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

Last Updated : 09 Sep, 2022

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# 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

Complexity Analysis:

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

Minimum increment/decrement operations required on Array to satisfy given conditions

Abdullah Aslam

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

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