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

• Difficulty Level : Medium
• Last Updated : 10 May, 2021

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:

## C++

 `// C++ program of the above approach``#include ``using` `namespace` `std;` `// Function that prints maximum``// equal elements``void` `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 Code``int` `main()``{``    ``// Given two arrays``    ``int` `A[] = { 6, 7, 3, 9, 5 };``    ``int` `B[] = { 7, 3, 9, 5, 6 };` `    ``int` `size = ``sizeof``(A) / ``sizeof``(A);` `    ``// Function Call``    ``maximumEqual(A, B, size);``    ``return` `0;``}`

## 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 + ``"\n"``);``}` `// 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`

## Python3

 `# Python3 program for the above approach` `# Function that prints maximum``# equal elements``def` `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 Code``if` `__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 approach``using` `System;``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``    ``Console.Write(finalans + ``"\n"``);``}` `// Driver Code``public` `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

 ``
Output:
`5`

Time Complexity: O(N)
Auxiliary Space: O(N)

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