that bit is more than no of 0s corresponding to that
> bit (counting that bit in all the integers) otherwise print a 0(if no
> of 0s corresponding to that bit are more).
>
> this you have to do for all bits in the integers.
>
> assumption:integers are of 32bits.
> no of integers in array are odd...(i.e. there is no case like no. of
> 1s=no. of 0s)
>
> i have  done this by counting the no of 1s and 0s for all bits.
>
> but can anyone suggest any other efficient approach (somewhat using
> bitwise operators).
>
> --
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[algogeeks] array problem

[himanshu kansal]

problem: There is an array containing integers.
for every bit in the integer,you have to print a 1 if no of 1s
corresponding to that bit is more than no of 0s corresponding to that
bit (counting that bit in all the integers) otherwise print a 0(if no
of 0s corresponding to that bit are more).

this you have to do for all bits in the integers.

assumption:integers are of 32bits.
no of integers in array are odd...(i.e. there is no case like no. of
1s=no. of 0s)

i have  done this by counting the no of 1s and 0s for all bits.

but can anyone suggest any other efficient approach (somewhat using
bitwise operators).

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Re: [algogeeks] Array Problem

[Yasir]

any O(nlogn) solution?

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Re: [algogeeks] Array problem

[Azazle simon]

@ps: ..:-)

On 5/22/11, Piyush Sinha  wrote:
> @MONSIEUR..
> someone once said"THE SECRET OF SUCCESS IS TO NEVER REVEAL YOUR
> SOURCES..." ;)...:P..:P
>
> On 5/22/11, MONSIEUR  wrote:
>> @piyush: excellent buddybtw what was the initial
>> spark...???.:-)
>>
>> On May 21, 1:01 pm, Piyush Sinha  wrote:
>>> @Amit JaspalThe algo given by me works for the given case..check
>>> it
>>>
>>> On 5/20/11, Anurag Bhatia  wrote:
>>>
>>>
>>>
>>> > Just need some clarification; sorry I joined the thread late. What are
>>> > we
>>> > trying maximize? A[j] -A[i] such that i>>
>>> > --Anurag
>>>
>>> > On Fri, May 20, 2011 at 12:34 AM, Kunal Patil 
>>> > wrote:
>>>
>>> >> @ Piyush: Excellent Solution...It appears both Correct and O(n)...Good
>>> >> work !![?]
>>>
>>> >> Just a minor correction in your algo.[?]
>>>
>>> >> while(B[i]>> >>  j++;
>>> >> must also check for J's bound as:
>>> >> while ( j < ( sizeof(A)/sizeof(A[0]) ) && B[i]>> >>  j++;
>>> >> Or it will crash when J goes out of bound and we try to reference
>>> >> C[j].
>>>
>>> >> Nywayz..thnx for the solution and algo !!
>>>
>>> >> --
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>>> *https://www.facebook.com/profile.php?id=10655377926*
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>>>  363.gif
>>> < 1KViewDownload
>>>
>>>  360.gif
>>> < 1KViewDownload
>>
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Re: [algogeeks] Array problem

[Piyush Sinha]

@Amit JaspalThe algo given by me works for the given case..check it

On 5/20/11, Anurag Bhatia  wrote:
> Just need some clarification; sorry I joined the thread late. What are we
> trying maximize? A[j] -A[i] such that i
> --Anurag
>
> On Fri, May 20, 2011 at 12:34 AM, Kunal Patil  wrote:
>>
>> @ Piyush: Excellent Solution...It appears both Correct and O(n)...Good
>> work !![?]
>>
>> Just a minor correction in your algo.[?]
>>
>> while(B[i]>  j++;
>> must also check for J's bound as:
>> while ( j < ( sizeof(A)/sizeof(A[0]) ) && B[i]>  j++;
>> Or it will crash when J goes out of bound and we try to reference C[j].
>>
>> Nywayz..thnx for the solution and algo !!
>>
>> --
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<<363.gif>><<360.gif>>

Re: [algogeeks] Array problem

[Anurag Bhatia]

Just need some clarification; sorry I joined the thread late. What are we
trying maximize? A[j] -A[i] such that i wrote:

> @ Piyush: Excellent Solution...It appears both Correct and O(n)...Good work
> !![?]
>
> Just a minor correction in your algo.[?]
>
> * while(B[i] * j++;
> must also check for J's bound as:
> **while ( j < ( sizeof(A)/sizeof(A[0]) )* && *B[i]  j++;
> Or it will crash when J goes out of bound and we try to reference C[j].
>
> Nywayz..thnx for the solution and algo !!
> *
>
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<<363.gif>><<360.gif>>

Re: [algogeeks] Array problem

[Amit Jaspal]

@ Above

I think your solution would not work for A[] = {9,6,3,2,8,1}

Ans is A[4]-A[1] > 0 i.e 4-1 = 3.

