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 operationslong 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 Codeint 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 approachimport java.util.*;class solution{// Function to find the minimum operationsstatic 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 Codepublic 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 operationsdef 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 Codeif __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 codelet 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)
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