Count pairs having bitwise XOR greater than K from a given array
Last Updated :
23 Jul, 2025
Given an array arr[]of size N and an integer K, the task is to count the number of pairs from the given array such that the Bitwise XOR of each pair is greater than K.
Examples:
Input: arr = {1, 2, 3, 5} , K = 2
Output: 4
Explanation:
Bitwise XOR of all possible pairs that satisfy the given conditions are:
arr[0] ^ arr[1] = 1 ^ 2 = 3
arr[0] ^ arr[3] = 1 ^ 5 = 4
arr[1] ^ arr[3] = 2 ^ 5 = 7
arr[0] ^ arr[3] = 3 ^ 5 = 6
Therefore, the required output is 4.
Input: arr[] = {3, 5, 6,8}, K = 2
Output: 6
Naive Approach: The simplest approach to solve this problem is to traverse the given array and generate all possible pairs of the given array and for each pair, check if bitwise XOR of the pair is greater than K or not. If found to be true, then increment the count of pairs having bitwise XOR greater than K. Finally, print the count of such pairs obtained.
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to Count pairs having
// bitwise XOR greater than K
// from a given array
int cntGreaterPairs(int arr[], int n, int k)
{
// Variable initialise to store the count
int count=0;
// Traverse in the array and
// generate all possible pairs
// of the given array
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
// For every pair, check if bitwise
// XOR of the pair is greater than
// K or not
if(arr[i]^arr[j]>k)
count++;
}
}
// return the count
return count;
}
//Driver code
int main()
{
int arr[] = {3, 5, 6, 8};
int K= 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout<<cntGreaterPairs(arr, N, K);
}
// This code is contributed by Utkarsh Kumar.
// Java program to implement
// the above approach
import java.util.*;
class Main {
// Function to Count pairs having
// bitwise XOR greater than K
// from a given array
static int cntGreaterPairs(int arr[], int n, int k)
{
// Variable initialise to store the count
int count = 0;
// Traverse in the array and
// generate all possible pairs
// of the given array
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// For every pair, check if bitwise
// XOR of the pair is greater than
// K or not
if ((arr[i] ^ arr[j]) > k)
count++;
}
}
// return the count
return count;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 3, 5, 6, 8 };
int K = 2;
int N = arr.length;
System.out.print(cntGreaterPairs(arr, N, K));
}
}
# Function to count pairs having bitwise XOR greater than K from a given array
def cntGreaterPairs(arr, n, k):
# Variable initialise to store the count
count = 0
# Traverse in the array and
# generate all possible pairs
# of the given array
for i in range(n):
for j in range(i+1, n):
# For every pair, check if bitwise
# XOR of the pair is greater than
# K or not
if arr[i]^arr[j] > k:
count += 1
# Return the count
return count
# Driver code
arr = [3, 5, 6, 8]
K = 2
N = len(arr)
print(cntGreaterPairs(arr, N, K))
// Function to count pairs having bitwise XOR greater than K from a given array
function cntGreaterPairs(arr, n, k) {
// Variable initialise to store the count
let count = 0;
// Traverse in the array and
// generate all possible pairs
// of the given array
for (let i = 0; i < n; i++) {
for (let j = i+1; j < n; j++) {
// For every pair, check if bitwise
// XOR of the pair is greater than
// K or not
if ((arr[i]^arr[j]) > k) {
count++;
}
}
}
// Return the count
return count;
}
// Driver code
const arr = [3, 5, 6, 8];
const K = 2;
const N = arr.length;
console.log(cntGreaterPairs(arr, N, K));
using System;
public class Program
{
// Function to Count pairs having
// bitwise XOR greater than K
// from a given array
public static int cntGreaterPairs(int[] arr, int n, int k)
{
// Variable initialise to store the count
int count = 0;
// Traverse in the array and
// generate all possible pairs
// of the given array
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
// For every pair, check if bitwise
// XOR of the pair is greater than
// K or not
if ((arr[i] ^ arr[j]) > k)
count++;
}
}
// return the count
return count;
}
//Driver code
public static void Main()
{
int[] arr = { 3, 5, 6, 8 };
int K = 2;
int N = arr.Length;
Console.WriteLine(cntGreaterPairs(arr, N, K));
}
}
Time Complexity:O(N2)
Auxiliary Space:O(1)
Efficient Approach: The problem can be solved using Trie. The idea is to iterate over the given array and for each array element, count the number of elements present in the Trie whose bitwise XOR with the current element is greater than K and insert the binary representation of the current element into the Trie. Finally, print the count of pairs having bitwise XOR greater than K. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of Trie
struct TrieNode
{
// Stores binary representation
// of numbers
TrieNode *child[2];
// Stores count of elements
// present in a node
int cnt;
// Function to initialize
// a Trie Node
TrieNode() {
child[0] = child[1] = NULL;
cnt = 0;
}
};
// Function to insert a number into Trie
void insertTrie(TrieNode *root, int N) {
// Traverse binary representation of X.