On Tue, May 17, 2011 at 9:57 PM, Kunal Patil  wrote:

> Ohh..If it is so...Sorry !![?]  I understood it the different way...[?]
> But if the question is as mentioned in your 2nd case then also I believe
> there is O(n) solution.[?]
>
> Maintain
> two pointers: *START* and *END*
> two variables: max1 and max2
> Assume arr[MAX_SIZE] to be the array containing the given elements.
>
> Algorithm:
> *1) Initially, make START point to zeroth element and END pointing to last
> element of the array.
> 2) Calculate max1 as:
> 2a) Compare arr[**START**] and arr[**END**].
>If arr[**START**] < arr[**END**]
>   {
>  max1 = **END** - **START**;
>  Jump to 3rd step
>   }
> 2b) If arr[**START**] >= arr[**END**]
>{
>  **END**-- ;
>  jump to step 2a and repeat this procedure till **
> END** != **START*
> *   }
> 3) Reset **END** so that it points to last element of the array.
> 4) Calculate max2 as:
> 4a) Compare arr[**START**] and arr[**END**].
>If arr[**START**] < arr[**END**]
>   {
>  max2 = **END** - **START**;
>  Jump to 5th step
>   }
> 4b) If arr**[START**] >= arr[**END**]
>{
> **START**++ ;
>  jump to step 4a and repeat this procedure till **
> START** != **END*
> *}
> 5) Return max( max1, max2)*
>
> Hope this algo is clear.
> This algo makes two passes over the array. Thus it is O(2n) = O(n)..[?]
> Let me know if this algo fails for any case you can think of..[?]
>
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-- 
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Men do less than they ought,
unless they do all they can

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Re: [algogeeks] Array problem

[Terence]

Try this case:
1000 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 1

On 2011-5-18 0:27, Kunal Patil wrote:

Ohh..If it is so...Sorry !!  I understood it the different way...
But if the question is as mentioned in your 2nd case then also I 
believe there is O(n) solution.


Maintain
two pointers: *START* and *END*
two variables: max1 and max2
Assume arr[MAX_SIZE] to be the array containing the given elements.

Algorithm:
*1) Initially, make START point to zeroth element and END pointing to 
last element of the array.

2) Calculate max1 as:
2a) Compare arr[**START**] and arr[**END**].
   If arr[**START**] < arr[**END**]
  {
 max1 = **END**- **START**;
 Jump to 3rd step
  }
2b) If arr[**START**] >= arr[**END**]
   {
**END**-- ;
 jump to step 2a and repeat this procedure 
till **END**!= **START*

*   }
3) Reset **END**so that it points to last element of the array.
4) Calculate max2 as:
4a) Compare arr[**START**] and arr[**END**].
   If arr[**START**] < arr[**END**]
  {
 max2 = **END**- **START**;
 Jump to 5th step
  }
4b) If arr**[START**] >= arr[**END**]
   {
**START**++ ;
 jump to step 4a and repeat this procedure 
till **START**!= **END*

*   }
5) Return max( max1, max2)*

Hope this algo is clear.
This algo makes two passes over the array. Thus it is O(2n) = O(n)..
Let me know if this algo fails for any case you can think of..
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<><><><><>

Re: [algogeeks] Array problem

[Kunal Patil]

@ Piyush: Excellent Solution...It appears both Correct and O(n)...Good work
!![?]

Just a minor correction in your algo.[?]
* while(B[i]http://groups.google.com/group/algogeeks?hl=en.

<<363.gif>><<360.gif>>

Re: [algogeeks] Array problem

[Piyush Sinha]

last night i was going through a similar kind of question and tried to
implement its algo in this question...If anyone finds any counter example
for it, please do comment..

