Minimum swaps required to sort an array in Java

Here, we will learn to get/find the minimum swaps that are required to sort an array using java program.
Submitted by Anamika Gupta, on August 08, 2018

Problem:

In this problem, we would have an unordered array with consecutive distinct natural numbers [1,2,3,..n], where n is the size of the array. We have to find the minimum number of swaps required to sort the array in ascending order.

Note: Think and try by yourself before looking to the solution...

Solution:

This problem can be solved easily by observing the actual position of elements and their current position , the actual position of element in sorted array will be the a[cur]-1 (element-1), by tracking the actual position of element if we come back to the current element then there exist a cycle , then count the size of that cycle , the number of swaps will be cycling size-1, do this for all the cycles and add them together.

Example:

Let an array: A =[2, 4, 5, 1, 3]

minimum swap required to sort an array in java

There exist two cycles:
Cycle 1: 2 → 4 → 1 → 2
Cycle 2: 5 → 3 → 5

Size of cycle = number of nodes:
Size of cycle 1=3
Size of cycle 2=2

Number of swaps: (3-1)+(2-1) = 3

Program:

import java.io.*;
import java.math.*;
import java.util.*;

public class Swap {
	static int minimumSwaps(int[] arr) {
		int swap=0;
		boolean visited[]=new boolean[arr.length];

		for(int i=0;i<arr.length;i++){
			int j=i,cycle=0;

			while(!visited[j]){
				visited[j]=true;
				j=arr[j]-1;
				cycle++;
			}
			
			if(cycle!=0)
				swap+=cycle-1;
		}
		return swap;
	}

	public static void main(String[] args) {
		
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] arr = new int[n];
		
		for (int i = 0; i < n; i++) {
			arr[i] = scanner.nextInt();
		}
		
		int res = minimumSwaps(arr);
		System.out.println(res);
		scanner.close();
	}
}

Output

    4
    4 3 2 1
    2

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