Minimum swaps to make two arrays identical

Difficulty Level : Hard
Last Updated : 22 Oct, 2020

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 <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
// 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:

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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