Skip to content
Related Articles
Get the best out of our app
GFG App
Open App
geeksforgeeks
Browser
Continue

Related Articles

Java Program to Maximize count of corresponding same elements in given Arrays by Rotation

Improve Article
Save Article
Like Article
Improve Article
Save Article
Like Article

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:
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:
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: 
 

  1. Store the position of all the elements of the array arr2[] in an array(say store[]).
  2. 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.
  3. 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 approach
import java.util.*;
class GFG{
  
// Function that prints maximum
// equal elements
static 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 Code
public 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


Output: 

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!


My Personal Notes arrow_drop_up
Last Updated : 25 Jan, 2022
Like Article
Save Article
Similar Reads
Related Tutorials