Count elements of same value placed at same indices of two given arrays
Last Updated :
04 Apr, 2023
Given two arrays A[] and B[] of N unique elements, the task is to find the maximum number of matched elements from the two given arrays.
Elements of the two arrays are matched if they are of the same value and can be placed at the same index (0-based indexing).(By right shift or left shift of the two arrays).
Examples:
Input: A[] = { 5, 3, 7, 9, 8 }, B[] = { 8, 7, 3, 5, 9 }
Output: 3
Explanation: Left shifting B[] by 1 index modifies B[] to { 7, 3, 5, 9, 8 }.
Therefore, elements in indices 1, 3 and 4 match. Therefore, the required count is 3.
Input: A[] = { 9, 5, 6, 2 }, B[] = { 6, 2, 9, 5 }
Output: 4
Naive Approach: The simplest approach to solve this problem is to observe that one right shift is the same as the (N-1) left shift, so perform only one type of shift, say right shift. Also, performing the right shift on A is the same as performing left shift B, so perform the right shift on only one array, say on A[]. Apply the right shift operations on A while keeping B as it is and compare all the values of A and B to find the total number of matches and keep track of the maximum of all.
Time complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using a Map to keep track of the difference between indices of equal elements present in arrays A[] and B[]. If the difference comes out to be negative, then change A[] by doing k( = N + difference) left shifts, which is equivalent to N - K right shifts.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count maximum matched
// elements from the arrays A[] and B[]
int maxMatch(int A[], int B[], int M, int N)
{
// Stores position of elements of
// array A[] in the array B[]
map<int,int> Aindex;
// Keep track of difference
// between the indices
map<int,int> diff;
// Traverse the array A[]
for(int i = 0; i < M; i++)
{
Aindex[A[i]] = i ;
}
// Traverse the array B[]
for(int i = 0; i < N; i++)
{
// If difference is negative, add N to it
if (i - Aindex[B[i]] < 0)
{
diff[M + i - Aindex[B[i]]] += 1;
}
// Keep track of the number of shifts
// required to place elements at same indices
else
{
diff[i - Aindex[B[i]]] += 1;
}
}
// Return the max matches
int max = 0;
for(auto ele = diff.begin(); ele != diff.end(); ele++)
{
if(ele->second > max)
{
max = ele->second;
}
}
return max;
}
// Driver code
int main()
{
int A[] = { 5, 3, 7, 9, 8 };
int B[] = { 8, 7, 3, 5, 9 };
int M = sizeof(A) / sizeof(A[0]);
int N = sizeof(B) / sizeof(B[0]);
// Returns the count
// of matched elements
cout << maxMatch(A, B, M, N);
return 0;
}
// This code is contributed by divyeshrabadiya07
// Java program for the above approach
import java.io.Console;
import java.util.HashMap;
import java.util.Map;
class GFG
{
// Function to count maximum matched
// elements from the arrays A[] and B[]
static int maxMatch(int[] A, int[] B)
{
// Stores position of elements of
// array A[] in the array B[]
HashMap<Integer, Integer> Aindex = new HashMap<Integer, Integer>();
// Keep track of difference
// between the indices
HashMap<Integer, Integer> diff = new HashMap<Integer, Integer>();
// Traverse the array A[]
for (int i = 0; i < A.length; i++)
{
Aindex.put(A[i], i);
}
// Traverse the array B[]
for (int i = 0; i < B.length; i++)
{
// If difference is negative, add N to it
if (i - Aindex.get(B[i]) < 0)
{
if (!diff.containsKey(A.length + i - Aindex.get(B[i])))
{
diff.put(A.length + i - Aindex.get(B[i]), 1);
} else {
diff.put(A.length + i - Aindex.get(B[i]), diff.get(A.length + i - Aindex.get(B[i])) + 1);
}
}
// Keep track of the number of shifts
// required to place elements at same indices
else {
if (!diff.containsKey(i - Aindex.get(B[i]))) {
diff.put(i - Aindex.get(B[i]), 1);
}
else
{
diff.put(i - Aindex.get(B[i]),
diff.get(i - Aindex.get(B[i])) + 1);
}
}
}
// Return the max matches
int max = 0;
for (Map.Entry<Integer, Integer> ele : diff.entrySet())
