# Minimum number of swaps required to sort an array

• Difficulty Level : Hard
• Last Updated : 03 Aug, 2022

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```

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 swaps to do so. `Graph for {4, 5, 2, 1, 3}`

Hence,

• ans = Σi = 1k(cycle_size – 1)

where, k is the number of cycles  Below is the implementation of the idea.

## C++

 `// C++ program to find ``// minimum number of swaps``// required to sort an array``#include` `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++;``        ``}` `        ``// Update answer by adding current cycle.``        ``if` `(cycle_size > 0)``        ``{``            ``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 > arrpos =``                  ``new` `ArrayList > ();``        ``for` `(``int` `i = ``0``; i < n; i++)``             ``arrpos.add(``new` `Pair (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>()``        ``{``            ``@Override``            ``public` `int` `compare(Pair o1,``                               ``Pair o2)``            ``{``                ``if` `(o1.getKey() > o2.getKey())``                    ``return` `-``1``;` `                ``// We can change this to make``                ``// it then look at the``                ``// words alphabetical order``                ``else` `if` `(o1.getKey().equals(o2.getKey()))``                    ``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++;``            ``}` `            ``// Update answer by adding current cycle.``            ``if``(cycle_size > ``0``)``            ``{``                ``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`

## Python3

 `# Python3 program to find ``# minimum number of swaps``# required to sort an array` `# Function returns the minimum``# number of swaps required to``# sort the array``def` `minSwaps(arr):``    ``n ``=` `len``(arr)``    ` `    ``# Create two arrays and use``    ``# as pairs where first array``    ``# is element and second array``    ``# is position of first element``    ``arrpos ``=` `[``*``enumerate``(arr)]``    ` `    ``# Sort the array by array element``    ``# values to get right position of``    ``# every element as the elements``    ``# of second array.``    ``arrpos.sort(key ``=` `lambda` `it : it[``1``])``    ` `    ``# To keep track of visited elements.``    ``# Initialize all elements as not``    ``# visited or false.``    ``vis ``=` `{k : ``False` `for` `k ``in` `range``(n)}``    ` `    ``# Initialize result``    ``ans ``=` `0``    ``for` `i ``in` `range``(n):``        ` `        ``# already swapped or``        ``# already present at``        ``# correct position``        ``if` `vis[i] ``or` `arrpos[i][``0``] ``=``=` `i:``            ``continue``            ` `        ``# find number of nodes``        ``# in this cycle and``        ``# add it to ans``        ``cycle_size ``=` `0``        ``j ``=` `i``        ` `        ``while` `not` `vis[j]:``            ` `            ``# mark node as visited``            ``vis[j] ``=` `True``            ` `            ``# move to next node``            ``j ``=` `arrpos[j][``0``]``            ``cycle_size ``+``=` `1``            ` `        ``# update answer by adding``        ``# current cycle``        ``if` `cycle_size > ``0``:``            ``ans ``+``=` `(cycle_size ``-` `1``)``            ` `    ``# return answer``    ``return` `ans` `# Driver Code    ``arr ``=` `[``1``, ``5``, ``4``, ``3``, ``2``]``print``(minSwaps(arr))` `# This code is contributed``# by Dharan Aditya`

## C#

 `// C# program to find ``// minimum number of swaps``// required to sort an array``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `public` `class` `GfG``{``  ` `  ``// Function returns the``  ``// minimum number of swaps``  ``// required to sort the array``  ``public`  `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``    ``List > arrpos =``      ``new` `List > ();``    ``for` `(``int` `i = 0; i < n; i++)``      ``arrpos.Add(``new` `KeyValuePair <``int``,``                 ``int``> (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((a,b)=> a.Key-b.Key);` `    ``// To keep track of visited elements. Initialize``    ``// all elements as not visited or false.``    ``Boolean[] vis = ``new` `Boolean[n];`  `    ``// 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].Value == 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[j].Value;``        ``cycle_size++;``      ``}` `      ``// Update answer by adding current cycle.``      ``if``(cycle_size > 0)``      ``{``        ``ans += (cycle_size - 1);``      ``}``    ``}` `    ``// Return result``    ``return` `ans;``  ``}``}` `// Driver class``public` `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();``    ``Console.WriteLine(g.minSwaps(a));``  ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`2`

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

Approach: As Pair class available in java from java 8 so we can use hashmap in older java version.

