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: 5
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: 2
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:
- 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:
- 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:
// 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 |
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!