# Maximum XOR of Two Numbers in an Array

Given an array Arr of non-negative integers of size N. The task is to find the maximum possible xor between two numbers present in the array.
Example

Input: Arr = {25, 10, 2, 8, 5, 3}
Output: 28
Explanation: The maximum result is 5 ^ 25 = 28

Input: Arr = {1, 2, 3, 4, 5, 6, 7}
Output:
Explanation: The maximum result is 1 ^ 6 = 7

Naive Approach: A Simple Solution is to generate all pairs of the given array and compute the XOR of the pairs. Finally, return the maximum XOR value. This solution takes time.
Below is the implementation of the above approach:

## C++

 `// C++ implementation` `#include ` `using` `namespace` `std;`   `// Function to return the` `// maximum xor` `int` `max_xor(``int` `arr[], ``int` `n)` `{`   `    ``int` `maxXor = 0;`   `    ``// Calculating xor of each pair` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``maxXor = max(maxXor,` `                         ``arr[i] ^ arr[j]);` `        ``}` `    ``}` `    ``return` `maxXor;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `arr[] = { 25, 10, 2, 8, 5, 3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << max_xor(arr, n) << endl;` `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `class` `GFG ` `{`   `    ``// Function to return the` `    ``// maximum xor` `    ``static` `int` `max_xor(``int` `arr[], ``int` `n) ` `    ``{` `        ``int` `maxXor = ``0``;`   `        ``// Calculating xor of each pair` `        ``for` `(``int` `i = ``0``; i < n; i++)` `        ``{` `            ``for` `(``int` `j = i + ``1``; j < n; j++) ` `            ``{` `                ``maxXor = Math.max(maxXor,` `                        ``arr[i] ^ arr[j]);` `            ``}` `        ``}` `        ``return` `maxXor;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `arr[] = {``25``, ``10``, ``2``, ``8``, ``5``, ``3``};` `        ``int` `n = arr.length;`   `        ``System.out.println(max_xor(arr, n));` `    ``}` `} `   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation`   `# Function to return the` `# maximum xor` `def` `max_xor(arr, n):`   `    ``maxXor ``=` `0``;`   `    ``# Calculating xor of each pair` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(i ``+` `1``, n):` `            ``maxXor ``=` `max``(maxXor,\` `                         ``arr[i] ^ arr[j]);`   `    ``return` `maxXor;`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``arr ``=` `[ ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `];` `    ``n ``=` `len``(arr);`   `    ``print``(max_xor(arr, n));`   `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG ` `{`   `// Function to return the` `// maximum xor` `static` `int` `max_xor(``int` `[]arr, ``int` `n) ` `{` `    ``int` `maxXor = 0;`   `    ``// Calculating xor of each pair` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `        ``{` `            ``maxXor = Math.Max(maxXor,` `                              ``arr[i] ^ arr[j]);` `        ``}` `    ``}` `    ``return` `maxXor;` `}`   `// Driver Code` `public` `static` `void` `Main() ` `{` `    ``int` `[]arr = {25, 10, 2, 8, 5, 3};` `    ``int` `n = arr.Length;`   `    ``Console.WriteLine(max_xor(arr, n));` `}` `}`   `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output