Below is the implementation of the idea.

## C++

 `#include ``using` `namespace` `std;``// Function returns the``// minimum number of swaps``// required to sort the array``int` `minSwaps(``int` `nums[], ``int` `n)``{``    ``int` `len = n;``    ``map<``int``, ``int``> map;``    ``for` `(``int` `i = 0; i < len; i++)``        ``map[nums[i]] = i;` `    ``sort(nums, nums + n);` `    ``// To keep track of visited elements. Initialize``    ``// all elements as not visited or false.``    ``bool` `visited[len] = { 0 };` `    ``// Initialize result``    ``int` `ans = 0;``    ``for` `(``int` `i = 0; i < len; i++) {` `        ``// already swapped and corrected or``        ``// already present at correct pos``        ``if` `(visited[i] || map[nums[i]] == i)``            ``continue``;` `        ``int` `j = i, cycle_size = 0;``        ``while` `(!visited[j]) {``            ``visited[j] = ``true``;` `            ``// move to next node``            ``j = map[nums[j]];``            ``cycle_size++;``        ``}` `        ``// Update answer by adding current cycle.``        ``if` `(cycle_size > 0) {``            ``ans += (cycle_size - 1);``        ``}``    ``}``    ``return` `ans;``}``int` `main()``{``    ``// Driver program to test the above function``    ``int` `a[] = { 1, 5, 4, 3, 2 };``    ``int` `n = 5;``    ``cout << minSwaps(a, n);``    ``return` `0;``}` `// This code is contributed by Harshal Khond`

## Java

 `import` `java.util.*;``import` `java.io.*;`` ` `class` `GfG``{``    ``// Function returns the``    ``// minimum number of swaps``    ``// required to sort the array``    ``public` `static` `int` `minSwaps(``int``[] nums)``    ``{``        ``int` `len = nums.length;``        ``HashMap map = ``new` `HashMap<>();``        ``for``(``int` `i=``0``;i ``0``) {``                ``ans += (cycle_size - ``1``);``            ``}``        ``}``        ``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 Saurabh Johari`

## Python3

 `# Function returns the``# minimum number of swaps``# required to sort the array``from` `functools ``import` `cmp_to_key` `def` `cmp``(a, b):``    ``return` `a ``-` `b` `def` `minSwaps(nums):``    ``Len` `=` `len``(nums)``    ``map` `=` `{}``    ``for` `i ``in` `range``(``Len``):``        ``map``[nums[i]] ``=` `i` `    ``nums ``=` `sorted``(nums, key ``=` `cmp_to_key(``cmp``))` `    ``# To keep track of visited elements. Initialize``    ``# all elements as not visited or false.``    ``visited ``=` `[``False` `for` `col ``in` `range``(``Len``)]``    ` `    ``# Initialize result``    ``ans ``=` `0``    ``for` `i ``in` `range``(``Len``):` `        ``# already swapped and corrected or``        ``# already present at correct pos``        ``if` `(visited[i] ``or` `map``[nums[i]] ``=``=` `i):``            ``continue` `        ``j,cycle_size ``=` `i, ``0``        ``while` `(visited[j] ``=``=` `False``):``            ``visited[j] ``=` `True` `            ``# move to next node``            ``j ``=` `map``[nums[j]]``            ``cycle_size ``+``=` `1` `        ``# Update answer by adding current cycle.``        ``if` `(cycle_size > ``0``):``            ``ans ``+``=` `(cycle_size ``-` `1``)` `    ``return` `ans` `# Driver program to test the above function``a ``=` `[ ``1``, ``5``, ``4``, ``3``, ``2` `]``print``(minSwaps(a))``    ` `# This code is contributed by shinjanpatra`