{
if (ele.getValue() > max)
{
max = ele.getValue();
}
}
return max;
}
// Driver Code
public static void main(String[] args)
{
int[] A = { 5, 3, 7, 9, 8 };
int[] B = { 8, 7, 3, 5, 9 };
// Returns the count
// of matched elements
System.out.println(maxMatch(A, B));
}
}
// This code is contributed by 29AjayKumar
# Python program for the above approach
# Function to count maximum matched
# elements from the arrays A[] and B[]
def maxMatch(A, B):
# Stores position of elements of
# array A[] in the array B[]
Aindex = {}
# Keep track of difference
# between the indices
diff = {}
# Traverse the array A[]
for i in range(len(A)):
Aindex[A[i]] = i
# Traverse the array B[]
for i in range(len(B)):
# If difference is negative, add N to it
if i-Aindex[B[i]] < 0:
if len(A)+i-Aindex[B[i]] not in diff:
diff[len(A)+i-Aindex[B[i]]] = 1
else:
diff[len(A)+i-Aindex[B[i]]] += 1
# Keep track of the number of shifts
# required to place elements at same indices
else:
if i-Aindex[B[i]] not in diff:
diff[i-Aindex[B[i]]] = 1
else:
diff[i-Aindex[B[i]]] += 1
# Return the max matches
return max(diff.values())
# Driver Code
A = [5, 3, 7, 9, 8]
B = [8, 7, 3, 5, 9]
# Returns the count
# of matched elements
print(maxMatch(A, B))
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to count maximum matched
// elements from the arrays A[] and B[]
static int maxMatch(int[] A, int[] B)
{
// Stores position of elements of
// array A[] in the array B[]
Dictionary<int,
int> Aindex = new Dictionary<int,
int>();
// Keep track of difference
// between the indices
Dictionary<int,
int> diff = new Dictionary<int,
int>();
// Traverse the array A[]
for(int i = 0; i < A.Length; i++)
{
Aindex[A[i]] = i ;
}
// Traverse the array B[]
for(int i = 0; i < B.Length; i++)
{
// If difference is negative, add N to it
if (i - Aindex[B[i]] < 0)
{
if (!diff.ContainsKey(A.Length + i -
Aindex[B[i]]))
{
diff[A.Length + i - Aindex[B[i]]] = 1;
}
else
{
diff[A.Length + i - Aindex[B[i]]] += 1;
}
}
// Keep track of the number of shifts
// required to place elements at same indices
else
{
if (!diff.ContainsKey(i - Aindex[B[i]]))
{
diff[i - Aindex[B[i]]] = 1;
}
else
{
diff[i - Aindex[B[i]]] += 1;
}
}
}
// Return the max matches
int max = 0;
foreach(KeyValuePair<int, int> ele in diff)
{
if (ele.Value > max)
{
max = ele.Value;
}
}
return max;
}
// Driver Code
static void Main()
{
int[] A = { 5, 3, 7, 9, 8 };
int[] B = { 8, 7, 3, 5, 9 };
// Returns the count
// of matched elements
Console.WriteLine(maxMatch(A, B));
}
}
// This code is contributed by divyesh072019
<script>
// JavaScript program for the above approach
// Function to count maximum matched
// elements from the arrays A[] and B[]
function maxMatch(A, B) {
// Stores position of elements of
// array A[] in the array B[]
var Aindex = {};
// Keep track of difference
// between the indices
var diff = {};
// Traverse the array A[]
for (var i = 0; i < A.length; i++) {
Aindex[A[i]] = i;
}
// Traverse the array B[]
for (var i = 0; i < B.length; i++) {
// If difference is negative, add N to it
if (i - Aindex[B[i]] < 0) {
if (!diff.hasOwnProperty(A.length + i - Aindex[B[i]])) {
diff[A.length + i - Aindex[B[i]]] = 1;
}
else {
diff[A.length + i - Aindex[B[i]]] += 1;
}
}
// Keep track of the number of shifts
// required to place elements at same indices
else {
if (!diff.hasOwnProperty(i - Aindex[B[i]])) {
diff[i - Aindex[B[i]]] = 1;
}
else {
diff[i - Aindex[B[i]]] += 1;
}
}
}
// Return the max matches
var max = 0;
for (const [key, value] of Object.entries(diff)) {
if (value > max) {
max = value;
}
}
return max;
}
// Driver Code
var A = [5, 3, 7, 9, 8];
var B = [8, 7, 3, 5, 9];
// Returns the count
// of matched elements
document.write(maxMatch(A, B));
</script>
Output:
3
Time Complexity: O(M+N), where M and N are the sizes of the input arrays A and B, respectively.
Auxiliary Space: O(M+N)
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