Maximum XOR of Two Numbers in an Array
Last Updated :
12 Jul, 2025
Given an array arr[] of non-negative integers. The task is to find the maximum possible XOR of any two elements of the array.
Example:
Input: arr[] = [26, 100, 25, 13, 4, 14]
Output: 126
Explanation: XOR of 26 ^ 100 = 126, which is the maximum possible XOR.
Input: arr[] = [1, 2, 3, 4, 5, 6, 7]
Output: 7
Explanation: XOR of 1 ^ 6 = 7, which is the maximum possible XOR.
[Naive Approach] Using 2 Nested Loops - O(n^2) Time and O(1) Space
The idea is to generate all possible pairs of elements of the array arr[] and find the maximum among them. To do so, use two nested loops to generate all the pairs and compute XOR of them. The maximum of all the pairs is the result.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum XOR
int maxXor(vector<int> &arr) {
int res = 0;
int fir = 0, sec = 0;
// Generate all possible pairs
for (int i = 0; i < arr.size(); i++) {
for (int j = i + 1; j < arr.size(); j++) {
if((arr[i]^arr[j]) > res){
res = max(res, arr[i] ^ arr[j]);
fir = arr[i], sec = arr[j];
}
}
}
return res;
}
int main() {
vector<int> arr = {26, 100, 25, 13, 4, 14};
cout << maxXor(arr);
return 0;
}
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0;
// Generate all possible pairs
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
res = Math.max(res, arr[i] ^ arr[j]);
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
System.out.println(maxXor(arr));
}
}
# Function to find the maximum XOR
def maxXor(arr):
res = 0
# Generate all possible pairs
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
res = max(res, arr[i] ^ arr[j])
return res
if __name__ == "__main__":
arr = [26, 100, 25, 13, 4, 14]
print(maxXor(arr))
using System;
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0;
// Generate all possible pairs
for (int i = 0; i < arr.Length; i++) {
for (int j = i + 1; j < arr.Length; j++) {
res = Math.Max(res, arr[i] ^ arr[j]);
}
}
return res;
}
static void Main(string[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
Console.WriteLine(maxXor(arr));
}
}
// Function to find the maximum XOR
function maxXor(arr) {
let res = 0;
// Generate all possible pairs
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
res = Math.max(res, arr[i] ^ arr[j]);
}
}
return res;
}
let arr = [26, 100, 25, 13, 4, 14];
console.log(maxXor(arr));
[Expected Approach - 1] - Using Bit Masking and HashSet - O(n * log m) Time and O(n * log m) Space
The idea is to find two numbers in an array arr[] such that their XOR equals a number X, where X is the maximum value we want to achieve at the current bit position (i-th bit). To achieve the largest XOR value, we aim to maximize the number of 1s in the XOR result, starting from the most significant bit (leftmost bit) to the least significant bit (rightmost bit).
To evaluate each bit position, we use a mask. A mask helps us focus on specific bits of the numbers in the array by keeping only the relevant bits up to the current position and ignoring the rest. Using the mask, we extract prefixes for all numbers in the array (i.e., the portions of the numbers defined by the mask). These prefixes help us determine if a pair of numbers exists in the array whose XOR can yield a maximum value at the current bit.
For each bit position:
- Apply the mask to extract prefixes from the numbers.
- Try to update the maximum XOR value by assuming the i-th bit of the result is 1.
- Using the set of prefixes, we check if any two prefixes can produce the desired XOR value (with the i-th bit set).
