http://codepad.org/Jui20xme
<http://codepad.org/Jui20xme>a little modification over anand's code.

On Sat, Sep 11, 2010 at 2:33 PM, ashita dadlani <[email protected]> wrote:

> @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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>>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Sahana Gururaj
>>>>>
>>>>>
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>>>>>
>>>>
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>>>
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>>
>>
>

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