# 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 = 2Output:2Explanation:

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 = 3Output:2Explanation:

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 <bits/stdc++.h>` `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[0];` ` ` `int` `largest = arr2[0];` ` ` `// 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[0]);` ` ` `int` `arr2[] = { 5, 1, 2 };` ` ` `int` `m = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]);` ` ` `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[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 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

`<script>` `// Javascript program to count elements in` `// first Array with absolute difference` `// greater than K with an element in` `// second Array ` `// Function to count the such elements` `function` `countDist(arr1, n, arr2, m, k)` `{` ` ` ` ` `// Store count of required elements in arr1` ` ` `var` `count = 0;` ` ` `// Initialise the smallest and the largest` ` ` `// value from the second array arr2` ` ` `var` `smallest = arr2[0];` ` ` `var` `largest = arr2[0];` ` ` `// Find the smallest and` ` ` `// the largest element in arr2` ` ` `for` `(i = 0; i < m; i++)` ` ` `{` ` ` `smallest = Math.max(smallest, arr2[i]);` ` ` `largest = Math.min(largest, arr1[i]);` ` ` `}` ` ` `for` `(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++;` ` ` `}` ` ` `// Prvar the final result` ` ` `document.write(count);` `}` `// Driver code` `var` `arr1 = [ 3, 1, 4 ];` `var` `n = arr1.length;` `var` `arr2 = [ 5, 1, 2 ];` `var` `m = arr2.length;` `var` `k = 2;` `countDist(arr1, n, arr2, m, k);` `// This code is contributed by umadevi9616` `</script>` |

**Output:**

2

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