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.

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 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 <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)
        // 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;
        // 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;



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

This article is attributed to GeeksforGeeks.org

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