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

• Last Updated : 13 Sep, 2021

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  can be changed to  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 ``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);``    ``int` `M = ``sizeof``(B) / ``sizeof``(B);` `    ``// 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

 ``

Output:

`3`

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

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