Make max elements in B[] equal to that of A[] by adding/subtracting integers in range [0, K]

Given two arrays A[] and B[] and an integer K, the task is to maximize the count of integers of array B[] that can be made equal with array A[] by adding or subtracting any integer in the range [0, K].

Examples:

Input: K=5, A[] = [100, 65, 35, 85, 55], B[] = [30, 60, 75, 95]
Output: 3
Explanation:
30 + 5, 60 + 5, 95 + 5 gives the values which are equal with few elements of array A[].

Input: K = 5, A[] = [10, 20, 30, 40, 50], B[] = [1, 20, 3]
Output: 1
Explanation:
Only the 2nd value can be made equal, Since its value [20] can be changed to [20] by  adding/subtracting 0 from it.

Approach: The idea is to check whether the absolute difference between elements of the array B[] with any element of the array A[] is less than or equals to K. If yes then include this in the count. Print the count all such elements after checking the above condition for all the elements in the array B[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
void countElement(int A[], int N,
                  int B[], int M, int K)
{

    // To store the count of element
    int cnt = 0;

    // Traverse the array B[]
    for (int i = 0; i < M; i++) {

        int currentElement = B[i];

        // Traverse the array A[]
        for (int j = 0; j < N; j++) {

            // Find the difference
            int diff
                = abs(currentElement - A[j]);

            // If difference is atmost
            // K then increment the cnt
            if (diff <= K) {
                cnt++;
                break;
            }
        }
    }

    // Print the count
    cout << cnt;
}

// Driver Code
int main()
{
    // Given array A[] and B[]
    int A[] = { 100, 65, 35, 85, 55 };
    int B[] = { 30, 60, 75, 95 };

    // Given K
    int K = 5;

    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(B) / sizeof(B[0]);

    // Function Call
    countElement(A, N, B, M, K);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
static void countElement(int A[], int N,
                         int B[], int M, int K)
{

    // To store the count of element
    int cnt = 0;

    // Traverse the array B[]
    for(int i = 0; i < M; i++)
    {
        int currentElement = B[i];

        // Traverse the array A[]
        for(int j = 0; j < N; j++)
        {

            // Find the difference
            int diff = Math.abs(
                       currentElement - A[j]);

            // If difference is atmost
            // K then increment the cnt
            if (diff <= K)
            {
                cnt++;
                break;
            }
        }
    }

    // Print the count
    System.out.print(cnt);
}

// Driver Code
public static void main(String[] args)
{

    // Given array A[] and B[]
    int A[] = { 100, 65, 35, 85, 55 };
    int B[] = { 30, 60, 75, 95 };

    // Given K
    int K = 5;

    int N = A.length;
    int M = B.length;

    // Function call
    countElement(A, N, B, M, K);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above approach

# Function that count the number of
# integers from array B such that
# subtracting element in the range
# [0, K] given any element in A
def countElement(A, N, B, M, K):

    # To store the count of element
    cnt = 0

    # Traverse the array B
    for i in range(M):
        currentElement = B[i]

        # Traverse the array A
        for j in range(N):

            # Find the difference
            diff = abs(currentElement - A[j])

            # If difference is atmost
            # K then increment the cnt
            if(diff <= K):
                cnt += 1
                break

    # Print the count
    print(cnt)

# Driver Code
if __name__ == '__main__':

    # Given array A and B
    A = [ 100, 65, 35, 85, 55 ]
    B = [ 30, 60, 75, 95 ]

    N = len(A)
    M = len(B)

    # Given K
    K = 5

    # Function call
    countElement(A, N, B, M, K)

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{

// Function that count the number of
// integers from array []B such that
// subtracting element in the range
// [0, K] given any element in []A
static void countElement(int []A, int N,
                         int []B, int M, int K)
{

    // To store the count of element
    int cnt = 0;

    // Traverse the array []B
    for(int i = 0; i < M; i++)
    {
        int currentElement = B[i];

        // Traverse the array []A
        for(int j = 0; j < N; j++)
        {

            // Find the difference
            int diff = Math.Abs(
                       currentElement - A[j]);

            // If difference is atmost
            // K then increment the cnt
            if (diff <= K)
            {
                cnt++;
                break;
            }
        }
    }

    // Print the count
    Console.Write(cnt);
}

// Driver Code
public static void Main(String[] args)
{

    // Given array []A and []B
    int []A = { 100, 65, 35, 85, 55 };
    int []B = { 30, 60, 75, 95 };

    // Given K
    int K = 5;

    int N = A.Length;
    int M = B.Length;

    // Function call
    countElement(A, N, B, M, K);
}
}

// This code is contributed by Rohit_ranjan

Javascript

<script>
// Javascript Script program to implement
// the above approach

// Function that count the number of
// integers from array B[] such that
// subtracting element in the range
// [0, K] given any element in A[]
function countElement(A, N, B, M, K)
{

    // To store the count of element
    let cnt = 0;

    // Traverse the array B[]
    for(let i = 0; i < M; i++)
    {
        let currentElement = B[i];

        // Traverse the array A[]
        for(let j = 0; j < N; j++)
        {

            // Find the difference
            let diff = Math.abs(
                       currentElement - A[j]);

            // If difference is atmost
            // K then increment the cnt
            if (diff <= K)
            {
                cnt++;
                break;
            }
        }
    }

    // Print the count
    document.write(cnt);
}


// Driver Code

    // Given array A[] and B[]
    let A = [ 100, 65, 35, 85, 55 ];
    let B = [ 30, 60, 75, 95 ];

    // Given K
    let K = 5;

    let N = A.length;
    let M = B.length;

    // Function call
    countElement(A, N, B, M, K);

</script>

Output:

3

Time Complexity: O(N*M), where N and M are the lengths of the arrays A[] and B[].
Auxiliary Space: O(1)