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

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)

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