# Minimum number of swaps required to sort an array

Given an array of n distinct elements, find the minimum number of swaps required to sort the array.

Examples:

```Input : {4, 3, 2, 1}
Output : 2
Explanation : Swap index 0 with 3 and 1 with 2 to
form the sorted array {1, 2, 3, 4}.

Input : {1, 5, 4, 3, 2}
Output : 2
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

This can be easily done by visualizing the problem as a graph. We will have n nodes and an edge directed from node i to node j if the element at i’th index must be present at j’th index in the sorted array.

```
Graph for {4, 3, 2, 1}
```

The graph will now contain many non-intersecting cycles. Now a cycle with 2 nodes will only require 1 swap to reach the correct ordering, similarly a cycle with 3 nodes will only require 2 swap to do so.

```
Graph for {4, 5, 2, 1, 5}
```

Hence,

• ans = Σi = 1k(cycle_size – 1)
• where k is the number of cycles

Below is the C++ implementation of the idea.

## C++

```// C++ program to find  minimum number of swaps
// required to sort an array
#include<bits/stdc++.h>
using namespace std;

// Function returns the minimum number of swaps
// required to sort the array
int minSwaps(int arr[], int n)
{
// Create an array of pairs where first
// element is array element and second element
// is position of first element
pair<int, int> arrPos[n];
for (int i = 0; i < n; i++)
{
arrPos[i].first = arr[i];
arrPos[i].second = i;
}

// Sort the array by array element values to
// get right position of every element as second
// element of pair.
sort(arrPos, arrPos + n);

// To keep track of visited elements. Initialize
// all elements as not visited or false.
vector<bool> vis(n, false);

// Initialize result
int ans = 0;

// Traverse array elements
for (int i = 0; i < n; i++)
{
// already swapped and corrected or
// already present at correct pos
if (vis[i] || arrPos[i].second == i)
continue;

// find out the number of  node in
// this cycle and add in ans
int cycle_size = 0;
int j = i;
while (!vis[j])
{
vis[j] = 1;

// move to next node
j = arrPos[j].second;
cycle_size++;
}

ans += (cycle_size - 1);
}

// Return result
return ans;
}

// Driver program to test the above function
int main()
{
int arr[] = {1, 5, 4, 3, 2};
int n = (sizeof(arr) / sizeof(int));
cout << minSwaps(arr, n);
return 0;
}
```

## Java

```// Java program to find  minimum number of swaps
// required to sort an array
import javafx.util.Pair;
import java.util.ArrayList;
import java.util.*;

class GfG
{
// Function returns the minimum number of swaps
// required to sort the array
public static int minSwaps(int[] arr)
{
int n = arr.length;

// Create two arrays and use as pairs where first
// array is element and second array
// is position of first element
ArrayList <Pair <Integer, Integer> > arrpos =
new ArrayList <Pair <Integer, Integer> > ();
for (int i = 0; i < n; i++)
arrpos.add(new Pair <Integer, Integer> (arr[i], i));

// Sort the array by array element values to
// get right position of every element as the
// elements of second array.
arrpos.sort(new Comparator<Pair<Integer, Integer>>()
{
@Override
public int compare(Pair<Integer, Integer> o1,
Pair<Integer, Integer> o2)
{
if (o1.getValue() > o2.getValue())
return -1;

// We can change this to make it then look at the
// words alphabetical order
else if (o1.getValue().equals(o2.getValue()))
return 0;

else
return 1;
}
});

// To keep track of visited elements. Initialize
// all elements as not visited or false.
Boolean[] vis = new Boolean[n];
Arrays.fill(vis, false);

// Initialize result
int ans = 0;

// Traverse array elements
for (int i = 0; i < n; i++)
{
// already swapped and corrected or
// already present at correct pos
if (vis[i] || arrpos.get(i).getValue() == i)
continue;

// find out the number of  node in
// this cycle and add in ans
int cycle_size = 0;
int j = i;
while (!vis[j])
{
vis[j] = true;

// move to next node
j = arrpos.get(j).getValue();
cycle_size++;
}

ans += (cycle_size - 1);
}

// Return result
return ans;
}
}

// Driver class
class MinSwaps
{
// Driver program to test the above function
public static void main(String[] args)
{
int []a = {1, 5, 4, 3, 2};
GfG g = new GfG();
System.out.println(g.minSwaps(a));
}
}
// This code is contributed by Saksham Seth
```

Output:

```2
```

Time Complexity: O(n*log(n))
Auxiliary Space: O(n)

Related Article :
Number of swaps to sort when only adjacent swapping allowed

This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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