## C#

 `using` `System;``using` `System.Collections.Generic;` ` ``class` `GfG`` ``{``   ` `    ``// Function returns the``    ``// minimum number of swaps``    ``// required to sort the array``    ``public`  `int` `minSwaps(``int``[] nums) {``        ``int` `len = nums.Length;``        ``Dictionary<``int``, ``int``> map = ``new` `Dictionary<``int``,``int``>();``        ``for` `(``int` `i = 0; i < len; i++)``            ``map.Add(nums[i], i);` `        ``Array.Sort(nums);` `        ``// To keep track of visited elements. Initialize``        ``// all elements as not visited or false.``        ``bool``[] visited = ``new` `bool``[len];``        `  `        ``// Initialize result``        ``int` `ans = 0;``        ``for` `(``int` `i = 0; i < len; i++) {` `            ``// already swapped and corrected or``            ``// already present at correct pos``            ``if` `(visited[i] || map[nums[i]] == i)``                ``continue``;` `            ``int` `j = i, cycle_size = 0;``            ``while` `(!visited[j]) {``                ``visited[j] = ``true``;` `                ``// move to next node``                ``j = map[nums[j]];``                ``cycle_size++;``            ``}` `            ``// Update answer by adding current cycle.``            ``if` `(cycle_size > 0) {``                ``ans += (cycle_size - 1);``            ``}``        ``}``        ``return` `ans;``    ``}``}` `// Driver class``public` `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();``        ``Console.WriteLine(g.minSwaps(a));``    ``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output

`2`

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

Straight forward solution(Greedy Solution):

While iterating over the array, check the current element, and if not in the correct place, replace that element with the index of the element which should have come in this place greedily which will give the optimal answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to find minimum number``// of swaps required to sort an array``#include ``using` `namespace` `std;` `void` `swap(vector<``int``> &arr, ``int` `i, ``int` `j)``{``    ``int` `temp = arr[i];``    ``arr[i] = arr[j];``    ``arr[j] = temp;``}` `int` `indexOf(vector<``int``> &arr, ``int` `ele)``{``    ``for``(``int` `i = 0; i < arr.size(); i++)``    ``{``        ``if` `(arr[i] == ele)``        ``{``            ``return` `i;``        ``}``    ``}``    ``return` `-1;``}` `// Return the minimum number``// of swaps required to sort the array``int` `minSwaps(vector<``int``> arr, ``int` `N)``{``    ``int` `ans = 0;``    ``vector<``int``> temp(arr.begin(),arr.end());``    ``sort(temp.begin(),temp.end());``    ` `    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// This is checking whether``        ``// the current element is``        ``// at the right place or not``        ``if` `(arr[i] != temp[i])``        ``{``            ``ans++;` `            ``// Swap the current element``            ``// with the right index``            ``// so that arr to arr[i] is sorted``            ``swap(arr, i, indexOf(arr, temp[i]));``        ``}``    ``}``    ``return` `ans;``}` `// Driver Code``int` `main()``{` `    ``vector<``int``> a = {101, 758, 315, 730,``                   ``472, 619, 460, 479};``    ` `    ``int` `n = a.size();``    ` `    ``// Output will be 5``    ``cout << minSwaps(a, n);``}` `// This code is contributed by mohit kumar 29`

## Java

 `// Java program to find``// minimum number of swaps``// required to sort an array``import` `java.util.*;``import` `java.io.*;` `class` `GfG``{` `    ``// Return the minimum number``    ``// of swaps required to sort the array``    ``public` `int` `minSwaps(``int``[] arr, ``int` `N)``    ``{``        ``int` `ans = ``0``;``        ``int``[] temp = Arrays.copyOfRange(arr, ``0``, N);``        ``Arrays.sort(temp);``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{` `            ``// This is checking whether``            ``// the current element is``            ``// at the right place or not``            ``if` `(arr[i] != temp[i])``            ``{``                ``ans++;` `                ``// Swap the current element``                ``// with the right index``                ``// so that arr to arr[i] is sorted``                ``swap(arr, i, indexOf(arr, temp[i]));``            ``}``        ``}``        ``return` `ans;``    ``}``    ``public` `void` `swap(``int``[] arr, ``int` `i, ``int` `j)``    ``{``        ``int` `temp = arr[i];``        ``arr[i] = arr[j];``        ``arr[j] = temp;``    ``}``    ``public` `int` `indexOf(``int``[] arr, ``int` `ele)``    ``{``        ``for` `(``int` `i = ``0``; i < arr.length; i++)``        ``{``            ``if` `(arr[i] == ele) {``                ``return` `i;``            ``}``        ``}``        ``return` `-``1``;``    ``}``}``// Driver class``class` `Main``{``    ` `    ``// Driver program to test``    ``// the above function``    ``public` `static` `void` `main(String[] args)``                             ``throws` `Exception``    ``{``        ``int``[] a``            ``= { ``101``, ``758``, ``315``, ``730``, ``472``,``                         ``619``, ``460``, ``479` `};``        ``int` `n = a.length;``        ``// Output will be 5``        ``System.out.println(``new` `GfG().minSwaps(a, n));``    ``}``}``// This code is contributed by Satvik Nema`