for (int i = 31; i >= 0; i--) {
// Stores ith bit of N
bool x = (N) & (1 << i);
// Check if an element already
// present in Trie having ith bit x.
if(!root->child[x]) {
// Create a new node of Trie.
root->child[x] = new TrieNode();
}
// Update count of elements
// whose ith bit is x
root->child[x]->cnt+= 1;
// Update root.
root= root->child[x];
}
}
// Function to count elements
// in Trie whose XOR with N
// exceeds K
int cntGreater(TrieNode * root,
int N, int K)
{
// Stores count of elements
// whose XOR with N exceeding K
int cntPairs = 0;
// Traverse binary representation
// of N and K in Trie
for (int i = 31; i >= 0 &&
root; i--) {
// Stores ith bit of N
bool x = N & (1 << i);
// Stores ith bit of K
bool y = K & (1 << i);
// If the ith bit of K is 1
if (y) {
// Update root.
root =
root->child[1 - x];
}
// If the ith bit of K is 0
else{
// If an element already
// present in Trie having
// ith bit (1 - x)
if (root->child[1 - x]) {
// Update cntPairs
cntPairs +=
root->child[1 - x]->cnt;
}
// Update root.
root = root->child[x];
}
}
return cntPairs;
}
// Function to count pairs that
// satisfy the given conditions.
int cntGreaterPairs(int arr[], int N,
int K) {
// Create root node of Trie
TrieNode *root = new TrieNode();
// Stores count of pairs that
// satisfy the given conditions
int cntPairs = 0;
// Traverse the given array.
for(int i = 0;i < N; i++){
// Update cntPairs
cntPairs += cntGreater(root,
arr[i], K);
// Insert arr[i] into Trie.
insertTrie(root, arr[i]);
}
return cntPairs;
}
//Driver code
int main()
{
int arr[] = {3, 5, 6, 8};
int K= 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout<<cntGreaterPairs(arr, N, K);
}
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Structure of Trie
static class TrieNode
{
// Stores binary representation
// of numbers
TrieNode []child = new TrieNode[2];
// Stores count of elements
// present in a node
int cnt;
// Function to initialize
// a Trie Node
TrieNode()
{
child[0] = child[1] = null;
cnt = 0;
}
};
// Function to insert a number
// into Trie
static void insertTrie(TrieNode root,
int N)
{
// Traverse binary representation
// of X.
for (int i = 31; i >= 0; i--)
{
// Stores ith bit of N
int x = (N) & (1 << i);
// Check if an element already
// present in Trie having ith
// bit x.
if (x < 2 && root.child[x] == null)
{
// Create a new node of Trie.
root.child[x] = new TrieNode();
}
// Update count of elements
// whose ith bit is x
if(x < 2 && root.child[x] != null)
root.child[x].cnt += 1;
// Update root.
if(x < 2)
root = root.child[x];
}
}
// Function to count elements
// in Trie whose XOR with N
// exceeds K
static int cntGreater(TrieNode root,
int N, int K)
{
// Stores count of elements
// whose XOR with N exceeding K
int cntPairs = 1;
// Traverse binary representation
// of N and K in Trie
for (int i = 31; i >= 0 &&
root!=null; i--)
{
// Stores ith bit of N
int x = N & (1 << i);
// Stores ith bit of K
int y = K & (1 << i);
// If the ith bit of K is 1
if (y == 1)
{
// Update root.
