Given an array arr[] consisting of N integers, the task is to find the length of the longest subsequence such that the prefix sum at each index of the subsequence is negative.
Examples:
Input: arr[] = {-1, -3, 3, -5, 8, 2}
Output: 5
Explanation: Longest subsequence satisfying the condition is {-1, -3, 3, -5, 2}.Input: arr[] = {2, -5, 2, -1, 5, 1, -9, 10}
Output: 6
Explanation: Longest subsequence satisfying the condition is {-1, -3, 3, -5, 2}.
Approach: The problem can be solved by using a Priority Queue. Follow the steps below to solve the problem:
- Initialize a priority queue, say pq, and a variable, say S as 0, to store the elements of the subsequence formed from elements up to an index i and to store the sum of the elements in the priority queue.
- Iterate over the range [0, N – 1] using the variable i and perform the following steps:
- Increment S by arr[i] and Push arr[i] into pq.
- Iterate until S is greater than 0 and in each iteration, decrement S by the top element of pq and then, pop the top element.
- Finally, after completing the above steps, print pq.size() as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum length// of a subsequence such that prefix sum// of any index is negativeint maxLengthSubsequence(int arr[], int N){ // Max priority Queue priority_queue<int> pq; // Stores the temporary sum of a // prefix of selected subsequence int S = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Increment S by arr[i] S += arr[i]; // Push arr[i] into pq pq.push(arr[i]); // Iterate until S // is greater than 0 while (S > 0) { // Decrement S by pq.top() S -= pq.top(); // Pop the top element pq.pop(); } } // Return the maxLength return pq.size();}// Driver Codeint main(){ // Given Input int arr[6] = { -1, -3, 3, -5, 8, 2 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call cout << maxLengthSubsequence(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.Collections;import java.util.PriorityQueue;public class GFG{ // Function to find the maximum length // of a subsequence such that prefix sum // of any index is negative static int maxLengthSubsequence(int arr[], int N) { // Max priority Queue PriorityQueue<Integer> pq = new PriorityQueue<>( Collections.reverseOrder()); // Stores the temporary sum of a // prefix of selected subsequence int S = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Increment S by arr[i] S += arr[i]; // Add arr[i] into pq pq.add(arr[i]); // Iterate until S // is greater than 0 while (S > 0) { // Decrement S by pq.peek() S -= pq.peek(); // Remove the top element pq.remove(); } } // Return the maxLength return pq.size(); } // Driver code public static void main(String[] args) { int arr[] = { -1, -3, 3, -5, 8, 2 }; int N = arr.length; // Function call System.out.println(maxLengthSubsequence(arr, N)); }}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to find the maximum length# of a subsequence such that prefix sum# of any index is negativedef maxLengthSubsequence(arr, N): # Max priority Queue pq = [] # Stores the temporary sum of a # prefix of selected subsequence S = 0 # Traverse the array arr[] for i in range(N): # Increment S by arr[i] S += arr[i] # Push arr[i] into pq pq.append(arr[i]) # Iterate until S # is greater than 0 pq.sort(reverse = False) while (S > 0): # Decrement S by pq.top() # pq.sort(reverse=False) S = S - max(pq) # Pop the top element pq = pq[1:] # print(len(pq)) # Return the maxLength return len(pq)# Driver Codeif __name__ == '__main__': # Given Input arr = [ -1, -3, 3, -5, 8, 2 ] N = len(arr) # Function call print(maxLengthSubsequence(arr, N)) # This code is contributed by SURENDRA_GANGWAR |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to find the maximum length// of a subsequence such that prefix sum// of any index is negativestatic int maxLengthSubsequence(int []arr, int N){ // Max priority Queue List<int> pq = new List<int>(); // Stores the temporary sum of a // prefix of selected subsequence int S = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // Increment S by arr[i] S += arr[i]; // Push arr[i] into pq pq.Add(arr[i]); pq.Sort(); // Iterate until S // is greater than 0 while (S > 0) { pq.Sort(); // Decrement S by pq.top() S -= pq[pq.Count-1]; // Pop the top element pq.RemoveAt(0); } } // Return the maxLength return pq.Count;}// Driver Codepublic static void Main(){ // Given Input int []arr = { -1, -3, 3, -5, 8, 2 }; int N = arr.Length; // Function call Console.Write(maxLengthSubsequence(arr, N)); }}// This code is contributed by ipg2016107. |
Javascript
<script>// JavaScript program for the above approach// Function to find the maximum length// of a subsequence such that prefix sum// of any index is negativefunction maxLengthSubsequence(arr, N) { // Max priority Queue let pq = new Array(); // Stores the temporary sum of a // prefix of selected subsequence let S = 0; // Traverse the array arr[] for (let i = 0; i < N; i++) { // Increment S by arr[i] S += arr[i]; // Push arr[i] into pq pq.push(arr[i]); pq.sort((a, b) => a - b); // Iterate until S // is greater than 0 while (S > 0) { pq.sort((a, b) => a - b); // Decrement S by pq.top() S -= pq[pq.length - 1]; // Pop the top element pq.shift(); } } // Return the maxLength return pq.length;}// Driver Code// Given Inputlet arr = [-1, -3, 3, -5, 8, 2];let N = arr.length;// Function calldocument.write(maxLengthSubsequence(arr, N));// This code is contributed by gfgking</script> |
Output:
5
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)
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