## Python3

 `# Python3 program to find``#minimum number of swaps``# required to sort an array` `# Return the minimum number``# of swaps required to sort``# the array``def` `minSwaps(arr, N):``    ` `    ``ans ``=` `0``    ``temp ``=` `arr.copy()``    ``temp.sort()``    ``for` `i ``in` `range``(N):``      ` `        ``# This is checking whether``        ``# the current element is``        ``# at the right place or not``        ``if` `(arr[i] !``=` `temp[i]):``            ``ans ``+``=` `1` `            ``# Swap the current element``            ``# with the right index``            ``# so that arr to arr[i]``            ``# is sorted``            ``swap(arr, i,``                 ``indexOf(arr, temp[i]))``  ` `    ``return` `ans``  ` `def` `swap(arr, i, j):``    ` `    ``temp ``=` `arr[i]``    ``arr[i] ``=` `arr[j]``    ``arr[j] ``=` `temp``    ` `def` `indexOf(arr, ele):``   ` `    ``for` `i ``in` `range``(``len``(arr)):       ``        ``if` `(arr[i] ``=``=` `ele):``                ``return` `i``    ``return` `-``1` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``      ``a ``=` `[``101``, ``758``, ``315``, ``730``,``           ``472``, ``619``, ``460``, ``479``]``      ``n ``=` `len``(a)``      ` `      ``# Output will be 5``      ``print``(minSwaps(a, n))` `# This code is contributed by Chitranayal`

## C#

 `// C# program to find``// minimum number of swaps``// required to sort an array``using` `System;``public` `class` `GFG``{``    ` `    ``// Return the minimum number``    ``// of swaps required to sort the array``    ``static` `int` `minSwaps(``int``[] arr, ``int` `N)``    ``{``        ``int` `ans = 0;``        ``int``[] temp = ``new` `int``[N];``        ``Array.Copy(arr, temp, N);``        ``Array.Sort(temp);``        ``for` `(``int` `i = 0; i < N; i++)``        ``{`` ` `            ``// This is checking whether``            ``// the current element is``            ``// at the right place or not``            ``if` `(arr[i] != temp[i])``            ``{``                ``ans++;`` ` `                ``// Swap the current element``                ``// with the right index``                ``// so that arr to arr[i] is sorted``                ``swap(arr, i, indexOf(arr, temp[i]));``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``static` `void` `swap(``int``[] arr, ``int` `i, ``int` `j)``    ``{``        ``int` `temp = arr[i];``        ``arr[i] = arr[j];``        ``arr[j] = temp;``    ``}``    ``static` `int` `indexOf(``int``[] arr, ``int` `ele)``    ``{``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``if` `(arr[i] == ele) {``                ``return` `i;``            ``}``        ``}``        ``return` `-1;``    ``}``    ` `    ``// Driver program to test``    ``// the above function``    ``static` `public` `void` `Main (){``        ``int``[] a``            ``= { 101, 758, 315, 730, 472,``                         ``619, 460, 479 };``        ``int` `n = a.Length;``        ``// Output will be 5``        ``Console.WriteLine(minSwaps(a, n));``    ``}``}` `// This code is contributed by rag2127`

## Javascript

 ``

Output

`5`

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

We can still improve the complexity by using a hashmap. The main operation here is the indexOf method inside the loop, which costs us n*n. We can improve this section to O(n), by using a hashmap to store the indexes. Still, we use the sort method, so the complexity cannot improve beyond O(n Log n)

Method using HashMap:

Same as before, make a new array (called temp), which is the sorted form of the input array. We know that we need to transform the input array to the new array (temp) in the minimum number of swaps. Make a map that stores the elements and their corresponding index, of the input array.

