# Minimize sum of numbers required to convert an array into a permutation of first N natural numbers

Given an array A[] of size N, the task is to find the minimum sum of numbers required to be added to array elements to convert the array into a permutation of 1 to N. If the array can not be converted to desired permutation, print -1.

Examples:

Input: A[] = {1, 1, 1, 1, 1}
Output: 10
Explanation: Increment A[1] by 1, A[2] by 2, A[3] by 3, A[4] by 4, thus A[] becomes {1, 2, 3, 4, 5}.
Minimum additions required = 1 + 2 + 3 + 4 = 10

Input: A[] = {2, 2, 3}
Output: -1

Approach: The idea is to use sorting. Follow these steps to solve this problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the minimum additions required ` `// to convert the array into a permutation of 1 to N` `int` `minimumAdditions(``int` `a[], ``int` `n)` `{` `    ``// Sort the array in increasing order` `    ``sort(a, a + n);` `    ``int` `ans = 0;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If a[i] > i + 1, then return -1` `        ``if` `((i + 1) - a[i] < 0) {` `            ``return` `-1;` `        ``}` `        ``if` `((i + 1) - a[i] > 0) {`   `            ``// Update answer` `            ``ans += (i + 1 - a[i]);` `        ``}` `    ``}`   `    ``// Return the required result` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Input` `    ``int` `A[] = { 1, 1, 1, 1, 1 };` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);`   `    ``// Function Call` `    ``cout << minimumAdditions(A, n);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.Arrays;`   `public` `class` `GFG {`   `    ``// Function to find the minimum additions required ` `    ``// to convert the array into a permutation of 1 to N` `    ``static` `int` `minimumAdditions(``int` `a[], ``int` `n)` `    ``{` `        ``// Sort the array in increasing order` `        ``Arrays.sort(a);` `        ``int` `ans = ``0``;`   `        ``// Traverse the array` `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// If a[i] > i + 1, then return -1` `            ``if` `((i + ``1``) - a[i] < ``0``) {` `                ``return` `-``1``;` `            ``}` `            ``if` `((i + ``1``) - a[i] > ``0``) {`   `                ``// Update answer` `                ``ans += (i + ``1` `- a[i]);` `            ``}` `        ``}`   `        ``// Return the required result` `        ``return` `ans;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{ ` `      `  `      ``// Given Input` `        ``int` `A[] = { ``1``, ``1``, ``1``, ``1``, ``1` `};` `        ``int` `n = A.length;`   `        ``// Function Call` `        ``System.out.println(minimumAdditions(A, n));` `    ``}` `}`   `// This code is contributed by abhinavjain194`

## Python3

 `# Python3 program for the above approach`   `# Function to find the minimum additions` `# required to convert the array into a ` `# permutation of 1 to N` `def` `minimumAdditions(a, n):` `    `  `    ``# Sort the array in increasing order` `    ``a ``=` `sorted``(a)` `    ``ans ``=` `0`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(n):`   `        ``# If a[i] > i + 1, then return -1` `        ``if` `((i ``+` `1``) ``-` `a[i] < ``0``):` `            ``return` `-``1`   `        ``if` `((i ``+` `1``) ``-` `a[i] > ``0``):`   `            ``# Update answer` `            ``ans ``+``=` `(i ``+` `1` `-` `a[i])`   `    ``# Return the required result` `    ``return` `ans`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given Input` `    ``A ``=` `[ ``1``, ``1``, ``1``, ``1``, ``1` `]` `    ``n ``=` `len``(A)`   `    ``# Function Call` `    ``print``(minimumAdditions(A, n))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find the minimum additions` `// required to convert the array into a` `// permutation of 1 to N` `static` `int` `minimumAdditions(``int` `[]a, ``int` `n)` `{` `    `  `    ``// Sort the array in increasing order` `    ``Array.Sort(a);` `    ``int` `ans = 0;`   `    ``// Traverse the array` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{` `        `  `        ``// If a[i] > i + 1, then return -1` `        ``if` `((i + 1) - a[i] < 0) ` `        ``{` `            ``return` `-1;` `        ``}` `        `  `        ``if` `((i + 1) - a[i] > 0) ` `        ``{` `            `  `            ``// Update answer` `            ``ans += (i + 1 - a[i]);` `        ``}` `    ``}`   `    ``// Return the required result` `    ``return` `ans;` `}`   `// Driver code` `static` `void` `Main()` `{ ` `    `  `    ``// Given Input` `    ``int``[] A = { 1, 1, 1, 1, 1 };` `    ``int` `n = A.Length;` `    `  `    ``// Function Call` `    ``Console.Write(minimumAdditions(A, n));` `}` `}`   `// This code is contributed by SoumikMondal`

## Javascript

 ``

Output:

`10`

Time Complexity: O(N* log(N))
Auxiliary Space: O(1)

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