# Count elements in first Array with absolute difference greater than K with an element in second Array

Given two arrays arr1[] and arr2[] and an integer K, our task is to find the number elements in the first array, for an element x, in arr1[], there exists at least one element y, in arr2[] such that absolute difference of x and y is greater than the integer K.

Examples:

Input: arr1 = {3, 1, 4}, arr2 = {5, 1, 2}, K = 2
Output:
Explanation:
Such elements are 1 and 4.
For 1, arr2[] has 5 and abs(1 – 5) = 4 which is greater than 2.
For 4, arr2[] has 1 and abs(4 – 1) = 3 which again is greater than 2.

Input: arr1 = {1, 2}, arr2 = {4, 6}, K = 3
Output:
Explanation:
Such elements are 1 and 2.
For 1, arr2[] has 6 and abs(1 – 6) = 5 which is greater than 3.
For 2, arr2[] has 6 and abs(2 – 6) = 4 which is greater than 3.

Naive Approach: Iterate for each element in arr1[] and check whether or not there exists an element in arr2 such that their absolute difference is greater than the value K.

Time complexity: O(N * M) where N and M are the sizes of the arrays 1 and 2 respectively.

Efficient Approach: To optimize the above method we have to observe that for each element in arr1[], we need only the smallest and largest element of arr2[] to check if it is distant or not. For each element x, in arr1, if the absolute difference of smallest or the largest value and x is greater than K then that element is distant.

Below is the implementation of the above approach:

## C++

 `// C++ program to count elements in first Array` `// with absolute difference greater than K` `// with an element in second Array`   `#include ` `using` `namespace` `std;`   `// Function to count the such elements` `void` `countDist(``int` `arr1[], ``int` `n, ``int` `arr2[],` `               ``int` `m, ``int` `k)` `{` `    ``// Store count of required elements in arr1` `    ``int` `count = 0;`   `    ``// Initialise the smallest and the largest` `    ``// value from the second array arr2[]` `    ``int` `smallest = arr2;` `    ``int` `largest = arr2;`   `    ``// Find the smallest and` `    ``// the largest element in arr2` `    ``for` `(``int` `i = 0; i < m; i++) {` `        ``smallest = max(smallest, arr2[i]);` `        ``largest = min(largest, arr1[i]);` `    ``}` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Check if absolute difference of smallest` `        ``// and arr1[i] or largest and arr1[i] is > K` `        ``// then arr[i] is a required element` `        ``if` `(``abs``(arr1[i] - smallest) > k` `            ``|| ``abs``(arr1[i] - largest) > k)` `            ``count++;` `    ``}`   `    ``// Print the final result` `    ``cout << count;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr1[] = { 3, 1, 4 };` `    ``int` `n = ``sizeof``(arr1) / ``sizeof``(arr1);` `    ``int` `arr2[] = { 5, 1, 2 };` `    ``int` `m = ``sizeof``(arr2) / ``sizeof``(arr2);` `    ``int` `k = 2;`   `    ``countDist(arr1, n, arr2, m, k);`   `    ``return` `0;` `}`