So at each i starting from 0 to N in the given array, where N is the size of the array:

1. 1. If i is not in its correct position according to the sorted array, then
2. 2. We will fill this position with the correct element from the hashmap we built earlier. We know the correct element which should come here is temp[i], so we look up the index of this element from the hashmap.
3. 3. After swapping the required elements, we update the content of the hashmap accordingly, as temp[i] to the ith position, and arr[i] to where temp[i] was earlier.

Below is the implementation of the above approach:

## C++

 `// C++ program to find``// minimum number of swaps``// required to sort an array``#include``using` `namespace` `std;` `void` `swap(vector<``int``> &arr,``          ``int` `i, ``int` `j)``{``  ``int` `temp = arr[i];``  ``arr[i] = arr[j];``  ``arr[j] = temp;``}``// Return the minimum number``// of swaps required to sort``// the array``int` `minSwaps(vector<``int``>arr,``             ``int` `N)``{``  ``int` `ans = 0;``  ``vector<``int``>temp = arr;` `  ``// Hashmap which stores the``  ``// indexes of the input array``  ``map <``int``, ``int``> h;` `  ``sort(temp.begin(), temp.end());``  ``for` `(``int` `i = 0; i < N; i++)``  ``{``    ``h[arr[i]] = i;``  ``}``  ``for` `(``int` `i = 0; i < N; i++)``  ``{``    ``// This is checking whether``    ``// the current element is``    ``// at the right place or not``    ``if` `(arr[i] != temp[i])``    ``{``      ``ans++;``      ``int` `init = arr[i];` `      ``// If not, swap this element``      ``// with the index of the``      ``// element which should come here``      ``swap(arr, i, h[temp[i]]);` `      ``// Update the indexes in``      ``// the hashmap accordingly``      ``h[init] = h[temp[i]];``      ``h[temp[i]] = i;``    ``}``  ``}``  ``return` `ans;``}` `// Driver class``int` `main()``{``  ``// Driver program to``  ``// test the above function``  ``vector <``int``> a = {101, 758, 315,``                    ``730, 472, 619,``                    ``460, 479};``  ``int` `n = a.size();``  ` `  ``// Output will be 5``  ``cout << minSwaps(a, n);``}` `// This code is contributed by Stream_Cipher`

## Java

 `// Java program to find``// minimum number of swaps``// required to sort an array``import` `java.util.*;``import` `java.io.*;` `class` `GfG``{` `    ``// Return the minimum number``    ``// of swaps required to sort the array``    ``public` `int` `minSwaps(``int``[] arr, ``int` `N)``    ``{` `        ``int` `ans = ``0``;``        ``int``[] temp = Arrays.copyOfRange(arr, ``0``, N);` `        ``// Hashmap which stores the``        ``// indexes of the input array``        ``HashMap h``            ``= ``new` `HashMap();` `        ``Arrays.sort(temp);``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{``            ``h.put(arr[i], i);``        ``}``        ``for` `(``int` `i = ``0``; i < N; i++)``        ``{` `            ``// This is checking whether``            ``// the current element is``            ``// at the right place or not``            ``if` `(arr[i] != temp[i])``            ``{``                ``ans++;``                ``int` `init = arr[i];` `                ``// If not, swap this element``                ``// with the index of the``                ``// element which should come here``                ``swap(arr, i, h.get(temp[i]));` `                ``// Update the indexes in``                ``// the hashmap accordingly``                ``h.put(init, h.get(temp[i]));``                ``h.put(temp[i], i);``            ``}``        ``}``        ``return` `ans;``    ``}``    ``public` `void` `swap(``int``[] arr, ``int` `i, ``int` `j)``    ``{``        ``int` `temp = arr[i];``        ``arr[i] = arr[j];``        ``arr[j] = temp;``    ``}``}` `// Driver class``class` `Main``{` `    ``// Driver program to test the above function``    ``public` `static` `void` `main(String[] args)``                           ``throws` `Exception``    ``{``        ``int``[] a``            ``= { ``101``, ``758``, ``315``, ``730``, ``472``,``                        ``619``, ``460``, ``479` `};``        ``int` `n = a.length;``        ``// Output will be 5``        ``System.out.println(``new` `GfG().minSwaps(a, n));``    ``}``}``// This code is contributed by Satvik Nema`

