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 pairsint 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 codeint 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
// Java implementation of the approachimport java.util.*;class solution{// Function to return the count of required pairsstatic 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 codepublic 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
# 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#
// C# implementation of the approachusing System;class GFG{ // Function to return the count of required pairsstatic 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 codestatic 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. |
Javascript
<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> |
Output:
3
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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