## Java

 `// Java program to count elements in first Array` `// with absolute difference greater than K` `// with an element in second Array` `class` `GFG{`   `// Function to count the such elements` `static` `void` `countDist(``int` `arr1[], ``int` `n, ` `                      ``int` `arr2[], ``int` `m,` `                      ``int` `k)` `{` `    ``// Store count of required elements in arr1` `    ``int` `count = ``0``;`   `    ``// Initialise the smallest and the largest` `    ``// value from the second array arr2[]` `    ``int` `smallest = arr2[``0``];` `    ``int` `largest = arr2[``0``];`   `    ``// Find the smallest and` `    ``// the largest element in arr2` `    ``for``(``int` `i = ``0``; i < m; i++) ` `    ``{` `       ``smallest = Math.max(smallest, arr2[i]);` `       ``largest = Math.min(largest, arr1[i]);` `    ``}` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{`   `       ``// Check if absolute difference ` `       ``// of smallest and arr1[i] or ` `       ``// largest and arr1[i] is > K ` `       ``// then arr[i] is a required element` `       ``if` `(Math.abs(arr1[i] - smallest) > k ||` `           ``Math.abs(arr1[i] - largest) > k)` `           ``count++;` `    ``}`   `    ``// Print the final result` `    ``System.out.print(count);` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr1[] = { ``3``, ``1``, ``4` `};` `    ``int` `n = arr1.length;` `    ``int` `arr2[] = { ``5``, ``1``, ``2` `};` `    ``int` `m = arr2.length;` `    ``int` `k = ``2``;`   `    ``countDist(arr1, n, arr2, m, k);` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to count elements in the first Array` `# with an absolute difference greater than K` `# with an element in the second Array`   `# Function to count the such elements` `def` `countDist(arr1, n, arr2, m, k):` `    `  `    ``# Store count of required elements in arr1` `    ``count ``=` `0`   `    ``# Initialise the smallest and the largest` `    ``# value from the second array arr2[]` `    ``smallest ``=` `arr2[``0``]` `    ``largest ``=` `arr2[``0``]`   `    ``# Find the smallest and` `    ``# the largest element in arr2` `    ``for` `i ``in` `range``(m):` `        ``smallest ``=` `max``(smallest, arr2[i])` `        ``largest ``=` `min``(largest, arr1[i])`   `    ``for` `i ``in` `range``(n):`   `        ``# Check if absolute difference of smallest` `        ``# and arr1[i] or largest and arr1[i] is > K` `        ``# then arr[i] is a required element` `        ``if` `(``abs``(arr1[i] ``-` `smallest) > k` `            ``or` `abs``(arr1[i] ``-` `largest) > k):` `            ``count ``+``=` `1`   `    ``# Print final result` `    ``print``(count)`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``arr1``=` `[ ``3``, ``1``, ``4` `]` `    ``n ``=` `len``(arr1)` `    ``arr2``=` `[ ``5``, ``1``, ``2` `]` `    ``m ``=` `len``(arr2)` `    ``k ``=` `2`   `    ``countDist(arr1, n, arr2, m, k)` `    `  `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to count elements in first array` `// with absolute difference greater than K` `// with an element in second Array` `using` `System;`   `class` `GFG{`   `// Function to count the such elements` `static` `void` `countDist(``int` `[]arr1, ``int` `n, ` `                      ``int` `[]arr2, ``int` `m,` `                      ``int` `k)` `{` `    ``// Store count of required elements in arr1` `    ``int` `count = 0;`   `    ``// Initialise the smallest and the largest` `    ``// value from the second array arr2[]` `    ``int` `smallest = arr2;` `    ``int` `largest = arr2;`   `    ``// Find the smallest and` `    ``// the largest element in arr2` `    ``for``(``int` `i = 0; i < m; i++) ` `    ``{` `       ``smallest = Math.Max(smallest, arr2[i]);` `       ``largest = Math.Min(largest, arr1[i]);` `    ``}` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `       `  `       ``// Check if absolute difference ` `       ``// of smallest and arr1[i] or ` `       ``// largest and arr1[i] is > K ` `       ``// then arr[i] is a required element` `       ``if` `(Math.Abs(arr1[i] - smallest) > k ||` `           ``Math.Abs(arr1[i] - largest) > k)` `           ``count++;` `    ``}`   `    ``// Print the readonly result` `    ``Console.Write(count);` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `[]arr1 = { 3, 1, 4 };` `    ``int` `n = arr1.Length;` `    ``int` `[]arr2 = { 5, 1, 2 };` `    ``int` `m = arr2.Length;` `    ``int` `k = 2;`   `    ``countDist(arr1, n, arr2, m, k);` `}` `}`   `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N + M), where N and M are the sizes of the given arrays.
Auxiliary Space: O(1).

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