# Minimum swaps to make two arrays identical

Given two arrays that have the same values but in a different order, we need to make a second array the same as a first array using the minimum number of swaps.
Examples:

```Input  : arrA[] = {3, 6, 4, 8},
arrB[] = {4, 6, 8, 3}
Output : 2
we can make arrB to same as arrA in 2 swaps
which are shown below,
swap 4 with 8,   arrB = {8, 6, 4, 3}
swap 8 with 3,   arrB = {3, 6, 4, 8}```

This problem can be solved by modifying the array B. We save the index of array A elements in array B i.e. if ith element of array A is at jth position in array B, then we will make arrB[i] = j
For above given example, modified array B will be, arrB = {3, 1, 0, 2}. This modified array represents the distribution of array A element in array B and our goal is to sort this modified array in a minimum number of swaps because after sorting only array B element will be aligned with array A elements.
Now count of minimum swaps for sorting an array can be found by visualizing the problem as a graph, this problem is already explained in previous article
So we count these swaps in a modified array and that will be our final answer.
Please see the below code for a better understanding.

## C++

 `// C++ program to make an array same to another` `// using minimum number of swap` `#include ` `using` `namespace` `std;`   `// Function returns the minimum number of swaps` `// required to sort the array` `// This method is taken from below post` `// https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/` `int` `minSwapsToSort(``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.` `        ``ans += (cycle_size - 1);` `    ``}`   `    ``// Return result` `    ``return` `ans;` `}`   `// method returns minimum number of swap to make` `// array B same as array A` `int` `minSwapToMakeArraySame(``int` `a[], ``int` `b[], ``int` `n)` `{` `    ``// map to store position of elements in array B` `    ``// we basically store element to index mapping.` `    ``map<``int``, ``int``> mp;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``mp[b[i]] = i;`   `    ``// now we're storing position of array A elements` `    ``// in array B.` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``b[i] = mp[a[i]];`   `    ``/* We can uncomment this section to print modified` `      ``b array` `    ``for (int i = 0; i < N; i++)` `        ``cout << b[i] << " ";` `    ``cout << endl; */`   `    ``// returing minimum swap for sorting in modified` `    ``// array B as final answer` `    ``return` `minSwapsToSort(b, n);` `}`   `//    Driver code to test above methods` `int` `main()` `{` `    ``int` `a[] = {3, 6, 4, 8};` `    ``int` `b[] = {4, 6, 8, 3};`   `    ``int` `n = ``sizeof``(a) / ``sizeof``(``int``);` `    ``cout << minSwapToMakeArraySame(a, b, n);` `    ``return` `0;` `}`

## Python3

 `# Python3 program to make ` `# an array same to another` `# using minimum number of swap`   `# Function returns the minimum ` `# number of swaps required to ` `# sort the array` `# This method is taken from below post` `# https: // www.geeksforgeeks.org/` `# minimum-number-swaps-required-sort-array/` `def` `minSwapsToSort(arr, n):`   `    ``# Create an array of pairs ` `    ``# where first element is ` `    ``# array element and second` `    ``# element is position of ` `    ``# first element` `    ``arrPos ``=` `[[``0` `for` `x ``in` `range``(``2``)] ` `                 ``for` `y ``in` `range``(n)]` `    `  `    ``for` `i ``in` `range``(n):    ` `        ``arrPos[i][``0``] ``=` `arr[i]` `        ``arrPos[i][``1``] ``=` `i`   `    ``# Sort the array by array ` `    ``# element values to get right` `    ``# position of every element ` `    ``# as second element of pair.` `    ``arrPos.sort()`   `    ``# To keep track of visited ` `    ``# elements. Initialize all ` `    ``# elements as not visited ` `    ``# or false.` `    ``vis ``=` `[``False``] ``*` `(n)`   `    ``# Initialize result` `    ``ans ``=` `0`   `    ``# Traverse array elements` `    ``for` `i ``in` `range``(n):` `    `  `        ``# Already swapped and corrected or` `        ``# already present at correct pos` `        ``if` `(vis[i] ``or` `arrPos[i][``1``] ``=``=` `i):` `            ``continue`   `        ``# Find out the number of  node in` `        ``# this cycle and add in ans` `        ``cycle_size ``=` `0` `        ``j ``=` `i` `        `  `        ``while` `(``not` `vis[j]):        ` `            ``vis[j] ``=` `1`   `            ``# Move to next node` `            ``j ``=` `arrPos[j][``1``]` `            ``cycle_size``+``=` `1` `       `  `        ``# Update answer by ` `        ``# adding current cycle.` `        ``ans ``+``=` `(cycle_size ``-` `1``)  `   `    ``# Return result` `    ``return` `ans`   `# Method returns minimum ` `# number of swap to mak` `# array B same as array A` `def` `minSwapToMakeArraySame(a, b, n):` `        `  `    ``# map to store position ` `    ``# of elements in array B` `    ``# we basically store ` `    ``# element to index mapping.` `    ``mp ``=` `{}` `    ``for` `i ``in` `range``(n):` `        ``mp[b[i]] ``=` `i`   `    ``# now we're storing position ` `    ``# of array A elements` `    ``# in array B.` `    ``for` `i ``in` `range``(n):` `        ``b[i] ``=` `mp[a[i]]`   `    ``# Returing minimum swap ` `    ``# for sorting in modified` `    ``# array B as final answer` `    ``return` `minSwapsToSort(b, n)`   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:`   `    ``a ``=` `[``3``, ``6``, ``4``, ``8``]` `    ``b ``=` `[``4``, ``6``, ``8``, ``3``]` `    ``n ``=` `len``(a)` `    ``print``(minSwapToMakeArraySame(a, b, n))`   `# This code is contributed by Chitranayal`

Output:

```2

```

This article is contributed by Utkarsh Trivedi. 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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Improved By : chitranayal

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