Algo:-

*Let the array be A[].
We can keep two arrays B[] and C[] which will do the following work..
B[i] will store the minimum value in A[] till ith index
C[i] will store the maximum value (starting from the end) in A[i] till ith
index.*

*Lets say taking amit's example...*

*A[] = { 5 3 4 5 3 3 }*

*then B[] = {5,3,3,3,3,3}; //starting from the beginning.
and C[] = {5,5,5,5,3,3}; //starting from the end*

*the we can take two pointers i and j and a max_diff (all initialised to 0)
and run the following loop*
*while(j<(sizeof(A)/sizeof(A[0])))
{
while(B[i]=0;i--)
{
   C[i] = ((A[i]>B[i+1])?A[i]:B[i+1]);
}*

For the given example, answer is coming to be j = 4,i=1,max_diff = 4-1-1 = 2

I hope I am clear... [?]

On Wed, May 18, 2011 at 4:52 PM, Piyush Sinha wrote:

> I think it can be done in O(n) but the auxilliary space required will be
> more... in the solution which i have got its in the order of 2n
>
>
> On Wed, May 18, 2011 at 4:44 PM, Kunal Patil  wrote:
>
>> @Amit: Ohh..Your test case is correct but not my solution..[?]
>> It only works if it is guaranteed that one end will be at the extreme of
>> the array ! (UseLess ! [?])
>> Sorry folks...
>> So can anybody prove that O(n) solution does not exist for this problem?
>> [?]
>>
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>
>
>
> --
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> *IIIT, Allahabad*
> *+91-8792136657*
> *+91-7483122727*
> *https://www.facebook.com/profile.php?id=10655377926 *
>
>


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Re: [algogeeks] Array problem

[Piyush Sinha]

I think it can be done in O(n) but the auxilliary space required will be
more... in the solution which i have got its in the order of 2n

On Wed, May 18, 2011 at 4:44 PM, Kunal Patil  wrote:

> @Amit: Ohh..Your test case is correct but not my solution..[?]
> It only works if it is guaranteed that one end will be at the extreme of
> the array ! (UseLess ! [?])
> Sorry folks...
> So can anybody prove that O(n) solution does not exist for this problem?
> [?]
>
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Re: [algogeeks] Array problem

[Kunal Patil]

@Amit: Ohh..Your test case is correct but not my solution..[?]
It only works if it is guaranteed that one end will be at the extreme of the
array ! (UseLess ! [?])
Sorry folks...
So can anybody prove that O(n) solution does not exist for this problem? [?]

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Re: [algogeeks] Array problem

[amit kumar]

i dnt htink a o(n) soln exists for this problem.

On Wed, May 18, 2011 at 3:47 PM, amit kumar  wrote:

> @kunal patil
> your soln does not work for
> 5 3 4 5 3 3
>
> On Tue, May 17, 2011 at 9:57 PM, Kunal Patil  wrote:
>
>> Ohh..If it is so...Sorry !![?]  I understood it the different way...[?]
>> But if the question is as mentioned in your 2nd case then also I believe
>> there is O(n) solution.[?]
>>
>> Maintain
>> two pointers: *START* and *END*
>> two variables: max1 and max2
>> Assume arr[MAX_SIZE] to be the array containing the given elements.
>>
>> Algorithm:
>> *1) Initially, make START point to zeroth element and END pointing to
>> last element of the array.
>> 2) Calculate max1 as:
>> 2a) Compare arr[**START**] and arr[**END**].
>>If arr[**START**] < arr[**END**]
>>   {
>>  max1 = **END** - **START**;
>>  Jump to 3rd step
>>   }
>> 2b) If arr[**START**] >= arr[**END**]
>>{
>>  **END**-- ;
>>  jump to step 2a and repeat this procedure till *
>> *END** != **START*
>> *   }
>> 3) Reset **END** so that it points to last element of the array.
>> 4) Calculate max2 as:
>> 4a) Compare arr[**START**] and arr[**END**].
>>If arr[**START**] < arr[**END**]
>>   {
>>  max2 = **END** - **START**;
>>  Jump to 5th step
>>   }
>> 4b) If arr**[START**] >= arr[**END**]
>>{
>> **START**++ ;
>>  jump to step 4a and repeat this procedure till *
>> *START** != **END*
>> *}
>> 5) Return max( max1, max2)*
>>
>> Hope this algo is clear.
>> This algo makes two passes over the array. Thus it is O(2n) = O(n)..[?]
>> Let me know if this algo fails for any case you can think of..[?]
>>
>> --
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>
>

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Re: [algogeeks] Array problem

