Minimum operations to change given array with at most 1 duplicate into a permutation of 1 to N

Given an array arr[] having N integers in the range [1, N] with at most one repeated element, the task is to find the minimum number of increment or decrement operations required to make the given array a permutation of numbers from 1 to N.

Examples:

Input: arr[] = {1, 2, 5, 3, 2} 
Output: 2 
Explanation: The given array contains 2 twice and 4 is missing from the array. Therefore, a 2 can be changed into 4 using two increment operations which is the minimum possible. 

Input: arr[] = {1, 2, 5, 3, 4} 
Output: 0 
Explanation: The given array already represents the permutation of integers from 1 to 5. 

Naive Approach: The given problem can be solved simply by finding the duplicate element and the missing element in the given array. This can be easily done by sorting the given array and checking for the duplicate and the missing element.

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

Efficient Approach: The given problem can be solved using a simple observation that the sum of all elements of a permutation of N integers is always equal to N*(N+1)/ 2. Hence, the number of required operations can simply be calculated by the formula | sum of array elements – N*(N+1)/ 2)|.

Below is the implementation of the above approach:

CPP

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum number of
// increment/decrement operations to change
// the given array into a permutation
int minCost(int arr[], int N)
{
    // Stores the sum of array elements
    int sumOfArray = 0;

    // Loop to iterate through the array
    for (int i = 0; i < N; i++) {
        sumOfArray += arr[i];
    }

    // Finding sum of first
    // N natural numbers
    int sumOfN = N * (N + 1) / 2;

    // Find the absolute difference
    int diff = sumOfArray - sumOfN;
    diff = diff < 0 ? -1 * diff : diff;

    // Return Answer
    return diff;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = sizeof(arr) / sizeof(int);

    cout << minCost(arr, n);

    return 0;
}

Java

// Java program of the above approach

public class GFG {

    // Function to find the minimum number of
    // increment/decrement operations to change
    // the given array into a permutation
    static int minCost(int []arr, int N)
    {
        // Stores the sum of array elements
        int sumOfArray = 0;

        // Loop to iterate through the array
        for (int i = 0; i < N; i++) {
            sumOfArray += arr[i];
        }

        // Finding sum of first
        // N natural numbers
        int sumOfN = N * (N + 1) / 2;

        // Find the absolute difference
        int diff = sumOfArray - sumOfN;
        diff = diff < 0 ? -1 * diff : diff;

        // Return Answer
        return diff;
    }

    // Driver code
    public static void main (String[] args) {

        int arr[] = { 1, 2, 5, 3, 2 };
        int n = arr.length;

        System.out.println(minCost(arr, n));
    }
}

// This code is contributed by AnkThon

Python3

# python program of the above approach

# Function to find the minimum number of
# increment/decrement operations to change
# the given array into a permutation
def minCost(arr, N):

    # Stores the sum of array elements
    sumOfArray = 0

    # Loop to iterate through the array
    for i in range(0, N):
        sumOfArray += arr[i]

        # Finding sum of first
        # N natural numbers
    sumOfN = N * (N + 1) // 2

    # Find the absolute difference
    diff = sumOfArray - sumOfN
    if diff < 0:
        diff = -1 * diff

        # Return Answer
    return diff

# Driver code
if __name__ == "__main__":

    arr = [1, 2, 5, 3, 2]
    n = len(arr)

    print(minCost(arr, n))

    # This code is contributed by rakeshsahni

C#

// C# program of the above approach
using System;
class GFG {

    // Function to find the minimum number of
    // increment/decrement operations to change
    // the given array into a permutation
    static int minCost(int []arr, int N)
    {
        // Stores the sum of array elements
        int sumOfArray = 0;

        // Loop to iterate through the array
        for (int i = 0; i < N; i++) {
            sumOfArray += arr[i];
        }

        // Finding sum of first
        // N natural numbers
        int sumOfN = N * (N + 1) / 2;

        // Find the absolute difference
        int diff = sumOfArray - sumOfN;
        diff = diff < 0 ? -1 * diff : diff;

        // Return Answer
        return diff;
    }

    // Driver code
    public static void Main () {

        int []arr = { 1, 2, 5, 3, 2 };
        int n = arr.Length;

        Console.Write(minCost(arr, n));
    }
}
// This code is contributed by Samim Hossain Mondal

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to find the minimum number of
        // increment/decrement operations to change
        // the given array into a permutation
        function minCost(arr, N)
        {

            // Stores the sum of array elements
            let sumOfArray = 0;

            // Loop to iterate through the array
            for (let i = 0; i < N; i++) {
                sumOfArray += arr[i];
            }

            // Finding sum of first
            // N natural numbers
            let sumOfN = N * (N + 1) / 2;

            // Find the absolute difference
            let diff = sumOfArray - sumOfN;
            diff = diff < 0 ? -1 * diff : diff;

            // Return Answer
            return diff;
        }

        // Driver code
        let arr = [1, 2, 5, 3, 2];
        let n = arr.length;

        document.write(minCost(arr, n));

    // This code is contributed by Potta Lokesh
    </script>

Output

2

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