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

Last Updated : 23 Jul, 2025

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:

Initialize a variable ans to store the required result.
Sort the given array A[] in increasing order.
Traverse the array, A[] using the variable i
- If the value of A[i] > i+1, update ans to -1 and break out of the loop.
- Else increment ans by the difference of i+1 and A[i].
Print the value of ans as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
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# 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

<script>
// Javascript program for the above approach

// Function to find the minimum additions required 
    // to convert the array into a permutation of 1 to N
function minimumAdditions(a, n)
{

    // Sort the array in increasing order
        a.sort(function(a, b){return a-b;});
        let ans = 0;
 
        // Traverse the array
        for (let 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
// Given Input
let A = [ 1, 1, 1, 1, 1 ];
let n = A.length;

// Function Call
document.write(minimumAdditions(A, n));

// This code is contributed by unknown2108
</script>

Output:

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

Analysis of Algorithms

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