@anand:
the maximum sum obtained from your solution is correct.
However,the subarray printed is not correct for the following case:
{-2,3,4,17,-8}
-8 is also getting printed which is not a part of thw subsequence.

On Sat, Sep 11, 2010 at 2:00 PM, ashita dadlani <[email protected]> wrote:

> @ashish:
> what if the array is {-2,3,4,17,-8,9}?
>
>
> On Wed, Sep 8, 2010 at 8:52 AM, Anand <[email protected]> wrote:
>
>> Maximum Value Contiguous Subsequence problem in O(n).
>> http://codepad.org/BhYxSlp4
>>
>>
>> On Tue, Sep 7, 2010 at 2:40 PM, ashish agarwal <
>> [email protected]> wrote:
>>
>>> yeah..it will be i=j+1;
>>> it was misprinted
>>>
>>>
>>> On Tue, Sep 7, 2010 at 10:57 AM, Sahana Gururaj <[email protected]>wrote:
>>>
>>>> In the else if condition, the increment of the end index i should be
>>>> i=j+1, not i=j+i; Otherwise the algo is right, follows the principles of
>>>> Kadane's algo :
>>>> http://en.wikipedia.org/wiki/Maximum_subarray_problem#Kadane.27s_algorithm
>>>>
>>>> On Mon, Sep 6, 2010 at 3:50 PM, ashish agarwal <
>>>> [email protected]> wrote:
>>>>
>>>>> int max=0,sum=0,begin=0,end=0,i=0;
>>>>> for(int j=0 to n){
>>>>> sum+=a[j];
>>>>> if(max<sum){
>>>>>     max=sum;
>>>>>     begin=i;
>>>>>     end=j;
>>>>> }
>>>>> else if(sum<0){
>>>>> i=j+i;
>>>>> sum=0;
>>>>> }
>>>>>
>>>>> return sum;
>>>>> i will tell the starting index and j will tell ending index;
>>>>> o(n);
>>>>> correct me if i am wrong
>>>>>
>>>>>
>>>>>
>>>>> On Mon, Sep 6, 2010 at 1:42 PM, bittu <[email protected]>wrote:
>>>>>
>>>>>> Given a sequence of integers, find a continuous subsequence which
>>>>>> maximizes the sum of its elements, that is, the elements of no other
>>>>>> single subsequence add up to a value larger than this one. An empty
>>>>>> subsequence is considered to have the sum 0; thus if all elements are
>>>>>> negative, the result must be the empty sequence.
>>>>>>
>>>>>>
>>>>>> Solution:O(n^2)   i want O(nlogn).......???????????????????
>>>>>>
>>>>>>
>>>>>>
>>>>>> #include <stdio.h>
>>>>>>  #include<conio.h>
>>>>>> #include<iostream.h>
>>>>>> #include<stdlib.h>
>>>>>> int main()
>>>>>> {
>>>>>>        int a[] = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1};
>>>>>>        int length = 11;
>>>>>>
>>>>>>        int begin, end, beginmax, endmax, maxsum, sum, i;
>>>>>>
>>>>>>        sum = 0;
>>>>>>        beginmax = 0;
>>>>>>        endmax = -1;
>>>>>>        maxsum = 0;
>>>>>>
>>>>>>
>>>>>>        for (begin=0; begin<length; begin++) {
>>>>>>                sum = 0;
>>>>>>                for(end=begin; end<length; end++) {
>>>>>>                        sum += a[end];
>>>>>>                        if(sum > maxsum) {
>>>>>>                                maxsum = sum;
>>>>>>                                beginmax = begin;
>>>>>>                                endmax = end;
>>>>>>                        }
>>>>>>
>>>>>>                }
>>>>>>                 cout<<sum<<"\t";
>>>>>>        }
>>>>>>  cout<<endl;
>>>>>>        for(i=beginmax; i<=endmax; i++) {
>>>>>>                printf("%d\n", a[i]);
>>>>>>        }
>>>>>>
>>>>>>        getch();
>>>>>> }
>>>>>>
>>>>>>
>>>>>> its has time complexity O(n^2) ..does better solution exist
>>>>>>
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>>>>
>>>>
>>>> --
>>>> Sahana Gururaj
>>>>
>>>>
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>

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