This process is repeated for all 32 bits, starting from the leftmost bit, to compute the largest possible XOR value step by step.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum XOR
int maxXor(vector<int> &arr) {
int res = 0, mask = 0;
// to store all unique bits
unordered_set<int> s;
for (int i = 30; i >= 0; i--) {
// set the i-th bit in mask
mask |= (1 << i);
for (auto value: arr) {
// keep prefix of all elements
// till the i-th bit
s.insert(value & mask);
}
int cur = res | (1 << i);
for (int prefix : s) {
if (s.count(cur ^ prefix)) {
res = cur;
break;
}
}
s.clear();
}
return res;
}
int main() {
vector<int> arr = {26, 100, 25, 13, 4, 14};
cout << maxXor(arr);
return 0;
}
import java.util.HashSet;
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0, mask = 0;
// to store all unique bits
HashSet<Integer> s = new HashSet<>();
for (int i = 30; i >= 0; i--) {
// set the i-th bit in mask
mask |= (1 << i);
for (int value : arr) {
// keep prefix of all elements
// till the i-th bit
s.add(value & mask);
}
int cur = res | (1 << i);
for (int prefix : s) {
if (s.contains(cur ^ prefix)) {
res = cur;
break;
}
}
s.clear();
}
return res;
}
public static void main(String[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
System.out.println(maxXor(arr));
}
}
# Function to find the maximum XOR
def maxXor(arr):
res = 0
mask = 0
# to store all unique bits
s = set()
for i in range(30, -1, -1):
# set the i-th bit in mask
mask |= (1 << i)
for num in arr:
# keep prefix of all elements
# till the i-th bit
s.add(num & mask)
cur = res | (1 << i)
for prefix in s:
if cur ^ prefix in s:
res = cur
break
s.clear()
return res
if __name__ == "__main__":
arr = [26, 100, 25, 13, 4, 14]
print(maxXor(arr))
using System;
using System.Collections.Generic;
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0, mask = 0;
// to store all unique bits
HashSet<int> s = new HashSet<int>();
for (int i = 30; i >= 0; i--) {
// set the i-th bit in mask
mask |= (1 << i);
foreach (int value in arr) {
// keep prefix of all elements
// till the i-th bit
s.Add(value & mask);
}
int cur = res | (1 << i);
foreach (int prefix in s) {
if (s.Contains(cur ^ prefix)) {
res = cur;
break;
}
}
s.Clear();
}
return res;
}
static void Main(string[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
Console.WriteLine(maxXor(arr));
}
}
// Function to find the maximum XOR
function maxXor(arr) {
let res = 0, mask = 0;
// to store all unique bits
let s = new Set();
for (let i = 30; i >= 0; i--) {
// set the i-th bit in mask
mask |= (1 << i);
for (let num of arr) {
// keep prefix of all elements
// till the i-th bit
s.add(num & mask);
}
let cur = res | (1 << i);
for (let prefix of s) {
if (s.has(cur ^ prefix)) {
res = cur;
break;
}
}
s.clear();
}
return res;
}
let arr = [26, 100, 25, 13, 4, 14];
console.log(maxXor(arr));
Time Complexity: O(n * log m), where n is the size of array and m is the maximum element.
Auxiliary Space: O(n * log m)
[Expected Approach - 2] - Using Trie - O(n * log m) Time and O(n * log m) Space
The idea is to use Trie data structure to effectively store and search bits of each element of array arr[]. For each element arr[i], check if an element with opposite bits is present in the Trie or not i.e. if current bit of arr[i] is set (1), then check if unset bit (0) at current index is present in Trie.
Note: Please refer to Trie article to know more about insert and search operation.
C++
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
Node* one;
Node* zero;
Node() {
one = nullptr;
zero = nullptr;
}
};
class Trie {
public:
Node* root;
Trie() {
root = new Node();
}
// Function to insert in Trie
void insert(int n) {
Node* curr = root;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// Check if the bit is 0
if (bit == 0) {
if (curr->zero == NULL) {
curr->zero = new Node();
}
curr = curr->zero;
}
// Else if bit is 1
else {
if (curr->one == NULL) {
curr->one = new Node();
}
curr = curr->one;
}
}
}
// Function to find element having
// the maximum XOR with value n
int findXOR(int n) {
Node* curr = root;
int res = 0;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// if the bit is 0
if (bit == 0) {
// if set bit is present
if (curr->one) {
curr = curr->one;
res += (1 << i);
}
else {
curr = curr->zero;
}
}
// Else if bit is 1
else {
// if unset bit is present
if (curr->zero) {
curr = curr->zero;
res += (1 << i);
}
else {
curr = curr->one;
}
}
}
return res;
}
};
// Function to find the maximum XOR
int maxXor(vector<int> &arr) {
int res = 0;
Trie *t = new Trie();
// insert the first element in trie
t->insert(arr[0]);
for (int i = 1; i < arr.size(); i++) {
res = max(t->findXOR(arr[i]), res);
t->insert(arr[i]);
}
return res;
}
int main() {
vector<int> arr = {26, 100, 25, 13, 4, 14};
cout << maxXor(arr);
return 0;
}
// Java program to find the maximum
// XOR n of two elements in an array
// using Trie data structure
import java.util.*;
class Node {
Node one;
Node zero;
Node() {
one = null;
zero = null;
}
}
class Trie {
Node root;
Trie() {
root = new Node();
}
// Function to insert in Trie
void insert(int n) {
Node curr = root;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// Check if the bit is 0