[amit kumar]

@kunal patil
your soln does not work for
5 3 4 5 3 3

On Tue, May 17, 2011 at 9:57 PM, Kunal Patil  wrote:

> Ohh..If it is so...Sorry !![?]  I understood it the different way...[?]
> But if the question is as mentioned in your 2nd case then also I believe
> there is O(n) solution.[?]
>
> Maintain
> two pointers: *START* and *END*
> two variables: max1 and max2
> Assume arr[MAX_SIZE] to be the array containing the given elements.
>
> Algorithm:
> *1) Initially, make START point to zeroth element and END pointing to last
> element of the array.
> 2) Calculate max1 as:
> 2a) Compare arr[**START**] and arr[**END**].
>If arr[**START**] < arr[**END**]
>   {
>  max1 = **END** - **START**;
>  Jump to 3rd step
>   }
> 2b) If arr[**START**] >= arr[**END**]
>{
>  **END**-- ;
>  jump to step 2a and repeat this procedure till **
> END** != **START*
> *   }
> 3) Reset **END** so that it points to last element of the array.
> 4) Calculate max2 as:
> 4a) Compare arr[**START**] and arr[**END**].
>If arr[**START**] < arr[**END**]
>   {
>  max2 = **END** - **START**;
>  Jump to 5th step
>   }
> 4b) If arr**[START**] >= arr[**END**]
>{
> **START**++ ;
>  jump to step 4a and repeat this procedure till **
> START** != **END*
> *}
> 5) Return max( max1, max2)*
>
> Hope this algo is clear.
> This algo makes two passes over the array. Thus it is O(2n) = O(n)..[?]
> Let me know if this algo fails for any case you can think of..[?]
>
> --
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Re: [algogeeks] Array problem

[Kunal Patil]

Ohh..If it is so...Sorry !![?]  I understood it the different way...[?]
But if the question is as mentioned in your 2nd case then also I believe
there is O(n) solution.[?]

Maintain
two pointers: *START* and *END*
two variables: max1 and max2
Assume arr[MAX_SIZE] to be the array containing the given elements.

Algorithm:
*1) Initially, make START point to zeroth element and END pointing to last
element of the array.
2) Calculate max1 as:
2a) Compare arr[**START**] and arr[**END**].
   If arr[**START**] < arr[**END**]
  {
 max1 = **END** - **START**;
 Jump to 3rd step
  }
2b) If arr[**START**] >= arr[**END**]
   {
 **END**-- ;
 jump to step 2a and repeat this procedure till **
END** != **START*
*   }
3) Reset **END** so that it points to last element of the array.
4) Calculate max2 as:
4a) Compare arr[**START**] and arr[**END**].
   If arr[**START**] < arr[**END**]
  {
 max2 = **END** - **START**;
 Jump to 5th step
  }
4b) If arr**[START**] >= arr[**END**]
   {
**START**++ ;
 jump to step 4a and repeat this procedure till **
START** != **END*
*}
5) Return max( max1, max2)*

Hope this algo is clear.
This algo makes two passes over the array. Thus it is O(2n) = O(n)..[?]
Let me know if this algo fails for any case you can think of..[?]

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Re: [algogeeks] Array problem

[Piyush Sinha]

@kunal

i think we both understood the problem differently...thats what i
asked from amit..that whether in the window is it neccessary the
elements in between should also be in increasing order or only the end
elements should have the relation of one being greater than the
other...I too thought for the same kind of algo had it been the first
case...but i think its the second case amit is asking forand if it
is so..ur answer is wrong...it should be 6(j-i=6-0=6)

On 5/17/11, Kunal Patil  wrote:
> @Anuj & Piyush:
> You didn't get the algo. It works on unsorted array also. You might have
> missed the statement
> *"else // next element is smaller than or equal to current element
>  reset curr_max to 1;*"
> Here, the comment itself shows unsorted elements have been taken into
> consideration.
> If you still have the doubt let me know.
> Here is code for you:
>
> #include#define MAX_SIZE 10
> int maxinterval(int a[],int size){
> int curr_max=1, prev_max=1;
>
> for(int k=0;k {
> if(a[k] < a[k+1])
> {
>   curr_max++;
>
>   if(curr_max > prev_max)
>   {
> prev_max=curr_max;
>   }
> }
> else
>   curr_max=1;
> }
> return prev_max;}
> int main(){
> int arr[MAX_SIZE] = {19,18,17,21,22,23,24};
> printf("%d\n",maxinterval(arr,7) );
> getchar();}
>
> As expected it gives output 5.
>
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Re: [algogeeks] Array problem

