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}
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) 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
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