Java Program to Maximize count of corresponding same elements in given Arrays by Rotation
Last Updated :
25 Jan, 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[].Â
Examples:Â
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Input: arr1[] = { 6, 7, 3, 9, 5 }, arr2[] = { 7, 3, 9, 5, 6 }Â
Output: 5Â
Explanation:Â
By performing cyclic left shift on array arr1[] by 1.Â
Updated array arr1[] = {7, 3, 9, 5, 6}.Â
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}Â
Output: 2Â
Explanation:Â
By performing cyclic left shift on array arr1[] by 1.Â
Updated array arr1[] = {3, 2, 4, 1}Â
This rotation contains a maximum number of equal elements between array arr1[] and arr2[].Â
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Approach: This problem can be solved using Greedy Approach. Below are the steps:Â
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- Store the position of all the elements of the array arr2[] in an array(say store[]).
- For each element in the array arr1[], do the following:Â
- 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.
- 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:Â
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Java
import java.util.*;
class GFG{
static void maximumEqual( int a[],
int b[], int n)
{
int store[] = new int [( int ) 1e5];
for ( int i = 0 ; i < n; i++)
{
store[b[i]] = i + 1 ;
}
int ans[] = new int [( int ) 1e5];
for ( int i = 0 ; i < n; i++)
{
int d = Math.abs(store[a[i]] - (i + 1 ));
if (store[a[i]] < i + 1 )
{
d = n - d;
}
ans[d]++;
}
int finalans = 0 ;
for ( int i = 0 ; i < 1e5; i++)
finalans = Math.max(finalans,
ans[i]);
System.out.print(finalans + "
");
}
public static void main(String[] args)
{
int A[] = { 6 , 7 , 3 , 9 , 5 };
int B[] = { 7 , 3 , 9 , 5 , 6 };
int size = A.length;
maximumEqual(A, B, size);
}
}
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Time Complexity: O(N)Â
Auxiliary Space: O(N)
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