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:
// C++ program of the above approach #include <bits/stdc++.h> 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 << "
"; } // 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[0]);
// Function Call
maximumEqual(A, B, size);
return 0;
} |
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!