Count pairs in Array such that their bitwise XOR has Kth right bit set
Last Updated :
23 Jul, 2025
Given an array arr[] of size N and an integer K, the task is to find the number of pairs in the array such that the Kth bit from the right of their bitwise XOR is set.
Examples:
Input: arr[] = {3, 13, 2, 9}, N = 4, K = 3
Output: 4
Explanation: The pairs that have the kth bit set in bitwise xor value are {3, 13}, {3, 9}, {2, 13}, {2, 9}
Input: arr[] = {7, 6, 5, 3}, N = 4, K = 2
Output: 3
Explanation: The pairs that have a kth bit set in bitwise xor value are {7, 3}, {6, 3}, {5, 3}
Naive Approach: The basic way to solve the problem is as follows:
The most basic way to solve this problem is to consider each possible pair of elements from the array and check whether the number formed from the bitwise xor of both elements has a kth bit set.
C++
#include <iostream>
using namespace std;
int countPairs(int arr[], int n, int k) {
int count = 0;
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
if ((arr[i]^arr[j]) & (1 << (k))) {
count++;
}
}
}
return count;
}
int main() {
// TestCase 1
int arr1[] = { 3, 13, 2, 9 };
int K = 1;
int N = sizeof(arr1) / sizeof(arr1[0]);
// Function Call
cout << countPairs(arr1, N, K) << endl;
// Testcase 2
int arr2[] = { 7, 6, 5, 3 };
K = 2;
N = sizeof(arr2) / sizeof(arr2[0]);
// Function Call
cout << countPairs(arr2, N, K) << endl;
return 0;
}
import java.util.Arrays;
public class GFG {
// Function to check kth bit
public static int countPairs(int[] arr, int n, int k) {
int count = 0;
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
// Chekcing bit is set of not
if (((arr[i]^arr[j]) & (1 << (k))) != 0) {
count++;
}
}
}
// Returning total pair
return count;
}
public static void main(String[] args) {
int[] arr1 = { 3, 13, 2, 9 };
int K = 1;
int N = arr1.length;
System.out.println(countPairs(arr1, N, K));
int[] arr2 = { 7, 6, 5, 3 };
K = 2;
N = arr2.length;
System.out.println(countPairs(arr2, N, K));
}
}
def countPairs(arr, n, k):
count = 0
for i in range(n-1):
for j in range(i+1, n):
if (arr[i]^arr[j]) & (1 << k):
count += 1
return count
# TestCase 1
arr1 = [3, 13, 2, 9]
K = 1
N = len(arr1)
# Function Call
print(countPairs(arr1, N, K))
# Testcase 2
arr2 = [7, 6, 5, 3]
K = 2
N = len(arr2)
# Function Call
print(countPairs(arr2, N, K))
using System;
public class GFG {
// Function to check kth bit
public static int countPairs(int[] arr, int n, int k) {
int count = 0;
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
// Chekcing bit is set of not
if (((arr[i]^arr[j]) & (1 << (k))) != 0) {
count++;
}
}
}
// Returning total pair
return count;
}
// Driver code
public static void Main(string[] args) {
int[] arr1 = { 3, 13, 2, 9 };
int K = 1;
int N = arr1.Length;
Console.WriteLine(countPairs(arr1, N, K));
int[] arr2 = { 7, 6, 5, 3 };
K = 2;
N = arr2.Length;
Console.WriteLine(countPairs(arr2, N, K));
}
}
function countPairs(arr, n, k) {
let count = 0;
for (let i = 0; i < n - 1; i++) {
for (let j = i + 1; j < n; j++) {
if ((arr[i] ^ arr[j]) & (1 << k)) {
count += 1;
}
}
}
return count;
}
// TestCase 1
const arr1 = [3, 13, 2, 9];
const K = 1;
const N1 = arr1.length;
// Function Call
console.log(countPairs(arr1, N1, K));
// Testcase 2
const arr2 = [7, 6, 5, 3];
const K1 = 2;
const N = arr2.length;
// Function Call
console.log(countPairs(arr2, N, K1));
Time Complexity: O(N2 * log(Max_ele)) where Max_ele is the largest element of the array.
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea:
Consider two numbers X and Y where X is having its Kth bit set. Only the following two scenarios can happen :
- If Y also has Kth bit set then X^Y will have Kth bit unset.
- If the Kth bit of Y is unset then X^Y will have the Kth bit set.
Follow the below-mentioned steps to implement the above approach:
- Count the number of elements having Kth right bit set.
- For each element arr[i] we know if it has Kth right bit set or not.
- If the Kth right bit is set in arr[i] then take a count of all those elements that have bit unset at the Kth position from right.
- If the Kth right bit is unset in arr[i] then consider all those elements that have set the bit at the Kth position from right.
- Ultimately, divide the final answer by 2 as the calculation considers each pair twice.
Below is the implementation of the above approach.
