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 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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