[Kunal Patil]

@Anuj & Piyush:
You didn't get the algo. It works on unsorted array also. You might have
missed the statement
*"else // next element is smaller than or equal to current element
 reset curr_max to 1;*"
Here, the comment itself shows unsorted elements have been taken into
consideration.
If you still have the doubt let me know.
Here is code for you:

#include#define MAX_SIZE 10
int maxinterval(int a[],int size){
int curr_max=1, prev_max=1;

for(int k=0;k prev_max)
  {
prev_max=curr_max;
  }
}
else
  curr_max=1;
}
return prev_max;}
int main(){
int arr[MAX_SIZE] = {19,18,17,21,22,23,24};
printf("%d\n",maxinterval(arr,7) );
getchar();}

As expected it gives output 5.

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Re: [algogeeks] Array problem

[Piyush Sinha]

@kunal..

anuj is right..ur code works only for sorted array...and I missed the
part of doing it in O(n) time...I don't think there is way of doing it
in O(n) time...if its there and if amit knows the solution, he should
provide some hints...

On 5/16/11, anuj agarwal  wrote:
> Kunal,
> Your solution runs in O(n) time but it is a wrong solution. It will run fine
> if the array is sorted.
>
> Anuj Agarwal
>
> Engineering is the art of making what you want from things you can get.
>
>
> On Mon, May 16, 2011 at 7:17 PM, Kunal Patil  wrote:
>
>> @Piyush Sinha: I doubt correctness of your solution. And even if it gets
>> out to be correct It is not O(n)
>> My approach:
>> Maintain 2 variables: curr_max and prev_max to keep knowledge about
>> current
>> maximum length and previous maximum length.
>>
>> Algorithm:
>>
>> *initialize curr_max and prev_max to 1
>>
>> for i=0 to size-2
>>if next element of array is greater than current element
>>  {
>>   increment curr_max;
>>   check whether curr_max is greater than prev_max, if yes,
>> assign   curr_max to prev_max;
>>  }
>>else // next element is smaller than or equal to current element
>>  reset curr_max to 1;
>> //End for
>>
>> Finally return prev_max*
>>
>> This is clearly O(n) as it iterates through array only once.
>>
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Re: [algogeeks] Array problem

[anuj agarwal]

Kunal,
Your solution runs in O(n) time but it is a wrong solution. It will run fine
if the array is sorted.

Anuj Agarwal

Engineering is the art of making what you want from things you can get.


On Mon, May 16, 2011 at 7:17 PM, Kunal Patil  wrote:

> @Piyush Sinha: I doubt correctness of your solution. And even if it gets
> out to be correct It is not O(n)
> My approach:
> Maintain 2 variables: curr_max and prev_max to keep knowledge about current
> maximum length and previous maximum length.
>
> Algorithm:
>
> *initialize curr_max and prev_max to 1
>
> for i=0 to size-2
>if next element of array is greater than current element
>  {
>   increment curr_max;
>   check whether curr_max is greater than prev_max, if yes,
> assign   curr_max to prev_max;
>  }
>else // next element is smaller than or equal to current element
>  reset curr_max to 1;
> //End for
>
> Finally return prev_max*
>
> This is clearly O(n) as it iterates through array only once.
>
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Re: [algogeeks] Array problem

[Kunal Patil]

@Piyush Sinha: I doubt correctness of your solution. And even if it gets out
to be correct It is not O(n)
My approach:
Maintain 2 variables: curr_max and prev_max to keep knowledge about current
maximum length and previous maximum length.

Algorithm:

*initialize curr_max and prev_max to 1

for i=0 to size-2
   if next element of array is greater than current element
 {
  increment curr_max;
  check whether curr_max is greater than prev_max, if yes,
assign   curr_max to prev_max;
 }
   else // next element is smaller than or equal to current element
 reset curr_max to 1;
//End for

Finally return prev_max*

This is clearly O(n) as it iterates through array only once.

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Re: [algogeeks] Array problem

[Piyush Sinha]

how about this??

