# Minimum swaps to make two arrays identical

Given two arrays which have same values but in different order, we need to make second array same as first array using 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}
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 distribution of array A element in array B and our goal is to sort this modified array in 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 modified array and that will be our final answer.
Please see below code for better understanding.

 `// 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://tutorialspoint.dev/slugresolver/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; ` `} `

Output:

```2
```

## tags:

Arrays Sorting Arrays Sorting