Count pairs with absolute difference equal to k
Last Updated :
23 Jul, 2025
Given an array arr[] and a positive integer k, the task is to count all pairs (i, j) such that i < j and absolute value of (arr[i] - arr[j]) is equal to k.
Examples:
Input: arr[] = [1, 4, 1, 4, 5], k = 3
Output: 4
Explanation: There are 4 pairs with absolute difference 3, the pairs are [1, 4], [1, 4], [1, 4] and [4, 1]
Input: arr[] = [8, 16, 12, 16, 4, 0], k = 4
Output: 5
Explanation: There are 5 pairs with absolute difference 4, the pairs are [8, 12], [8, 4], [16, 12], [12, 16], [4, 0].
[Naive Approach] Generating all pairs - O(n^2) Time and O(1) Space
The basic idea is to use two nested loops to generate all pairs in arr[]. For each pair, if the absolute difference is equal to k, increment the count by 1.
C++
// C++ Program to count all pairs with difference equal to k
// by generating all pairs
#include <iostream>
#include <vector>
using namespace std;
int countPairs(vector<int> &arr, int k) {
int n = arr.size();
int cnt = 0;
// generate all possible pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If absolute difference = k, then increment
// count by 1
if (abs(arr[i] - arr[j]) == k) {
cnt += 1;
}
}
}
return cnt;
}
int main() {
vector<int> arr = {1, 4, 1, 4, 5};
int k = 3;
cout << countPairs(arr, k);
return 0;
}
// C Program to count all pairs with difference equal to k
// by generating all pairs
#include <stdio.h>
int countPairs(int arr[], int n, int k) {
int cnt = 0;
// generate all possible pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If absolute difference = k, then increment
// count by 1
if (abs(arr[i] - arr[j]) == k) {
cnt += 1;
}
}
}
return cnt;
}
int main() {
int arr[] = {1, 4, 1, 4, 5};
int k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d", countPairs(arr, n, k));
return 0;
}
// Java Program to count all pairs with difference equal to k
// by generating all pairs
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.length;
int cnt = 0;
// generate all possible pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If absolute difference = k, then increment
// count by 1
if (Math.abs(arr[i] - arr[j]) == k) {
cnt += 1;
}
}
}
return cnt;
}
public static void main(String[] args) {
int[] arr = {1, 4, 1, 4, 5};
int k = 3;
System.out.println(countPairs(arr, k));
}
}
# Python Program to count all pairs with difference equal to k
# by generating all pairs
def countPairs(arr, k):
n = len(arr)
cnt = 0
# generate all possible pairs
for i in range(n):
for j in range(i + 1, n):
# If absolute difference = k, then increment
# count by 1
if abs(arr[i] - arr[j]) == k:
cnt += 1
return cnt
if __name__ == "__main__":
arr = [1, 4, 1, 4, 5]
k = 3
print(countPairs(arr, k))
// C# Program to count all pairs with difference equal to k
// by generating all pairs
using System;
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.Length;
int cnt = 0;
// generate all possible pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// If absolute difference = k, then increment
// count by 1
if (Math.Abs(arr[i] - arr[j]) == k) {
cnt += 1;
}
}
}
return cnt;
}
static void Main() {
int[] arr = {1, 4, 1, 4, 5};
int k = 3;
Console.WriteLine(countPairs(arr, k));
}
}
// JavaScript Program to count all pairs with difference equal
// to k by generating all pairs
function countPairs(arr, k) {
const n = arr.length;
let cnt = 0;
// generate all possible pairs
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
// If absolute difference = k, then increment
// count by 1
if (Math.abs(arr[i] - arr[j]) === k) {
cnt += 1;
}
}
}
return cnt;
}
// Driver Code
const arr = [1, 4, 1, 4, 5];
const k = 3;
console.log(countPairs(arr, k));
[Better Approach] Sorting and Two Pointer Technique - O(n*logn) Time and O(1) Space
The idea is to first sort the array and then use the two-pointer technique by maintaining two pointers, say i and j and initialize them to the beginning of the array. According to the sum of the elements, we can have three cases:
- arr[i] - arr[j] < target: We need to increase the difference between the elements, move the j pointer towards right.
- arr[i] - arr[j] > target: We need to decrease the difference between the elements, move the i pointer towards right.
- arr[i] - arr[j] = target: We have found a pair whose difference is equal to target. We can find the product of the count of both the elements and add them to the result.