*int maxinterval(int a[],int i,int j)
{
 if(i==j)
  return 0;
 int max1 = 0,max2;
 max2 = maxinterval(a,i+1,j);
 while(imax1)
max1 =j-i;
  }
  i++;
 }
 return(max1>max2?max1:max2);
}
*
On Mon, May 16, 2011 at 11:36 AM, anuj agarwal wrote:

> How about create a BST and then, for each node find the difference between
> the node and its child and do this for all except leaf nodes.
> If u want i will write the code for the same.
>
> Anuj Agarwal
>
> Engineering is the art of making what you want from things you can get.
>
>
>  On Mon, May 16, 2011 at 11:20 AM, anshu mishra  > wrote:
>
>> @amit ur code is wrong. just check it for this {5, 4, 1, 8, 4, 4};
>>
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Re: [algogeeks] Array problem

[anuj agarwal]

How about create a BST and then, for each node find the difference between
the node and its child and do this for all except leaf nodes.
If u want i will write the code for the same.

Anuj Agarwal

Engineering is the art of making what you want from things you can get.


On Mon, May 16, 2011 at 11:20 AM, anshu mishra wrote:

> @amit ur code is wrong. just check it for this {5, 4, 1, 8, 4, 4};
>
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Re: [algogeeks] Array problem

[anshu mishra]

@amit ur code is wrong. just check it for this {5, 4, 1, 8, 4, 4};

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Re: [algogeeks] Array problem

[Piyush Sinha]

@amit jaspal

I have doubt with your question...in the maximum window found i.e.
A[i..j]...the elements should be in increasing order or only the end
elements should have the relation of A[i] wrote:

> Given an array A[i..j] find out maximum j-i such that A[i] O(n) time.
>
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[algogeeks] Array problem

[amit]

Given an array A[i..j] find out maximum j-i such that A[i]http://groups.google.com/group/algogeeks?hl=en.

[algogeeks] array problem

[jalaj jaiswal]

sort the input array. only following operations on array is allowed:
1)get(index) -gets the element at that index
2)reverse(int start,int end) - example reverse(1,3) for the array
[1,2,3,4,5] will return [1,4,3,2,5]

better then nlogn

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Final Year Undergraduate,
IIIT ALLAHABAD

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Re: [algogeeks] Array Problem

[Chonku]

How about combining sum and multiplication in the first step. As in perform
both sum and multiplication or may be sum of squares.

On Wed, Aug 18, 2010 at 8:50 PM, Rais Khan  wrote:

> @Chonku: Your algo seems to fail with following input.
> Arr1[]= {1,6}
> Arr2[]={7}
>
>
>
>
> On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan wrote:
>
>> @Nikhil: Your algo seems to fail with following input. What do you say?
>> Arr1[]= {1,2,3}
>> Arr2[]={6}
>>
>>
>>
>>
>> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal <
>> nikhil.bhoja...@gmail.com> wrote:
>>
>>> Sum all the elements of both the arrays..let it be s1 and s2
>>> Multiply the elements and call as m1 and m2
>>> if(s1==s2) &&(m1==m2)
>>> return 1;else
>>> return 0;
>>>
>>> O(n)
>>>
>>> On Tue, Aug 17, 2010 at 11:33 PM, amit  wrote:
>>>
 Given two arrays of numbers, find if each of the two arrays have the
 same set of integers ? Suggest an algo which can run faster than NlogN
 without extra space?

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>>>
>>>
>>> --
>>> Thanks & Regards
>>> Nikhil Agarwal
>>> Senior Undergraduate
>>> 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] Array Problem

[Nikhil Agarwal]

1 more correction .While multiplying we won't multiply by 0(if it is an
element).


On Wed, Aug 18, 2010 at 8:50 PM, Rais Khan  wrote:

> @Chonku: Your algo seems to fail with following input.
> Arr1[]= {1,6}
> Arr2[]={7}
>
>
>
> On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan wrote:
>
>> @Nikhil: Your algo seems to fail with following input. What do you say?
>> Arr1[]= {1,2,3}
>> Arr2[]={6}
>>
>>
>>
>>
>> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal <
>> nikhil.bhoja...@gmail.com> wrote:
>>
>>> Sum all the elements of both the arrays..let it be s1 and s2
>>> Multiply the elements and call as m1 and m2
>>> if(s1==s2) &&(m1==m2)
>>> return 1;else
>>> return 0;
>>>
>>> O(n)
>>>
>>> On Tue, Aug 17, 2010 at 11:33 PM, amit  wrote:
>>>
 Given two arrays of numbers, find if each of the two arrays have the
 same set of integers ? Suggest an algo which can run faster than NlogN
 without extra space?