`28`

Time Complexity: O(N^{2})           , where N is the size of the array
Auxiliary Space: O(1)
Efficient Approach: The approach is similar to this article where the task is to find the maximum AND value pair
So the idea is to change the problem statement from finding the maximum xor of two numbers in an array to -> find two numbers in an array, such that xor of which equals to a number X. In this case, X will be the maximum number we want to achieve till i-th bit.
To find the largest value of an XOR operation, the value of xor should have every bit to be a set bit i.e 1. In a 32-bit number, the goal is to get the most 1 set starting left to right.
To evaluate each bit, there is a mask needed for that bit. A mask defines which bit should be present in the answer and which bit is not. Here we will use a mask to keep the prefix for every number ( means by taking the ans with the mask how many bits are remaining from the number ) in the input till the i-th bit then with the list of possible numbers in our set, after inserting the number we will evaluate if we can update the max for that bit position to be 1.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach`   `#include ` `using` `namespace` `std;`   `// Function to return the` `// maximum xor` `int` `max_xor(``int` `arr[], ``int` `n)` `{` `    ``int` `maxx = 0, mask = 0;`   `    ``set<``int``> se;`   `    ``for` `(``int` `i = 30; i >= 0; i--) {`   `        ``// set the i'th bit in mask` `        ``// like 100000, 110000, 111000..` `        ``mask |= (1 << i);`   `        ``for` `(``int` `i = 0; i < n; ++i) {`   `            ``// Just keep the prefix till` `            ``// i'th bit neglecting all` `            ``// the bit's after i'th bit` `            ``se.insert(arr[i] & mask);` `        ``}`   `        ``int` `newMaxx = maxx | (1 << i);`   `        ``for` `(``int` `prefix : se) {`   `            ``// find two pair in set` `            ``// such that a^b = newMaxx` `            ``// which is the highest` `            ``// possible bit can be obtained` `            ``if` `(se.count(newMaxx ^ prefix)) {` `                ``maxx = newMaxx;` `                ``break``;` `            ``}` `        ``}`   `        ``// clear the set for next` `        ``// iteration` `        ``se.clear();` `    ``}`   `    ``return` `maxx;` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `arr[] = { 25, 10, 2, 8, 5, 3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << max_xor(arr, n) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the above approach` `import` `java.util.*;` `class` `GFG` `{`   `// Function to return the` `// maximum xor` `static` `int` `max_xor(``int` `arr[], ``int` `n)` `{` `    ``int` `maxx = ``0``, mask = ``0``;`   `    ``HashSet se = ``new` `HashSet();`   `    ``for` `(``int` `i = ``30``; i >= ``0``; i--) ` `    ``{`   `        ``// set the i'th bit in mask` `        ``// like 100000, 110000, 111000..` `        ``mask |= (``1` `<< i);`   `        ``for` `(``int` `j = ``0``; j < n; ++j) ` `        ``{`   `            ``// Just keep the prefix till` `            ``// i'th bit neglecting all` `            ``// the bit's after i'th bit` `            ``se.add(arr[j] & mask);` `        ``}`   `        ``int` `newMaxx = maxx | (``1` `<< i);`   `        ``for` `(``int` `prefix : se)` `        ``{`   `            ``// find two pair in set` `            ``// such that a^b = newMaxx` `            ``// which is the highest` `            ``// possible bit can be obtained` `            ``if` `(se.contains(newMaxx ^ prefix))` `            ``{` `                ``maxx = newMaxx;` `                ``break``;` `            ``}` `        ``}`   `        ``// clear the set for next` `        ``// iteration` `        ``se.clear();` `    ``}` `    ``return` `maxx;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `arr[] = { ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `};` `    ``int` `n = arr.length;`   `    ``System.out.println(max_xor(arr, n));` `}` `} `   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation of the above approach `   `# Function to return the ` `# maximum xor ` `def` `max_xor( arr , n):` `    `  `    ``maxx ``=` `0` `    ``mask ``=` `0``; `   `    ``se ``=` `set``()` `    `  `    ``for` `i ``in` `range``(``30``, ``-``1``, ``-``1``):` `        `  `        ``# set the i'th bit in mask ` `        ``# like 100000, 110000, 111000..` `        ``mask |``=` `(``1` `<< i)` `        ``newMaxx ``=` `maxx | (``1` `<< i)` `    `  `        ``for` `i ``in` `range``(n):` `            `  `            ``# Just keep the prefix till ` `            ``# i'th bit neglecting all ` `            ``# the bit's after i'th bit ` `            ``se.add(arr[i] & mask)`   `        ``for` `prefix ``in` `se:` `            `  `            ``# find two pair in set ` `            ``# such that a^b = newMaxx ` `            ``# which is the highest ` `            ``# possible bit can be obtained` `            ``if` `(newMaxx ^ prefix) ``in` `se:` `                ``maxx ``=` `newMaxx` `                ``break` `                `  `        ``# clear the set for next ` `        ``# iteration ` `        ``se.clear()` `    ``return` `maxx`   `# Driver Code ` `arr ``=` `[ ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `]` `n ``=` `len``(arr)` `print``(max_xor(arr, n)) `   `# This code is contributed by ANKITKUMAR34`

## C#

 `// C# implementation of the above approach` `using` `System;` `using` `System.Collections.Generic;` `    `  `class` `GFG` `{`   `// Function to return the` `// maximum xor` `static` `int` `max_xor(``int` `[]arr, ``int` `n)` `{` `    ``int` `maxx = 0, mask = 0;`   `    ``HashSet<``int``> se = ``new` `HashSet<``int``>();`   `    ``for` `(``int` `i = 30; i >= 0; i--) ` `    ``{`   `        ``// set the i'th bit in mask` `        ``// like 100000, 110000, 111000..` `        ``mask |= (1 << i);`   `        ``for` `(``int` `j = 0; j < n; ++j) ` `        ``{`   `            ``// Just keep the prefix till` `            ``// i'th bit neglecting all` `            ``// the bit's after i'th bit` `            ``se.Add(arr[j] & mask);` `        ``}`   `        ``int` `newMaxx = maxx | (1 << i);`   `        ``foreach` `(``int` `prefix ``in` `se)` `        ``{`   `            ``// find two pair in set` `            ``// such that a^b = newMaxx` `            ``// which is the highest` `            ``// possible bit can be obtained` `            ``if` `(se.Contains(newMaxx ^ prefix))` `            ``{` `                ``maxx = newMaxx;` `                ``break``;` `            ``}` `        ``}`   `        ``// clear the set for next` `        ``// iteration` `        ``se.Clear();` `    ``}` `    ``return` `maxx;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `[]arr = { 25, 10, 2, 8, 5, 3 };` `    ``int` `n = arr.Length;`   `    ``Console.WriteLine(max_xor(arr, n));` `}` `}`   `// This code is contributed by Princi Singh`

