Given an array arr[], the task is to find the maximum sum subset containing the equal number of positive and negative elements.
Examples:
Input: arr[] = {1, -2, 3, 4, -5, 8}
Output: 6
Explanation:
Maximum sum subset with equal number of positive and negative elements {8, -2}Input: arr[] = {-1, -2, -3, -4, -5}
Output: 0
Explanation:
As there are no positive element in the array, Maximum sum subset will be {}
Approach: The idea is to store negative and positive elements into two different arrays and then sort them individually in increasing order. Then use two pointers starting from the highest element of each array and include those pairs whose sum is greater than 0. Otherwise, If the sum of the pair is less than 0 then stop finding more elements because there will be no such pair possible with a sum greater than 0 in the left pairs.
Below is the implementation of the above approach:
C++
// C++ implementation to find the// maximum sum subset having equal// number of positive and negative // elements in the subset#include <bits/stdc++.h>using namespace std;// Function to find maximum sum// subset with equal number of // positive and negative elementsint findMaxSum(int* arr, int n){ vector<int> a; vector<int> b; // Loop to store the positive // and negative elements in // two different array for (int i = 0; i < n; i++) { if (arr[i] > 0) { a.push_back(arr[i]); } else if (arr[i] < 0) { b.push_back(arr[i]); } } // Sort both the array sort(a.begin(), a.end()); sort(b.begin(), b.end()); // Pointers starting from // the highest elements int p = a.size() - 1; int q = b.size() - 1; int s = 0; // Find pairs having sum // greater than zero while (p >= 0 && q >= 0) { if (a[p] + b[q] > 0) { s = s + a[p] + b[q]; } else { break; } p = p - 1; q = q - 1; } return s;}// Driver codeint main(){ int arr1[] = { 1, -2, 3, 4, -5, 8 }; int n1 = sizeof(arr1) / sizeof(arr1[0]); cout << findMaxSum(arr1, n1) << endl; return 0;} |
Java
// Java implementation to find the// maximum sum subset having equal// number of positive and negative // elements in the subsetimport java.util.*;class GFG{// Function to find maximum sum// subset with equal number of // positive and negative elementsstatic int findMaxSum(int []arr, int n){ Vector<Integer> a = new Vector<Integer>(); Vector<Integer> b = new Vector<Integer>(); // Loop to store the positive // and negative elements in // two different array for(int i = 0; i < n; i++) { if (arr[i] > 0) { a.add(arr[i]); } else if (arr[i] < 0) { b.add(arr[i]); } } // Sort both the array Collections.sort(a); Collections.sort(b); // Pointers starting from // the highest elements int p = a.size() - 1; int q = b.size() - 1; int s = 0; // Find pairs having sum // greater than zero while (p >= 0 && q >= 0) { if (a.get(p) + b.get(q) > 0) { s = s + a.get(p) + b.get(q); } else { break; } p = p - 1; q = q - 1; } return s;}// Driver codepublic static void main(String[] args){ int arr1[] = { 1, -2, 3, 4, -5, 8 }; int n1 = arr1.length; System.out.print( findMaxSum(arr1, n1) + "\n");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to find the# maximum sum subset having equal# number of positive and negative # elements in the subset# Function to find maximum sum# subset with equal number of # positive and negative elementsdef findMaxSum(arr, n): a = [] b = [] # Loop to store the positive # and negative elements in # two different array for i in range(n): if (arr[i] > 0): a.append(arr[i]) elif (arr[i] < 0): b.append(arr[i]) # Sort both the array a.sort() b.sort() # Pointers starting from # the highest elements p = len(a) - 1 q = len(b) - 1 s = 0 # Find pairs having sum # greater than zero while (p >= 0 and q >= 0): if (a[p] + b[q] > 0): s = s + a[p] + b[q] else: break p = p - 1 q = q - 1 return s # Driver codearr1 = [ 1, -2, 3, 4, -5, 8 ]n1 = len(arr1)print(findMaxSum(arr1, n1))# This code is contributed by shubhamsingh10 |
C#
// C# implementation to find the// maximum sum subset having equal// number of positive and negative // elements in the subsetusing System;using System.Collections.Generic;class GFG{// Function to find maximum sum// subset with equal number of // positive and negative elementsstatic int findMaxSum(int []arr, int n){ List<int> a = new List<int>(); List<int> b = new List<int>(); // Loop to store the positive // and negative elements in // two different array for(int i = 0; i < n; i++) { if (arr[i] > 0) { a.Add(arr[i]); } else if (arr[i] < 0) { b.Add(arr[i]); } } // Sort both the array a.Sort(); b.Sort(); // Pointers starting from // the highest elements int p = a.Count - 1; int q = b.Count - 1; int s = 0; // Find pairs having sum // greater than zero while (p >= 0 && q >= 0) { if (a[p] + b[q] > 0) { s = s + a[p] + b[q]; } else { break; } p = p - 1; q = q - 1; } return s;}// Driver codepublic static void Main(String[] args){ int []arr1 = { 1, -2, 3, 4, -5, 8 }; int n1 = arr1.Length; Console.Write(findMaxSum(arr1, n1) + "\n");}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript implementation to find the// maximum sum subset having equal// number of positive and negative // elements in the subset// Function to find maximum sum// subset with equal number of // positive and negative elementsfunction findMaxSum(arr, n){ var a = []; var b = []; // Loop to store the positive // and negative elements in // two different array for(var i = 0; i < n; i++) { if (arr[i] > 0) { a.push(arr[i]); } else if (arr[i] < 0) { b.push(arr[i]); } } // Sort both the array a.sort((a, b) => a - b) b.sort((a, b) => a - b) // Pointers starting from // the highest elements var p = a.length - 1; var q = b.length - 1; var s = 0; // Find pairs having sum // greater than zero while (p >= 0 && q >= 0) { if (a[p] + b[q] > 0) { s = s + a[p] + b[q]; } else { break; } p = p - 1; q = q - 1; } return s;}// Driver codevar arr1 = [ 1, -2, 3, 4, -5, 8 ];var n1 = arr1.length;document.write( findMaxSum(arr1, n1));// This code is contributed by rrrtnx</script> |
Output:
6
Performance Analysis:
- Time Complexity: O (N*log N) as the inbuilt sort function takes N log N time to complete all operations, hence the overall time taken by the algorithm is N log N
- Auxiliary Space: O(N) since an extra array is used and in the worst case, all elements will be inserted inside the vector hence algorithm takes up linear space
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