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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