C++
// C++ Program to count all pairs with difference equal to k
// using Two Pointer Technique
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int countPairs(vector<int> &arr, int k) {
int n = arr.size();
int cnt = 0;
sort(arr.begin(), arr.end());
int i = 0, j = 0;
while(j < n) {
// If the difference is greater than k, increase
// the difference by moving pointer j towards right
if(arr[j] - arr[i] < k)
j++;
// If difference is greater than k, decrease the
// difference by moving pointer i towards right
else if(arr[j] - arr[i] > k)
i++;
// If difference is equal to k, count all such pairs
else {
int ele1 = arr[i], ele2 = arr[j];
int cnt1 = 0, cnt2 = 0;
// Count frequency of first element of the pair
while(j < n && arr[j] == ele2) {
j++;
cnt2++;
}
// Count frequency of second element of the pair
while(i < n && arr[i] == ele1) {
i++;
cnt1++;
}
// If both the elements are same, then count of
// pairs = the number of ways to choose 2
// elements among cnt1 elements
if(ele1 == ele2)
cnt += (cnt1 * (cnt1 - 1))/2;
// If the elements are different, then count of
// pairs = product of the count of both elements
else
cnt += (cnt1 * cnt2);
}
}
return cnt;
}
int main() {
vector<int> arr = {1, 4, 1, 4, 5};
int k = 3;
cout << countPairs(arr, k);
return 0;
}
// C Program to count all pairs with difference equal to k
// using Two Pointer Technique
#include <stdio.h>
int compare(const void *a, const void *b) {
return (*(int *)a - *(int *)b);
}
int countPairs(int arr[], int n, int k) {
int cnt = 0;
qsort(arr, n, sizeof(int), compare);
int i = 0, j = 0;
while (j < n) {
if (arr[j] - arr[i] < k)
j++;
// If difference is greater than k, decrease the
// difference by moving i pointer towards right
else if (arr[j] - arr[i] > k)
i++;
// If difference is equal to k, increase the difference
// by moving j pointer towards right
else {
int ele1 = arr[i], ele2 = arr[j];
int cnt1 = 0, cnt2 = 0;
// Count frequency of first element of the pair
while (j < n && arr[j] == ele2) {
j++;
cnt2++;
}
// Count frequency of second element of the pair
while (i < n && arr[i] == ele1) {
i++;
cnt1++;
}
// If both the elements are same, then count of
// pairs = the number of ways to choose 2
// elements among cnt1 elements
if (ele1 == ele2)
cnt += (cnt1 * (cnt1 - 1)) / 2;
// If the elements are different, then count of
// pairs = product of the count of both elements
else
cnt += (cnt1 * cnt2);
}
}
return cnt;
}
int main() {
int arr[] = {1, 4, 1, 4, 5};
int k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", countPairs(arr, n, k));
return 0;
}
// Java Program to count all pairs with difference equal to k
// using Two Pointer Technique
import java.util.Arrays;
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.length;
int cnt = 0;
Arrays.sort(arr);
int i = 0, j = 0;
while (j < n) {
if (arr[j] - arr[i] < k)
j++;
// If difference is greater than k, decrease the
// difference by moving i pointer towards right
else if (arr[j] - arr[i] > k)
i++;
// If difference is equal to k, increase the difference
// by moving j pointer towards right
else {
int ele1 = arr[i], ele2 = arr[j];
int cnt1 = 0, cnt2 = 0;
// Count frequency of first element of the pair
while (j < n && arr[j] == ele2) {
j++;
cnt2++;
}
// Count frequency of second element of the pair
while (i < n && arr[i] == ele1) {
i++;
cnt1++;
}
// If both the elements are same, then count of
// pairs = the number of ways to choose 2
// elements among cnt1 elements
if (ele1 == ele2)
cnt += (cnt1 * (cnt1 - 1)) / 2;
// If the elements are different, then count of
// pairs = product of the count of both elements
else
cnt += (cnt1 * cnt2);
}
}
return cnt;
}
public static void main(String[] args) {
int[] arr = {1, 4, 1, 4, 5};
int k = 3;
System.out.println(countPairs(arr, k));
}
}
# Python Program to count all pairs with difference equal to k
# using Two Pointer Technique
def countPairs(arr, k):
n = len(arr)
cnt = 0
arr.sort()
i = 0
j = 0
while j < n:
# If the difference is greater than k, increase
# the difference by moving j pointer towards right
if arr[j] - arr[i] < k:
