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Length of longest subsequence such that prefix sum at every element remains greater than zero

  • Difficulty Level : Medium
  • Last Updated : 12 Nov, 2021

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:

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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++




// C++ implementation for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate longest length
// of subsequence such that its prefix sum
// at every element stays greater than zero
int maxScore(int arr[], int N)
{
    // Variable to store the answer
    int score = 0;
 
    // Min heap implementation
    // using a priority queue
    priority_queue<int, vector<int>,
                   greater<int> >
        pq;
 
    // Variable to store the sum
    int sum = 0;
    for (int i = 0; i < N; i++) {
 
        // Add the current element
        // to the sum
        sum += arr[i];
 
        // Push the element in
        // the min-heap
        pq.push(arr[i]);
 
        // If the sum becomes less than
        // zero pop the top element of
        // the min-heap and subtract it
        // from the sum
        if (sum < 0) {
            int a = pq.top();
            sum -= a;
            pq.pop();
        }
    }
 
    // Return the answer
    return pq.size();
}
 
// Driver Code
int main()
{
    int arr[] = { -2, 3, 3, -7, -5, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxScore(arr, N);
 
    return 0;
}

Java




// Java code for the above approach
import java.io.*;
import java.util.PriorityQueue;
class GFG
{
   
    // Function to calculate longest length
    // of subsequence such that its prefix sum
    // at every element stays greater than zero
    static int maxScore(int arr[], int N)
    {
       
        // Variable to store the answer
        int score = 0;
 
        // Min heap implementation
        // using a priority queue
 
        PriorityQueue<Integer> pq
            = new PriorityQueue<Integer>();
 
        // Variable to store the sum
        int sum = 0;
        for (int i = 0; i < N; i++) {
 
            // Add the current element
            // to the sum
            sum += arr[i];
 
            // Push the element in
            // the min-heap
            pq.add(arr[i]);
 
            // If the sum becomes less than
            // zero pop the top element of
            // the min-heap and subtract it
            // from the sum
            if (sum < 0) {
                int a = pq.poll();
                sum -= a;
            }
        }
 
        // Return the answer
        return pq.size();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { -2, 3, 3, -7, -5, 1 };
        int N = arr.length;
 
        System.out.println(maxScore(arr, N));
    }
}
 
// This code is contributed by Potta Lokesh
Output
4

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

 




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