root = root.child[1 - x];
}
// If the ith bit of K is 0
else
{
// If an element already
// present in Trie having
// ith bit (1 - x)
if (x < 2 &&
root.child[1 - x] != null)
{
// Update cntPairs
cntPairs += root.child[1 - x].cnt;
}
// Update root.
if(x < 2)
root = root.child[x];
}
}
return cntPairs;
}
// Function to count pairs that
// satisfy the given conditions.
static int cntGreaterPairs(int arr[],
int N, int K)
{
// Create root node of Trie
TrieNode root = new TrieNode();
// Stores count of pairs that
// satisfy the given conditions
int cntPairs = 0;
// Traverse the given array.
for (int i = 0; i < N; i++)
{
// Update cntPairs
cntPairs += cntGreater(root,
arr[i], K);
// Insert arr[i] into Trie.
insertTrie(root, arr[i]);
}
return cntPairs;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {3, 5, 6, 8};
int K = 2;
int N = arr.length;
System.out.print(cntGreaterPairs(arr,
N, K));
}
}
// This code is contributed by shikhasingrajput
# Python3 program to implement
# the above approach
# Structure of Trie
class TrieNode:
# Function to initialize
# a Trie Node
def __init__(self):
self.child = [None, None]
self.cnt = 0
# Function to insert a number into Trie
def insertTrie(root, N):
# Traverse binary representation of X.
for i in range(31, -1, -1):
# Stores ith bit of N
x = bool((N) & (1 << i))
# Check if an element already
# present in Trie having ith bit x.
if (root.child[x] == None):
# Create a new node of Trie.
root.child[x] = TrieNode()
# Update count of elements
# whose ith bit is x
root.child[x].cnt += 1
# Update root
root= root.child[x]
# Function to count elements
# in Trie whose XOR with N
# exceeds K
def cntGreater(root, N, K):
# Stores count of elements
# whose XOR with N exceeding K
cntPairs = 0
# Traverse binary representation
# of N and K in Trie
for i in range(31, -1, -1):
if (root == None):
break
# Stores ith bit of N
x = bool(N & (1 << i))
# Stores ith bit of K
y = K & (1 << i)
# If the ith bit of K is 1
if (y != 0):
# Update root
root = root.child[1 - x]
# If the ith bit of K is 0
else:
# If an element already
# present in Trie having
# ith bit (1 - x)
if (root.child[1 - x]):
# Update cntPairs
cntPairs += root.child[1 - x].cnt
# Update root
root = root.child[x]
return cntPairs
# Function to count pairs that
# satisfy the given conditions.
def cntGreaterPairs(arr, N, K):
# Create root node of Trie
root = TrieNode()
# Stores count of pairs that
# satisfy the given conditions
cntPairs = 0
# Traverse the given array.
for i in range(N):
# Update cntPairs
cntPairs += cntGreater(root, arr[i], K)
# Insert arr[i] into Trie.
insertTrie(root, arr[i])
return cntPairs
# Driver code
if __name__=='__main__':
arr = [ 3, 5, 6, 8 ]
K = 2
N = len(arr)
print(cntGreaterPairs(arr, N, K))
# This code is contributed by rutvik_56
// C# program to implement
// the above approach
using System;
class GFG{
// Structure of Trie
public class TrieNode
{
// Stores binary representation
// of numbers
public TrieNode []child = new TrieNode[2];
// Stores count of elements
// present in a node
public int cnt;
// Function to initialize
// a Trie Node
public TrieNode()
{
child[0] = child[1] = null;
cnt = 0;
}
};
// Function to insert a number
// into Trie
static void insertTrie(TrieNode root,
int N)
{
// Traverse binary representation
// of X.
for(int i = 31; i >= 0; i--)
{
// Stores ith bit of N
int x = (N) & (1 << i);
// Check if an element already
// present in Trie having ith
// bit x.
if (x < 2 && root.child[x] == null)
{
// Create a new node of Trie.
root.child[x] = new TrieNode();
}
// Update count of elements
// whose ith bit is x
if(x < 2 && root.child[x] != null)
root.child[x].cnt += 1;
// Update root.