j += 1
# If difference is greater than k, decrease the
# difference by moving i pointer towards right
elif arr[j] - arr[i] > k:
i += 1
# If difference is equal to k, count all such pairs
else:
ele1 = arr[i]
ele2 = arr[j]
cnt1 = 0
cnt2 = 0
# Count frequency of first element of the pair
while j < n and arr[j] == ele2:
j += 1
cnt2 += 1
# Count frequency of second element of the pair
while i < n and arr[i] == ele1:
i += 1
cnt1 += 1
# If both the elements are same, then count of
# pairs = the number of ways to choose 2
# elements among cnt1 elements
if ele1 == ele2:
cnt += (cnt1 * (cnt1 - 1)) // 2
# If the elements are different, then count of
# pairs = product of the count of both elements
else:
cnt += (cnt1 * cnt2)
return cnt
if __name__ == "__main__":
arr = [1, 4, 1, 4, 5]
k = 3
print(countPairs(arr, k))
// C# Program to count all pairs with difference equal to k
// using Two Pointer Technique
using System;
using System.Linq;
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.Length;
int cnt = 0;
Array.Sort(arr);
int i = 0, j = 0;
while (j < n) {
// If the difference is greater than k, increase
// the difference by moving j pointer towards right
if (arr[j] - arr[i] < k)
j++;
// If difference is greater than k, decrease the
// difference by moving i pointer towards right
else if (arr[j] - arr[i] > k)
i++;
// If difference is equal to k, count all such pairs
else {
int ele1 = arr[i], ele2 = arr[j];
int cnt1 = 0, cnt2 = 0;
// Count frequency of first element of the pair
while (j < n && arr[j] == ele2) {
j++;
cnt2++;
}
// Count frequency of second element of the pair
while (i < n && arr[i] == ele1) {
i++;
cnt1++;
}
// If both the elements are same, then count of
// pairs = the number of ways to choose 2
// elements among cnt1 elements
if (ele1 == ele2)
cnt += (cnt1 * (cnt1 - 1)) / 2;
// If the elements are different, then count of
// pairs = product of the count of both elements
else
cnt += (cnt1 * cnt2);
}
}
return cnt;
}
static void Main(string[] args) {
int[] arr = { 1, 4, 1, 4, 5 };
int k = 3;
Console.WriteLine(countPairs(arr, k));
}
}
// JavaScript Program to count all pairs with difference
// equal to k using Two Pointer Technique
function countPairs(arr, k) {
const n = arr.length;
let cnt = 0;
arr.sort((a, b) => a - b);
let i = 0, j = 0;
while (j < n) {
// If the difference is greater than k, increase
// the difference by moving j pointer towards right
if (arr[j] - arr[i] < k)
j++;
// If difference is greater than k, decrease the
// difference by moving i pointer towards right
else if (arr[j] - arr[i] > k)
i++;
// If difference is equal to k, count all such pairs
else {
const ele1 = arr[i], ele2 = arr[j];
let cnt1 = 0, cnt2 = 0;
// Count frequency of first element of the pair
while (j < n && arr[j] === ele2) {
j++;
cnt2++;
}
// Count frequency of second element of the pair
while (i < n && arr[i] === ele1) {
i++;
cnt1++;
}
// If both the elements are same, then count of
// pairs = the number of ways to choose 2
// elements among cnt1 elements
if (ele1 === ele2)
cnt += (cnt1 * (cnt1 - 1)) / 2;
// If the elements are different, then count of
// pairs = product of the count of both elements
else
cnt += (cnt1 * cnt2);
}
}
return cnt;
}
const arr = [1, 4, 1, 4, 5];
const k = 3;
console.log(countPairs(arr, k));
[Expected Approach] Using Hash Map or Dictionary – O(n) Time and O(n) Space
The idea is to count the frequency of each number in a hash map or dictionary as we go through the array Iterate over the array and for each element arr[i], we need another element say complement such that abs(arr[i] - complement) = k. Now, we can have two cases:
- (arr[i] - complement) is positive:
- arr[i] - complement = k
- So, complement = arr[i] - k
- (arr[i] - complement) is negative:
- (arr[i] - complement) = -k
- So, complement = arr[i] + k
So for each element arr[i], we can check if complement (arr[i] + k) or (arr[i] - k) is present in the hash map. If it is, increment the count variable by the occurrences of complement in map.
