# Pair formation such that maximum pair sum is minimized

Given an array of size 2 * N integers. Divide the array into N pairs, such that the maximum pair sum is minimized. In other words, the optimal division of array into N pairs should result into a maximum pair sum which is minimum of other maximum pair sum of all possibilities.

Examples:

Input : N = 2
arr[] = { 5, 8, 3, 9 }
Output : (3, 9) (5, 8)
Explanation:
Possible pairs are :
1. (8, 9) (3, 5) Maximum Sum of a Pair = 17
2. (5, 9) (3, 8) Maximum Sum of a Pair = 14
3. (3, 9) (5, 8) Maximum Sum of a Pair = 13
Thus, in case 3, the maximum pair sum is minimum of all the other cases. Hence, the answer is(3, 9) (5, 8).

Input : N = 2
arr[] = { 9, 6, 5, 1 }
Output : (1, 9) (5, 6)

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to first sort the given array and then iterate over the loop to form pairs (i, j) where i would start from 0 and j would start from end of array correspondingly. Increment i and Decrement j to form the next pair and so on.

Below is the implementation of above approach.

## C++

 `// CPP Program to divide the array into ` `// N pairs such that maximum pair is minimized ` `#include ` ` `  `using` `namespace` `std; ` ` `  `void` `findOptimalPairs(``int` `arr[], ``int` `N) ` `{ ` `    ``sort(arr, arr + N); ` ` `  `    ``// After Sorting Maintain two variables i and j  ` `    ``// pointing to start and end of array Such that  ` `    ``// smallest element of array pairs with largest ` `    ``// element ` `    ``for` `(``int` `i = 0, j = N - 1; i <= j; i++, j--) ` `        ``cout << ``"("` `<< arr[i] << ``", "` `<< arr[j] << ``")"` `<< ``" "``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 9, 6, 5, 1 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``findOptimalPairs(arr, N);  ` `    ``return` `0; ` `} `

## Java

 `// Java Program to divide the array into ` `// N pairs such that maximum pair is minimized ` `import` `java.io.*; ` `import` `java.util.Arrays; ` ` `  `class` `GFG { ` `     `  `static` `void` `findOptimalPairs(``int` `arr[], ``int` `N) ` `{ ` `    ``Arrays.sort(arr); ` ` `  `    ``// After Sorting Maintain two variables i and j  ` `    ``// pointing to start and end of array Such that  ` `    ``// smallest element of array pairs with largest ` `    ``// element ` `    ``for` `(``int` `i = ``0``, j = N - ``1``; i <= j; i++, j--) ` `        ``System.out.print( ``"("` `+ arr[i] + ``", "` `+ arr[j] + ``")"` `+ ``" "``); ` `} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `arr[] = {``9``, ``6``, ``5``, ``1``}; ` `        ``int` `N = arr.length; ` ` `  `        ``findOptimalPairs(arr, N); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## Python3

 `# Python 3 Program to divide the array into ` `# N pairs such that maximum pair is minimized ` ` `  `def` `findOptimalPairs(arr, N): ` `    ``arr.sort(reverse ``=` `False``) ` ` `  `    ``# After Sorting Maintain two variables  ` `    ``# i and j pointing to start and end of  ` `    ``# array Such that smallest element of  ` `    ``# array pairs with largest element ` `    ``i ``=` `0` `    ``j ``=` `N ``-` `1` `    ``while``(i <``=` `j): ` `        ``print``(``"("``, arr[i], ``","``,  ` `                   ``arr[j], ``")"``, end ``=` `" "``) ` `        ``i ``+``=` `1` `        ``j ``-``=` `1` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``9``, ``6``, ``5``, ``1``] ` `    ``N ``=` `len``(arr) ` ` `  `    ``findOptimalPairs(arr, N)  ` `     `  `# This code is contributed by ` `# Sahil_Shelangia `

## C#

 `// C# Program to divide the array into ` `// N pairs such that maximum pair is minimized ` ` `  `using` `System; ` ` `  `public` `class` `GFG{ ` `    ``static` `void` `findOptimalPairs(``int` `[]arr, ``int` `N) ` `{ ` `    ``Array.Sort(arr); ` ` `  `    ``// After Sorting Maintain two variables i and j  ` `    ``// pointing to start and end of array Such that  ` `    ``// smallest element of array pairs with largest ` `    ``// element ` `    ``for` `(``int` `i = 0, j = N - 1; i <= j; i++, j--) ` `        ``Console.Write( ``"("` `+ arr[i] + ``", "` `+ arr[j] + ``")"` `+ ``" "``); ` `} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main (){ ` `         `  `        ``int` `[]arr = {9, 6, 5, 1}; ` `        ``int` `N = arr.Length; ` `        ``findOptimalPairs(arr, N); ` ` `  ` `  `// This code is contributed by ajit. ` ` `  `    ``} ` `} `

## PHP

 ` `

Output:

```(1, 9) (5, 6)
```

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