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 in2swaps 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 <bits/stdc++.h>` `using` `namespace` `std;` `// Function returns the minimum number of swaps` `// required to sort the array` `// This method is taken from below post` `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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.