C++
// C++ Program to count all pairs with difference equal to k
// using Hash Map
#include <iostream>
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;
int countPairs(vector<int> &arr, int k) {
int n = arr.size();
unordered_map<int, int> freq;
int cnt = 0;
for (int i = 0; i < n; i++) {
// Check if the complement (arr[i] + k)
// exists in the map. If yes, increment count
if (freq.find(arr[i] + k) != freq.end())
cnt += freq[arr[i] + k];
// Check if the complement (arr[i] - k)
// exists in the map. If yes, increment count
if (freq.find(arr[i] - k) != freq.end())
cnt += freq[arr[i] - k];
// Increment the frequency of arr[i]
freq[arr[i]]++;
}
return cnt;
}
int main() {
vector<int> arr = {1, 4, 1, 4, 5};
int k = 3;
cout << countPairs(arr, k);
return 0;
}
// Java Program to count all pairs with difference equal to k
// using Hash Map
import java.util.HashMap;
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.length;
HashMap<Integer, Integer> freq = new HashMap<>();
int cnt = 0;
for (int i = 0; i < n; i++) {
// Check if the complement (arr[i] + k)
// exists in the map. If yes, increment count
if (freq.containsKey(arr[i] + k))
cnt += freq.get(arr[i] + k);
// Check if the complement (arr[i] - k)
// exists in the map. If yes, increment count
if (freq.containsKey(arr[i] - k))
cnt += freq.get(arr[i] - k);
// Increment the frequency of arr[i]
freq.put(arr[i], freq.getOrDefault(arr[i], 0) + 1);
}
return cnt;
}
public static void main(String[] args) {
int[] arr = {1, 4, 1, 4, 5};
int k = 3;
System.out.println(countPairs(arr, k));
}
}
# Python Program to count all pairs with difference equal to k
# using Hash Map
def countPairs(arr, k):
n = len(arr)
freq = {}
cnt = 0
for i in range(n):
# Check if the complement (arr[i] + k)
# exists in the map. If yes, increment count
if (arr[i] + k) in freq:
cnt += freq[arr[i] + k]
# Check if the complement (arr[i] - k)
# exists in the map. If yes, increment count
if (arr[i] - k) in freq:
cnt += freq[arr[i] - k]
# Increment the frequency of arr[i]
freq[arr[i]] = freq.get(arr[i], 0) + 1
return cnt
if __name__ == "__main__":
arr = [1, 4, 1, 4, 5]
k = 3
print(countPairs(arr, k))
// C# Program to count all pairs with difference equal to k
// using Hash Map
using System;
using System.Collections.Generic;
class GfG {
static int countPairs(int[] arr, int k) {
int n = arr.Length;
Dictionary<int, int> freq = new Dictionary<int, int>();
int cnt = 0;
for (int i = 0; i < n; i++) {
// Check if the complement (arr[i] + k)
// exists in the map. If yes, increment count
if (freq.ContainsKey(arr[i] + k))
cnt += freq[arr[i] + k];
// Check if the complement (arr[i] - k)
// exists in the map. If yes, increment count
if (freq.ContainsKey(arr[i] - k))
cnt += freq[arr[i] - k];
// Increment the frequency of arr[i]
if (freq.ContainsKey(arr[i]))
freq[arr[i]]++;
else
freq[arr[i]] = 1;
}
return cnt;
}
static void Main(string[] args) {
int[] arr = { 1, 4, 1, 4, 5 };
int k = 3;
Console.WriteLine(countPairs(arr, k));
}
}
// JavaScript Program to count all pairs with difference equal to k
// using Hash Map
function countPairs(arr, k) {
const n = arr.length;
const freq = {};
let cnt = 0;
for (let i = 0; i < n; i++) {
// Check if the complement (arr[i] + k)
// exists in the map. If yes, increment count
if ((arr[i] + k) in freq)
cnt += freq[arr[i] + k];
// Check if the complement (arr[i] - k)
// exists in the map. If yes, increment count
if ((arr[i] - k) in freq)
cnt += freq[arr[i] - k];
// Increment the frequency of arr[i]
freq[arr[i]] = (freq[arr[i]] || 0) + 1;
}
return cnt;
}
// Driver Code
const arr = [1, 4, 1, 4, 5];
const k = 3;
console.log(countPairs(arr, k));
Count pairs with absolute difference equal to k
Visit Course
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem