Minimum number of increment/decrement operations such that array contains all elements from 1 to N

Given an array of N elements, the task is to convert it into a permutation (Each number from 1 to N occurs exactly once) by using the following operations a minimum number of times: 

• Increment any number.
• Decrement any number.

Examples:

Input: arr[] = {1, 1, 4}
Output: 2
The array can be converted into [1, 2, 3]
by adding 1 to the 1st index i.e. 1 + 1 = 2
and decrementing 2nd index by 1 i.e. 4- 1 = 3

Input: arr[] = {3, 0}
Output: 2

The array can be converted into [2, 1]

Approach: To minimize the number of moves/operations, sort the given array and make a[i] = i+1 (0-based) which will take abs(i+1-a[i]) no. of operations for each element.

Below is the implementation of the above approach:  

C++

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

// Function to find the minimum operations
long long minimumMoves(int a[], int n)
{

    long long operations = 0;

    // Sort the given array
    sort(a, a + n);

    // Count operations by assigning a[i] = i+1
    for (int i = 0; i < n; i++)
        operations += abs(a[i] - (i + 1));

    return operations;
}

// Driver Code
int main()
{
    int arr[] = { 5, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << minimumMoves(arr, n);

    return 0;
}

Java

// Java implementation of the above approach

import java.util.*;
class solution
{
// Function to find the minimum operations
static long minimumMoves(int a[], int n)
{

    long operations = 0;

    // Sort the given array
    Arrays.sort(a);

    // Count operations by assigning a[i] = i+1
    for (int i = 0; i < n; i++)
        operations += (long)Math.abs(a[i] - (i + 1));

    return operations;
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 3, 2 };
    int n = arr.length;

    System.out.print(minimumMoves(arr, n));

}

}
//contributed by Arnab Kundu

Python3

# Python 3 implementation of the above approach

# Function to find the minimum operations
def minimumMoves(a, n):

    operations = 0
    # Sort the given array
    a.sort(reverse = False)

    # Count operations by assigning a[i] = i+1
    for i in range(0,n,1):
        operations = operations + abs(a[i] - (i + 1))

    return operations

# Driver Code
if __name__ == '__main__':
    arr = [ 5, 3, 2 ]
    n = len(arr)

    print(minimumMoves(arr, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;

class GFG
{
// Function to find the minimum operations
static long minimumMoves(int []a, int n)
{

    long operations = 0;

    // Sort the given array
    Array.Sort(a);

    // Count operations by assigning
    // a[i] = i+1
    for (int i = 0; i < n; i++)
        operations += (long)Math.Abs(a[i] - (i + 1));

    return operations;
}

// Driver Code
static public void Main ()
{
    int []arr = { 5, 3, 2 };
    int n = arr.Length;

    Console.WriteLine(minimumMoves(arr, n));
}
}

// This code is contributed by Sach_Code

PHP

<?php
// PHP implementation of the above approach
// Function to find the minimum operations

function minimumMoves($a, $n)
{
    $operations = 0;

    // Sort the given array
    sort($a);

    // Count operations by assigning
    // a[i] = i+1
    for ($i = 0; $i < $n; $i++)
        $operations += abs($a[$i] -
                          ($i + 1));

    return $operations;
}

// Driver Code
$arr = array( 5, 3, 2 );
$n = sizeof($arr);

echo minimumMoves($arr, $n);

// This code is contributed by ajit
?>

Javascript

<script>

// Javascript implementation of the above approach

// Function to find the minimum operations
function minimumMoves(a, n)
{
    let operations = 0;

    // Sort the given array
    a.sort();

    // Count operations by assigning
    // a[i] = i+1
    for(let i = 0; i < n; i++)
        operations += Math.abs(a[i] - (i + 1));

    return operations;
}

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

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

// This code is contributed by divyesh072019

</script>

Output

4

Complexity Analysis:

• Time Complexity: O(NlogN)
• Auxiliary Space: O(1)