if (bit == 0) {
if (curr.zero == null) {
curr.zero = new Node();
}
curr = curr.zero;
}
// Else if bit is 1
else {
if (curr.one == null) {
curr.one = new Node();
}
curr = curr.one;
}
}
}
// Function to find element having
// the maximum XOR with value n
int findXOR(int n) {
Node curr = root;
int res = 0;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// if the bit is 0
if (bit == 0) {
// if set bit is present
if (curr.one != null) {
curr = curr.one;
res += (1 << i);
}
else {
curr = curr.zero;
}
}
// Else if bit is 1
else {
// if unset bit is present
if (curr.zero != null) {
curr = curr.zero;
res += (1 << i);
}
else {
curr = curr.one;
}
}
}
return res;
}
}
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0;
Trie t = new Trie();
// insert the first element in trie
t.insert(arr[0]);
for (int i = 1; i < arr.length; i++) {
res = Math.max(t.findXOR(arr[i]), res);
t.insert(arr[i]);
}
return res;
}
public static void main(String[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
System.out.println(maxXor(arr));
}
}
# Python program to find the maximum
# XOR n of two elements in an array
# using Trie data structure
class Node:
def __init__(self):
self.one = None
self.zero = None
class Trie:
def __init__(self):
self.root = Node()
# Function to insert in Trie
def insert(self, n):
curr = self.root
for i in range(31, -1, -1):
bit = (n >> i) & 1
# Check if the bit is 0
if bit == 0:
if not curr.zero:
curr.zero = Node()
curr = curr.zero
# Else if bit is 1
else:
if not curr.one:
curr.one = Node()
curr = curr.one
# Function to find element having
# the maximum XOR with value n
def findXOR(self, n):
curr = self.root
res = 0
for i in range(31, -1, -1):
bit = (n >> i) & 1
# if the bit is 0
if bit == 0:
# if set bit is present
if curr.one:
curr = curr.one
res += (1 << i)
else:
curr = curr.zero
# Else if bit is 1
else:
# if unset bit is present
if curr.zero:
curr = curr.zero
res += (1 << i)
else:
curr = curr.one
return res
# Function to find the maximum XOR
def maxXor(arr):
res = 0
t = Trie()
# insert the first element in trie
t.insert(arr[0])
for i in range(1, len(arr)):
res = max(t.findXOR(arr[i]), res)
t.insert(arr[i])
return res
if __name__ == "__main__":
arr = [26, 100, 25, 13, 4, 14]
print(maxXor(arr))
// C# program to find the maximum
// XOR n of two elements in an array
// using Trie data structure
using System;
class Node {
public Node one;
public Node zero;
public Node() {
one = null;
zero = null;
}
}
class Trie {
public Node root;
public Trie() {
root = new Node();
}
// Function to insert in Trie
public void insert(int n) {
Node curr = root;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// Check if the bit is 0
if (bit == 0) {
if (curr.zero == null) {
curr.zero = new Node();
}
curr = curr.zero;
}
// Else if bit is 1
else {
if (curr.one == null) {
curr.one = new Node();
}
curr = curr.one;
}
}
}
// Function to find element having
// the maximum XOR with value n
public int findXOR(int n) {
Node curr = root;
int res = 0;
for (int i = 31; i >= 0; i--) {
int bit = (n >> i) & 1;
// if the bit is 0
if (bit == 0) {
// if set bit is present
if (curr.one != null) {
curr = curr.one;
res += (1 << i);
}
else {
curr = curr.zero;
}
}
// Else if bit is 1
else {
// if unset bit is present
if (curr.zero != null) {
curr = curr.zero;
res += (1 << i);
}
else {
curr = curr.one;
}
}
}
return res;
}
}
class GfG {
// Function to find the maximum XOR
static int maxXor(int[] arr) {
int res = 0;
Trie t = new Trie();
// insert the first element in trie
t.insert(arr[0]);
for (int i = 1; i < arr.Length; i++) {
res = Math.Max(t.findXOR(arr[i]), res);
t.insert(arr[i]);
}
return res;
}
static void Main(string[] args) {
int[] arr = {26, 100, 25, 13, 4, 14};
Console.WriteLine(maxXor(arr));
}
}
// JavaScript program to find the maximum
// XOR n of two elements in an array
// using Trie data structure
class Node {
constructor() {
this.one = null;
this.zero = null;
}
}
class Trie {
constructor() {
this.root = new Node();
}
// Function to insert in Trie
insert(n) {
let curr = this.root;
for (let i = 31; i >= 0; i--) {
let bit = (n >> i) & 1;
// Check if the bit is 0
if (bit === 0) {
if (!curr.zero) {
curr.zero = new Node();
}
curr = curr.zero;
}
// Else if bit is 1
else {
if (!curr.one) {
curr.one = new Node();
}
curr = curr.one;
}
}
}
// Function to find element having
// the maximum XOR with value n
findXOR(n) {
let curr = this.root;
let res = 0;
for (let i = 31; i >= 0; i--) {
let bit = (n >> i) & 1;
// if the bit is 0
if (bit === 0) {
// if set bit is present
if (curr.one) {
curr = curr.one;
res += (1 << i);
}
else {
curr = curr.zero;
}
}
// Else if bit is 1
else {
// if unset bit is present
if (curr.zero) {
curr = curr.zero;
res += (1 << i);
}
else {
curr = curr.one;
}
}
}
return res;
}
}
// Function to find the maximum XOR
function maxXOR(arr) {
let res = 0;
let t = new Trie();
// insert the first element in trie
t.insert(arr[0]);
for (let i = 1; i < arr.length; i++) {
res = Math.max(t.findXOR(arr[i]), res);
t.insert(arr[i]);
}
return res;
}
let arr = [26, 100, 25, 13, 4, 14];
console.log(maxXOR(arr));
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