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>>>
>>>
>>> --
>>> Thanks & Regards
>>> Nikhil Agarwal
>>> Senior Undergraduate
>>> 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] Array Problem

[Nikhil Agarwal]

On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan  wrote:

> @Nikhil: Your algo seems to fail with following input. What do you say?
> Arr1[]= {1,2,3}
> Arr2[]={6}
>

There is an obvious check. N1==N2 at the beginning.

>
>
>
> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal  > wrote:
>
>> Sum all the elements of both the arrays..let it be s1 and s2
>> Multiply the elements and call as m1 and m2
>> if(s1==s2) &&(m1==m2)
>> return 1;else
>> return 0;
>>
>> O(n)
>>
>> On Tue, Aug 17, 2010 at 11:33 PM, amit  wrote:
>>
>>> Given two arrays of numbers, find if each of the two arrays have the
>>> same set of integers ? Suggest an algo which can run faster than NlogN
>>> without extra space?
>>>
>>> --
>>> You received this message because you are subscribed to the Google Groups
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>>>
>>>
>>
>>
>> --
>> Thanks & Regards
>> Nikhil Agarwal
>> Senior Undergraduate
>> 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] Array Problem

[srinivas reddy]

add one more thing to the solution suggested by nikhil i.e;count the number
of elements in array 1 and number of elements in array2 if these two values
are equal then after follow the algo proposed by nikhil agarwal..

On Wed, Aug 18, 2010 at 8:50 PM, Rais Khan  wrote:

> @Chonku: Your algo seems to fail with following input.
> Arr1[]= {1,6}
> Arr2[]={7}
>
>
>
>
> On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan wrote:
>
>> @Nikhil: Your algo seems to fail with following input. What do you say?
>> Arr1[]= {1,2,3}
>> Arr2[]={6}
>>
>>
>>
>>
>> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal <
>> nikhil.bhoja...@gmail.com> wrote:
>>
>>> Sum all the elements of both the arrays..let it be s1 and s2
>>> Multiply the elements and call as m1 and m2
>>> if(s1==s2) &&(m1==m2)
>>> return 1;else
>>> return 0;
>>>
>>> O(n)
>>>
>>> On Tue, Aug 17, 2010 at 11:33 PM, amit  wrote:
>>>
 Given two arrays of numbers, find if each of the two arrays have the
 same set of integers ? Suggest an algo which can run faster than NlogN
 without extra space?

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>>>
>>>
>>> --
>>> Thanks & Regards
>>> Nikhil Agarwal
>>> Senior Undergraduate
>>> Computer Science & Engineering,
>>> National Institute Of Technology, Durgapur,India
>>> http://tech-nikk.blogspot.com
>>> http://beta.freshersworld.com/communities/nitd
>>>
>>>
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>>
>>
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Re: [algogeeks] Array Problem

[Rais Khan]

@Chonku: Your algo seems to fail with following input.
Arr1[]= {1,6}
Arr2[]={7}



On Wed, Aug 18, 2010 at 8:42 PM, Rais Khan  wrote:

> @Nikhil: Your algo seems to fail with following input. What do you say?
> Arr1[]= {1,2,3}
> Arr2[]={6}
>
>
>
>
> On Wed, Aug 18, 2010 at 7:17 AM, Nikhil Agarwal  > wrote:
>
>> Sum all the elements of both the arrays..let it be s1 and s2
>> Multiply the elements and call as m1 and m2
>> if(s1==s2) &&(m1==m2)
>> return 1;else
>> return 0;
>>
>> O(n)
>>
>> On Tue, Aug 17, 2010 at 11:33 PM, amit  wrote:
>>
>>> Given two arrays of numbers, find if each of the two arrays have the
>>> same set of integers ? Suggest an algo which can run faster than NlogN
>>> without extra space?
>>>
>>> --
>>> You received this message because you are subscribed to the Google Groups
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>>>
>>>
>>
>>
>> --
>> Thanks & Regards
>> Nikhil Agarwal
>> Senior Undergraduate
>> Computer Science & Engineering,
>> National Institute Of Technology, Durgapur,India
>> http://tech-nikk.blogspot.com
>> http://beta.freshersworld.com/communities/nitd
>>
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>
>

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