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

  • Last Updated : 04 Feb, 2022

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: 
 

Javascript




<script>
 
// JavaScript program of the above approach
 
// Function that prints maximum
// equal elements
function 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 + "
");
}
 
// 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)
 

Please refer complete article on Maximize count of corresponding same elements in given Arrays by Rotation for more details!
 


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