## Python3

 `# Python3 program to find``# minimum number of swaps``# required to sort an array` `# Return the minimum number``# of swaps required to sort``# the array``def` `minSwap(arr, n):``    ` `    ``ans ``=` `0``    ``temp ``=` `arr.copy()` `    ``# Dictionary which stores the``      ``# indexes of the input array``    ``h ``=` `{}` `    ``temp.sort()` `    ``for` `i ``in` `range``(n):``        ` `        ``#h.[arr[i]``        ``h[arr[i]] ``=` `i``        ` `    ``init ``=` `0``    ` `    ``for` `i ``in` `range``(n):` `        ``# This is checking whether``        ``# the current element is``        ``# at the right place or not``        ``if` `(arr[i] !``=` `temp[i]):``            ``ans ``+``=` `1``            ``init ``=` `arr[i]` `            ``# If not, swap this element``              ``# with the index of the``              ``# element which should come here``            ``arr[i], arr[h[temp[i]]] ``=` `arr[h[temp[i]]], arr[i]` `            ``# Update the indexes in``              ``# the hashmap accordingly``            ``h[init] ``=` `h[temp[i]]``            ``h[temp[i]] ``=` `i``            ` `    ``return` `ans` `# Driver code``a ``=` `[ ``101``, ``758``, ``315``, ``730``,``      ``472``, ``619``, ``460``, ``479` `]``n ``=` `len``(a)` `# Output will be 5``print``(minSwap(a, n))` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program to find``// minimum number of swaps``// required to sort an array``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;``public` `class` `GfG {` `  ``// Return the minimum number``  ``// of swaps required to sort the array``  ``public` `int` `minSwaps(``int``[] arr, ``int` `N) {` `    ``int` `ans = 0;``    ``int``[] temp = ``new` `int``[N];``    ``arr.CopyTo(temp,0);` `    ``// Hashmap which stores the``    ``// indexes of the input array``    ``Dictionary<``int``, ``int``> h = ``new` `Dictionary<``int``, ``int``>();` `    ``Array.Sort(temp);``    ``for` `(``int` `i = 0; i < N; i++) {``      ``h.Add(arr[i], i);``    ``}``    ``for` `(``int` `i = 0; i < N; i++) {` `      ``// This is checking whether``      ``// the current element is``      ``// at the right place or not``      ``if` `(arr[i] != temp[i]) {``        ``ans++;``        ``int` `init = arr[i];` `        ``// If not, swap this element``        ``// with the index of the``        ``// element which should come here``        ``swap(arr, i, h[temp[i]]);` `        ``// Update the indexes in``        ``// the hashmap accordingly``        ``h[init]= h[temp[i]];``        ``h[temp[i]]= i;``      ``}``    ``}``    ``return` `ans;``  ``}` `  ``public` `void` `swap(``int``[] arr, ``int` `i, ``int` `j) {``    ``int` `temp = arr[i];``    ``arr[i] = arr[j];``    ``arr[j] = temp;``  ``}``}` `// Driver class``public` `class` `GFG {` `  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int``[] a = { 101, 758, 315, 730, 472, 619, 460, 479 };``    ``int` `n = a.Length;``    ` `    ``// Output will be 5``    ``Console.WriteLine(``new` `GfG().minSwaps(a, n));``  ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`5`

Time Complexity: O(n Log n)
Auxiliary Space: O(n)

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

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