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: 2
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: 2
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:

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// 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;
}
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// 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
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# 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
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// 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
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Output:
2

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




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