C++
// C++ implementation of above program
#include <bits/stdc++.h>
using namespace std;
int solve(int arr[], int n, int k)
{
int bits[32];
// Initialize an array to store the
// count of elements having i'th bit set
memset(bits, 0, sizeof(bits));
// Count the number of elements that
// have jth bit set
for (int i = 0; i < n; i++) {
for (int j = 0; j < 32; j++) {
// check if jth bit is set or not
if ((arr[i] & (1 << j)) > 0) {
bits[j]++;
}
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
// If kth bit is set for an element
// then add all those numbers to the
// answer for which kth bit is unset.
if ((arr[i] & (1 << k)) > 0) {
ans += (n - bits[k]);
}
// Else do the opposite
else {
ans += (bits[k]);
}
}
// Since the above calculation considers
// each pair twice we need to divide the
// answer by 2
ans = ans / 2;
// Return the answer
return ans;
}
// Driver Code
int main()
{
// TestCase 1
int arr1[] = { 3, 13, 2, 9 };
int K = 1;
int N = sizeof(arr1) / sizeof(arr1[0]);
// Function Call
cout << solve(arr1, N, K) << endl;
// Testcase 2
int arr2[] = { 7, 6, 5, 3 };
K = 2;
N = sizeof(arr2) / sizeof(arr2[0]);
// Function Call
cout << solve(arr2, N, K) << endl;
return 0;
}
// Java code for the above approach
import java.util.Arrays;
public class Main {
public static int solve(int[] arr, int n, int k) {
int[] bits = new int[32];
// Initialize an array to store the
// count of elements having i'th bit set
Arrays.fill(bits, 0);
// Count the number of elements that
// have jth bit set
for (int i = 0; i < n; i++) {
for (int j = 0; j < 32; j++) {
// check if jth bit is set or not
if ((arr[i] & (1 << j)) > 0) {
bits[j]++;
}
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
// If kth bit is set for an element
// then add all those numbers to the
// answer for which kth bit is unset.
if ((arr[i] & (1 << k)) > 0) {
ans += (n - bits[k]);
}
// Else do the opposite
else {
ans += (bits[k]);
}
}
// Since the above calculation considers
// each pair twice we need to divide the
// answer by 2
ans = ans / 2;
// Return the answer
return ans;
}
public static void main(String[] args) {
// TestCase 1
int[] arr1 = {3, 13, 2, 9};
int K = 1;
int N = arr1.length;
// Function Call
System.out.println(solve(arr1, N, K));
// Testcase 2
int[] arr2 = {7, 6, 5, 3};
K = 2;
N = arr2.length;
// Function Call
System.out.println(solve(arr2, N, K));
}
}
// This code is contributed by lokeshpotta20.
# python implementation of above program
def solve(arr, n, k):
# Initialize an array to store the
# count of elements having i'th bit set
bits = [0] * 32
# Count the number of elements that
# have jth bit set
for i in range(n):
for j in range(32):
# check if jth bit is set or not
if arr[i] & (1 << j) > 0:
bits[j] += 1
ans = 0
for i in range(n):
# If kth bit is set for an element
# then add all those numbers to the
# answer for which kth bit is unset.
if arr[i] & (1 << k) > 0:
ans += (n - bits[k])
# Else do the opposite
else:
ans += bits[k]
# Since the above calculation considers
# each pair twice we need to divide the
# answer by 2
ans = ans // 2
# Return the answer
return ans
# Driver Code
# TestCase 1
arr1 = [3, 13, 2, 9]
K = 1
N = len(arr1)
# Function Call
print(solve(arr1, N, K))
# Testcase 2
arr2 = [7, 6, 5, 3]
K = 2
N = len(arr2)
# Function Call
print(solve(arr2, N, K))
#This code is contributed by rutikbhosale.
// C# implementation of above program
using System;
class Gfg{
static int solve(int[] arr, int n, int k)
{
int[] bits=new int[32];
// Initialize an array to store the
// count of elements having i'th bit set
for(int i=0; i<32; i++)
bits[i]=0;
// Count the number of elements that
// have jth bit set
for (int i = 0; i < n; i++) {
for (int j = 0; j < 32; j++) {
// check if jth bit is set or not
if ((arr[i] & (1 << j)) > 0) {
bits[j]++;
}
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
// If kth bit is set for an element
// then add all those numbers to the
// answer for which kth bit is unset.
if ((arr[i] & (1 << k)) > 0) {
ans += (n - bits[k]);
}
// Else do the opposite
else {
ans += (bits[k]);
}
}
// Since the above calculation considers
// each pair twice we need to divide the
// answer by 2
ans = ans / 2;
// Return the answer
return ans;
}
// Driver Code
public static void Main()
{
// TestCase 1
int[] arr1 = { 3, 13, 2, 9 };
int K = 1;
int N = arr1.Length;
// Function Call
Console.WriteLine(solve(arr1, N, K));
// Testcase 2
int[] arr2 = { 7, 6, 5, 3 };
K = 2;
N = arr2.Length;
// Function Call
Console.WriteLine(solve(arr2, N, K));
}
}
// Javascript implementation of above program
function solve( arr, n, k)
{
let bits=new Array(32).fill(0);
// Initialize an array to store the
// count of elements having i'th bit set
// Count the number of elements that
// have jth bit set
for (let i = 0; i < n; i++) {
for (let j = 0; j < 32; j++) {
// check if jth bit is set or not
if ((arr[i] & (1 << j)) > 0) {
bits[j]++;
}
}
}
let ans = 0;
for (let i = 0; i < n; i++) {
// If kth bit is set for an element
// then add all those numbers to the
// answer for which kth bit is unset.
if ((arr[i] & (1 << k)) > 0) {
ans += (n - bits[k]);
}
// Else do the opposite
else {
ans += (bits[k]);
}
}
// Since the above calculation considers
// each pair twice we need to divide the
// answer by 2
ans = Math.floor(ans / 2);
// Return the answer
return ans;
}
// Driver Code
// TestCase 1
let arr1 = [ 3, 13, 2, 9 ];
let K = 1;
let N = arr1.length;
// Function Call
console.log(solve(arr1, N, K));
// Testcase 2
let arr2 = [ 7, 6, 5, 3 ];
K = 2;
N = arr2.length;
// Function Call
console.log(solve(arr2, N, K));
// This code is contributed by ratiagrawal.
Time Complexity: O(N * log(Max_ele)) where Max_ele is the maximum value in the array
Auxiliary Space: O(1)
Related Articles:
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