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# Size of longest subsequence such that prefix sum at each factor stays better than zero

Given an array arr[] of size N and an integer X, the task is to find the length of the longest subsequence such that the prefix sum at every element of the subsequence remains greater than zero.

Example:

Input: arr[] = {-2, -1, 1, 2, -2}, N = 5
Output: 3
Explanation: The sequence can be made of elements at index 2, 3 and 4. The prefix sum at every element stays greater than zero: 1, 3, 1

Input: arr[] = {-2, 3, 3, -7, -5, 1}, N = 6
Output: 12

Approach: The given problem can be solved using a greedy approach. The idea is to create a min-heap priority queue and traverse the array from the left to right. Add the current element arr[i] to the sum and minheap, and if the sum becomes less than zero, remove the most negative element from the minheap and subtract it from the sum. The below approach can be followed to solve the problem:

• Initialize a min-heap with priority queue data structure
• Initialize a variable sum to calculate the prefix sum of the desired subsequence
• Iterate the array and at every element arr[i] and add the value to the sum and min-heap
• If the value of sum becomes less than zero, remove the most negative element from the min-heap and subtract that value from the sum
• Return the size of the min-heap as the length of the longest subsequence

Below is the implementation of the above approach:

## C++

 ` ` `#include ``using` `namespace` `std;`` ` `int` `maxScore(``int` `arr[], ``int` `N)``{``    ``    ``int` `score = 0;`` ` `    ``    ``    ``priority_queue<``int``, vector<``int``>,``                   ``greater<``int``> >``        ``pq;`` ` `    ``    ``int` `sum = 0;``    ``for` `(``int` `i = 0; i < N; i++) {`` ` `        ``        ``        ``sum += arr[i];`` ` `        ``        ``        ``pq.push(arr[i]);`` ` `        ``        ``        ``        ``        ``if` `(sum < 0) {``            ``int` `a = pq.top();``            ``sum -= a;``            ``pq.pop();``        ``}``    ``}`` ` `    ``    ``return` `pq.size();``}`` ` `int` `main()``{``    ``int` `arr[] = { -2, 3, 3, -7, -5, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`` ` `    ``cout << maxScore(arr, N);`` ` `    ``return` `0;``}`

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

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