# Maximum sum subset having equal number of positive and negative elements

• Last Updated : 29 Aug, 2022

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:
Explanation:
Maximum sum subset with equal number of positive and negative elements {8, -2}

Input: arr[] = {-1, -2, -3, -4, -5}
Output:
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 ` `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);` `    ``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 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 a = ``new` `Vector();``    ``Vector b = ``new` `Vector();``    ` `    ``// 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`

## 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 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`

## C#

 `// C# implementation to find the``// maximum sum subset having equal``// number of positive and negative``// elements in the subset``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find maximum sum``// subset with equal number of``// positive and negative elements``static` `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 code``public` `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

 ``

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

My Personal Notes arrow_drop_up