# Maximum sum of minimums of pairs in an array

Given an array **arr[]** of **N** integers where **N** is even, the task is to group the array elements in the pairs **(X1, Y1), (X2, Y2), (X3, Y3), …** such that the sum **min(X1, Y1) + min(X2, Y2) + min(X3, Y3) + …** is maximum.**Examples:**

Input:arr[] = {1, 5, 3, 2}Output:4

(1, 5) and (3, 2) -> 1 + 2 = 3

(1, 3) and (5, 2) -> 1 + 2 = 3

(1, 2) and (5, 3) -> 1 + 3 = 4Input:arr[] = {1, 3, 2, 1, 4, 5}Output:7

**Approach:** No matter how the pairs are formed, the maximum element from the array will always be ignored as it will be the maximum element in every pair it is put into. Same goes for the second maximum element unless it is paired with the maximum element. So, to maximize the sum an optimal approach will be to sort the array and start making pairs in order starting from the maximum element.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the maximum` `// required sum of the pairs` `int` `maxSum(` `int` `a[], ` `int` `n)` `{` ` ` `// Sort the array` ` ` `sort(a, a + n);` ` ` `// To store the sum` ` ` `int` `sum = 0;` ` ` `// Start making pairs of every two` ` ` `// consecutive elements as n is even` ` ` `for` `(` `int` `i = 0; i < n - 1; i += 2) {` ` ` `// Minimum element of the current pair` ` ` `sum += a[i];` ` ` `}` ` ` `// Return the maximum possible sum` ` ` `return` `sum;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 3, 2, 1, 4, 5 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << maxSum(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.Arrays;` `class` `GFG` `{` ` ` `// Function to return the maximum` `// required sum of the pairs` `static` `int` `maxSum(` `int` `a[], ` `int` `n)` `{` ` ` `// Sort the array` ` ` `Arrays.sort(a);` ` ` `// To store the sum` ` ` `int` `sum = ` `0` `;` ` ` `// Start making pairs of every two` ` ` `// consecutive elements as n is even` ` ` `for` `(` `int` `i = ` `0` `; i < n - ` `1` `; i += ` `2` `)` ` ` `{` ` ` `// Minimum element of the current pair` ` ` `sum += a[i];` ` ` `}` ` ` `// Return the maximum possible sum` ` ` `return` `sum;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `1` `, ` `3` `, ` `2` `, ` `1` `, ` `4` `, ` `5` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(maxSum(arr, n));` `}` `}` `// This code is contributed by Code_Mech` |

## Python3

`# Python3 implementation of the approach` `# Function to return the maximum` `# required sum of the pairs` `def` `maxSum(a, n) :` ` ` `# Sort the array` ` ` `a.sort();` ` ` `# To store the sum` ` ` `sum` `=` `0` `;` ` ` `# Start making pairs of every two` ` ` `# consecutive elements as n is even` ` ` `for` `i ` `in` `range` `(` `0` `, n ` `-` `1` `, ` `2` `) :` ` ` `# Minimum element of the current pair` ` ` `sum` `+` `=` `a[i];` ` ` `# Return the maximum possible sum` ` ` `return` `sum` `;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `1` `, ` `3` `, ` `2` `, ` `1` `, ` `4` `, ` `5` `];` ` ` `n ` `=` `len` `(arr);` ` ` `print` `(maxSum(arr, n));` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` ` ` `class` `GFG` `{` ` ` `// Function to return the maximum` `// required sum of the pairs` `static` `int` `maxSum(` `int` `[]a, ` `int` `n)` `{` ` ` `// Sort the array` ` ` `Array.Sort(a);` ` ` `// To store the sum` ` ` `int` `sum = 0;` ` ` `// Start making pairs of every two` ` ` `// consecutive elements as n is even` ` ` `for` `(` `int` `i = 0; i < n - 1; i += 2)` ` ` `{` ` ` `// Minimum element of the current pair` ` ` `sum += a[i];` ` ` `}` ` ` `// Return the maximum possible sum` ` ` `return` `sum;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]arr = { 1, 3, 2, 1, 4, 5 };` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(maxSum(arr, n));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the maximum` `// required sum of the pairs` `function` `maxSum(a, n) {` ` ` `// Sort the array` ` ` `a.sort((a, b) => a - b);` ` ` `// To store the sum` ` ` `let sum = 0;` ` ` `// Start making pairs of every two` ` ` `// consecutive elements as n is even` ` ` `for` `(let i = 0; i < n - 1; i += 2) {` ` ` `// Minimum element of the current pair` ` ` `sum += a[i];` ` ` `}` ` ` `// Return the maximum possible sum` ` ` `return` `sum;` `}` `// Driver code` `let arr = [1, 3, 2, 1, 4, 5];` `let n = arr.length;` `document.write(maxSum(arr, n));` `// This code is contributed by _saurabh_jaiswal` `</script>` |

**Output:**

7

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