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

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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!


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Last Updated : 25 Jan, 2022
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