Count pairs in an array such that the absolute difference between them is ≥ K

Given an array arr[] and an integer K, the task is to find the count of pairs (arr[i], arr[j]) from the array such that |arr[i] – arr[j]| ≥ K. Note that (arr[i], arr[j]) and arr[j], arr[i] will be counted only once.

Examples:

Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 3
All valid pairs are (1, 3), (1, 4) and (2, 4)



Input: arr[] = {7, 4, 12, 56, 123}, K = 50
Output: 5

Approach: Sort the given array. Now for every element arr[i], find the first element on the right arr[j] such that (arr[j] – arr[i]) ≥ K. This is because after this element, every element will satisfy the same condition with arr[i] as the array is sorted and the count of elements that will make a valid pair with arr[i] will be (N – j) where N is the size of the given array.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of required pairs
int count(int arr[], int n, int k)
{
  
    // Sort the given array
    sort(arr, arr + n);
  
    // To store the required count
    int cnt = 0;
    int i = 0, j = 1;
  
    while (i < n && j < n) {
  
        // Update j such that it is always > i
        j = (j <= i) ? (i + 1) : j;
  
        // Find the first element arr[j] such that
        // (arr[j] - arr[i]) >= K
        // This is because after this element, all
        // the elements will have absolute differecne
        // with arr[i] >= k and the count of
        // valid pairs will be (n - j)
        while (j < n && (arr[j] - arr[i]) < k)
            j++;
  
        // Update the count of valid pairs
        cnt += (n - j);
  
        // Get to the next element to repeat the steps
        i++;
    }
  
    // Return the count
    return cnt;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
  
    cout << count(arr, n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class solution
{
  
// Function to return the count of required pairs
static int count(int arr[], int n, int k)
{
  
    // Sort the given array
    Arrays.sort(arr);
  
    // To store the required count
    int cnt = 0;
    int i = 0, j = 1;
  
    while (i < n && j < n) {
  
        // Update j such that it is always > i
        j = (j <= i) ? (i + 1) : j;
  
        // Find the first element arr[j] such that
        // (arr[j] - arr[i]) >= K
        // This is because after this element, all
        // the elements will have absolute differecne
        // with arr[i] >= k and the count of
        // valid pairs will be (n - j)
        while (j < n && (arr[j] - arr[i]) < k)
            j++;
  
        // Update the count of valid pairs
        cnt += (n - j);
  
        // Get to the next element to repeat the steps
        i++;
    }
  
    // Return the count
    return cnt;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;
    int k = 2;
  
    System.out.println(count(arr, n, k));
  
}
}

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Python3

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# Python3 implementation of the approach 
  
# Function to return the count of required pairs 
def count(arr, n, k) :
  
    # Sort the given array 
    arr.sort(); 
  
    # To store the required count 
    cnt = 0
    i = 0; j = 1
  
    while (i < n and j < n) :
  
        # Update j such that it is always > i 
        if j <= i :
            j = i + 1
        else :
            j = j
  
        # Find the first element arr[j] such that 
        # (arr[j] - arr[i]) >= K 
        # This is because after this element, all 
        # the elements will have absolute differecne 
        # with arr[i] >= k and the count of 
        # valid pairs will be (n - j) 
        while (j < n and (arr[j] - arr[i]) < k) :
            j += 1
  
        # Update the count of valid pairs 
        cnt += (n - j); 
  
        # Get to the next element to repeat the steps 
        i += 1
  
    # Return the count 
    return cnt; 
  
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3, 4 ]; 
    n = len(arr); 
    k = 2
  
    print(count(arr, n, k)); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the count of required pairs
static int count(int []arr, int n, int k)
{
  
    // Sort the given array
    Array.Sort(arr);
  
    // To store the required count
    int cnt = 0;
    int i = 0, j = 1;
  
    while (i < n && j < n) 
    {
  
        // Update j such that it is always > i
        j = (j <= i) ? (i + 1) : j;
  
        // Find the first element arr[j] such that
        // (arr[j] - arr[i]) >= K
        // This is because after this element, all
        // the elements will have absolute differecne
        // with arr[i] >= k and the count of
        // valid pairs will be (n - j)
        while (j < n && (arr[j] - arr[i]) < k)
            j++;
  
        // Update the count of valid pairs
        cnt += (n - j);
  
        // Get to the next element to repeat the steps
        i++;
    }
  
    // Return the count
    return cnt;
}
  
// Driver code
static public void Main ()
{
      
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;
    int k = 2;
  
    Console.Write(count(arr, n, k));
  
}
}
  
// This code is contributed by jit_t.

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Output:

3


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