if(x < 2)
root = root.child[x];
}
}
// Function to count elements
// in Trie whose XOR with N
// exceeds K
static int cntGreater(TrieNode root,
int N, int K)
{
// Stores count of elements
// whose XOR with N exceeding K
int cntPairs = 1;
// Traverse binary representation
// of N and K in Trie
for(int i = 31; i >= 0 &&
root != null; i--)
{
// Stores ith bit of N
int x = N & (1 << i);
// Stores ith bit of K
int y = K & (1 << i);
// If the ith bit of K is 1
if (y == 1)
{
// Update root.
root = root.child[1 - x];
}
// If the ith bit of K is 0
else
{
// If an element already
// present in Trie having
// ith bit (1 - x)
if (x < 2 &&
root.child[1 - x] != null)
{
// Update cntPairs
cntPairs += root.child[1 - x].cnt;
}
// Update root.
if(x < 2)
root = root.child[x];
}
}
return cntPairs;
}
// Function to count pairs that
// satisfy the given conditions.
static int cntGreaterPairs(int []arr,
int N, int K)
{
// Create root node of Trie
TrieNode root = new TrieNode();
// Stores count of pairs that
// satisfy the given conditions
int cntPairs = 0;
// Traverse the given array.
for(int i = 0; i < N; i++)
{
// Update cntPairs
cntPairs += cntGreater(root,
arr[i], K);
// Insert arr[i] into Trie.
insertTrie(root, arr[i]);
}
return cntPairs;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 3, 5, 6, 8 };
int K = 2;
int N = arr.Length;
Console.Write(cntGreaterPairs(arr,
N, K));
}
}
// This code is contributed by gauravrajput1
<script>
// Javascript program to implement
// the above approach
// Structure of Trie
class TrieNode
{
constructor()
{
this.child = new Array(2);
this.child[0] = this.child[1] = null;
this.cnt = 0;
}
}
// Function to insert a number
// into Trie
function insertTrie(root,N)
{
// Traverse binary representation
// of X.
for (let i = 31; i >= 0; i--)
{
// Stores ith bit of N
let x = (N) & (1 << i);
// Check if an element already
// present in Trie having ith
// bit x.
if (x < 2 && root.child[x] == null)
{
// Create a new node of Trie.
root.child[x] = new TrieNode();
}
// Update count of elements
// whose ith bit is x
if(x < 2 && root.child[x] != null)
root.child[x].cnt += 1;
// Update root.
if(x < 2)
root = root.child[x];
}
}
// Function to count elements
// in Trie whose XOR with N
// exceeds K
function cntGreater(root, N, K)
{
// Stores count of elements
// whose XOR with N exceeding K
let cntPairs = 1;
// Traverse binary representation
// of N and K in Trie
for (let i = 31; i >= 0 &&
root!=null; i--)
{
// Stores ith bit of N
let x = N & (1 << i);
// Stores ith bit of K
let y = K & (1 << i);
// If the ith bit of K is 1
if (y == 1)
{
// Update root.
root = root.child[1 - x];
}
// If the ith bit of K is 0
else
{
// If an element already
// present in Trie having
// ith bit (1 - x)
if (x < 2 &&
root.child[1 - x] != null)
{
// Update cntPairs
cntPairs += root.child[1 - x].cnt;
}
// Update root.
if(x < 2)
root = root.child[x];
}
}
return cntPairs;
}
// Function to count pairs that
// satisfy the given conditions.
function cntGreaterPairs(arr,N,K)
{
// Create root node of Trie
let root = new TrieNode();
// Stores count of pairs that
// satisfy the given conditions
let cntPairs = 0;
// Traverse the given array.
for (let i = 0; i < N; i++)
{
// Update cntPairs
cntPairs += cntGreater(root,
arr[i], K);
// Insert arr[i] into Trie.
insertTrie(root, arr[i]);
}
return cntPairs;
}
// Driver code
let arr=[3, 5, 6, 8];
let K = 2;
let N = arr.length;
document.write(cntGreaterPairs(arr,N, K));
// This code is contributed by patel2127
</script>
Time Complexity:O(N * 32)
Auxiliary Space:O(N * 32)
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