Given two arrays arr1[] and arr2[] of N integers and array arr1[] has distinct elements. The task is to find the maximum count of corresponding same elements in the given arrays by performing cyclic left or right shift on array arr1[].
Examples:
Input: arr1[] = { 6, 7, 3, 9, 5 }, arr2[] = { 7, 3, 9, 5, 6 }
Output: 5
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {7, 3, 9, 5, 6}.
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
Input: arr1[] = {1, 3, 2, 4}, arr2[] = {4, 2, 3, 1}
Output: 2
Explanation:
By performing cyclic left shift on array arr1[] by 1.
Updated array arr1[] = {3, 2, 4, 1}
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].
Approach: This problem can be solved using Greedy Approach. Below are the steps:
- Store the position of all the elements of the array arr2[] in an array(say store[]).
- For each element in the array arr1[], do the following:
- Find the difference(say diff) between the position of the current element in arr2[] with the position in arr1[].
- If diff is less than 0 then update diff to (N – diff).
- Store the frequency of current difference diff in a map.
- After the above steps, the maximum frequency stored in map is the maximum number of equal elements after rotation on arr1[].
Below is the implementation of the above approach:
Java
// Java program of the above approachimport java.util.*;class GFG{ // Function that prints maximum// equal elementsstatic void maximumEqual(int a[], int b[], int n){ // Vector to store the index // of elements of array b int store[] = new int[(int) 1e5]; // Storing the positions of // array B for (int i = 0; i < n; i++) { store[b[i]] = i + 1; } // frequency array to keep count // of elements with similar // difference in distances int ans[] = new int[(int) 1e5]; // Iterate through all element in arr1[] for (int i = 0; i < n; i++) { // Calculate number of // shift required to // make current element // equal int d = Math.abs(store[a[i]] - (i + 1)); // If d is less than 0 if (store[a[i]] < i + 1) { d = n - d; } // Store the frequency // of current diff ans[d]++; } int finalans = 0; // Compute the maximum frequency // stored for (int i = 0; i < 1e5; i++) finalans = Math.max(finalans, ans[i]); // Printing the maximum number // of equal elements System.out.print(finalans + "");} // Driver Codepublic static void main(String[] args){ // Given two arrays int A[] = { 6, 7, 3, 9, 5 }; int B[] = { 7, 3, 9, 5, 6 }; int size = A.length; // Function Call maximumEqual(A, B, size);}} // This code is contributed by sapnasingh4991 |
5
Time Complexity: O(N)
Auxiliary Space: O(N)
Please refer complete article on Maximize count of corresponding same elements in given Arrays by Rotation for more details!
