Maximum sum subset having equal number of positive and negative elements

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 elements
int 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 code
int main()
{
    int arr1[] = { 1, -2, 3, 4, -5, 8 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
  
    cout << findMaxSum(arr1, n1) << endl;
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to find the
// maximum sum subset having equal
// number of positive and negative 
// elements in the subset
import java.util.*;
  
class GFG{
  
// Function to find maximum sum
// subset with equal number of 
// positive and negative elements
static 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 code
public 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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 elements
def 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 code
arr1 = [ 1, -2, 3, 4, -5, 8 ]
n1 = len(arr1)
  
print(findMaxSum(arr1, n1))
  
# This code is contributed by shubhamsingh10

chevron_right


Output:

6

Performance Analysis:

  • Time Complexity: O(N*logN)
  • Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.