Count pairs in an array such that the absolute difference between them is ≥ K
Last Updated :
07 Jun, 2022
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++
// 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 difference
// 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;
}
// 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 difference
// 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));
}
}
# 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 difference
# 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
// 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 difference
// 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.
<script>
// JavaScript implementation of the approach
// Function to return the count of required pairs
function count(arr, n, k) {
// Sort the given array
arr.sort();
// To store the required count
var cnt = 0;
var i = 0;
var j = 1;
while (i < n && 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 difference
// with arr[i] >= k and the count of
// valid pairs will be (n - j)
while (j < n && (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
var arr = [ 1, 2, 3, 4 ];
var n = arr.length;
var k = 2;
document.write(count(arr, n, k));
// This code is contributed by AnkThon
</script>
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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