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:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;// Function that prints maximum// equal elementsvoid maximumEqual(int a[], int b[], int n){ // Vector to store the index // of elements of array b vector<int> store(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 vector<int> ans(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 = 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 = max(finalans, ans[i]); // Printing the maximum number // of equal elements cout << finalans << "\n";}// Driver Codeint main(){ // Given two arrays int A[] = { 6, 7, 3, 9, 5 }; int B[] = { 7, 3, 9, 5, 6 }; int size = sizeof(A) / sizeof(A[0]); // Function Call maximumEqual(A, B, size); return 0;} |
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 + "\n");}// 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 |
Python3
# Python3 program for the above approach# Function that prints maximum# equal elementsdef maximumEqual(a, b, n): # List to store the index # of elements of array b store = [0] * 10 ** 5 # Storing the positions of # array B for i in range(n): store[b[i]] = i + 1 # Frequency array to keep count # of elements with similar # difference in distances ans = [0] * 10 ** 5 # Iterate through all element # in arr1[] for i in range(n): # Calculate number of shift # required to make current # element equal d = 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] += 1 finalans = 0 # Compute the maximum frequency # stored for i in range(10 ** 5): finalans = max(finalans, ans[i]) # Printing the maximum number # of equal elements print(finalans)# Driver Codeif __name__ == '__main__': # Given two arrays A = [ 6, 7, 3, 9, 5 ] B = [ 7, 3, 9, 5, 6 ] size = len(A) # Function Call maximumEqual(A, B, size)# This code is contributed by Shivam Singh |
C#
// C# program of the above approachusing System;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 Console.Write(finalans + "\n");}// Driver Codepublic static void Main(){ // 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 chitranayal |
Javascript
<script>// JavaScript program of the above approach// Function that prints maximum// equal elementsfunction maximumEqual(a, b, n){ // Vector to store the index // of elements of array b let store = Array.from({length: 1e5}, (_, i) => 0); // Storing the positions of // array B for (let i = 0; i < n; i++) { store[b[i]] = i + 1; } // frequency array to keep count // of elements with similar // difference in distances let ans = Array.from({length: 1e5}, (_, i) => 0); // Iterate through all element in arr1[] for (let i = 0; i < n; i++) { // Calculate number of // shift required to // make current element // equal let 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]++; } let finalans = 0; // Compute the maximum frequency // stored for (let i = 0; i < 1e5; i++) finalans = Math.max(finalans, ans[i]); // Printing the maximum number // of equal elements document.write(finalans + "\n");}// Driver Code // Given two arrays let A = [ 6, 7, 3, 9, 5 ]; let B = [ 7, 3, 9, 5, 6 ]; let size = A.length; // Function Call maximumEqual(A, B, size); </script> |
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Info to that Topic: geeksforgeeks.org/maximize-count-of-corresponding-same-elements-in-given-arrays-by-rotation/ […]