## Javascript

 ``

Output

`28`

Time Complexity: , where N is the size of the array and M is the maximum number present in the array
Auxiliary Space: O(logM)

Better Approach: Another approach would be to use a Trie structure to store the bit representation of the numbers and for N terms compute the maximum XOR each can produce by going through the Trie.

## C++

 `#include `   `using` `namespace` `std;`   `class` `Node {` `public``:` `    ``Node* one;` `    ``Node* zero;` `};`   `class` `trie {` `    ``Node* root;`   `public``:` `    ``trie() { root = ``new` `Node(); }`   `    ``void` `insert(``int` `n)` `    ``{` `        ``Node* temp = root;` `        ``for` `(``int` `i = 31; i >= 0; i--) {` `            ``int` `bit = (n >> i) & 1;` `            ``if` `(bit == 0) {` `                ``if` `(temp->zero == NULL) {` `                    ``temp->zero = ``new` `Node();` `                ``}` `                ``temp = temp->zero;` `            ``}` `            ``else` `{` `                ``if` `(temp->one == NULL) {` `                    ``temp->one = ``new` `Node();` `                ``}` `                ``temp = temp->one;` `            ``}` `        ``}` `    ``}`   `    ``int` `max_xor_helper(``int` `value)` `    ``{` `        ``Node* temp = root;` `        ``int` `current_ans = 0;`   `        ``for` `(``int` `i = 31; i >= 0; i--) {` `            ``int` `bit = (value >> i) & 1;` `            ``if` `(bit == 0) {` `                ``if` `(temp->one) {` `                    ``temp = temp->one;` `                    ``current_ans += (1 << i);` `                ``}` `                ``else` `{` `                    ``temp = temp->zero;` `                ``}` `            ``}` `            ``else` `{` `                ``if` `(temp->zero) {` `                    ``temp = temp->zero;` `                    ``current_ans += (1 << i);` `                ``}` `                ``else` `{` `                    ``temp = temp->one;` `                ``}` `            ``}` `        ``}` `        ``return` `current_ans;` `    ``}`   `    ``int` `max_xor(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `max_val = 0;` `        ``insert(arr[0]);` `        ``for` `(``int` `i = 1; i < n; i++) {` `            ``max_val = max(max_xor_helper(arr[i]), max_val);` `            ``insert(arr[i]);` `        ``}` `        ``return` `max_val;` `    ``}` `};`   `int` `main()` `{` `    ``int` `input[] = { 25, 10, 2, 8, 5, 3 };` `    ``int` `n = ``sizeof``(input) / ``sizeof``(``int``);` `    ``trie t;` `    ``cout << t.max_xor(input, n);`   `    ``return` `0;` `}`

## Java

 `// Java code for the above approach` `import` `java.io.*;`   `class` `GFG {`   `  ``static` `class` `Node {` `    ``public` `Node one, zero;` `  ``}`   `  ``static` `class` `trie {` `    ``Node root;` `    ``public` `trie() { root = ``new` `Node(); }`   `    ``public` `void` `insert(``int` `n)` `    ``{` `      ``Node temp = root;` `      ``for` `(``int` `i = ``31``; i >= ``0``; i--) {` `        ``int` `bit = (n >> i) & ``1``;` `        ``if` `(bit == ``0``) {` `          ``if` `(temp.zero == ``null``) {` `            ``temp.zero = ``new` `Node();` `          ``}` `          ``temp = temp.zero;` `        ``}` `        ``else` `{` `          ``if` `(temp.one == ``null``) {` `            ``temp.one = ``new` `Node();` `          ``}` `          ``temp = temp.one;` `        ``}` `      ``}` `    ``}`   `    ``public` `int` `max_xor_helper(``int` `value)` `    ``{` `      ``Node temp = root;` `      ``int` `current_ans = ``0``;` `      ``for` `(``int` `i = ``31``; i >= ``0``; i--) {` `        ``int` `bit = (value >> i) & ``1``;` `        ``if` `(bit == ``0``) {` `          ``if` `(temp.one != ``null``) {` `            ``temp = temp.one;` `            ``current_ans += (``1` `<< i);` `          ``}` `          ``else` `{` `            ``temp = temp.zero;` `          ``}` `        ``}` `        ``else` `{` `          ``if` `(temp.zero != ``null``) {` `            ``temp = temp.zero;` `            ``current_ans += (``1` `<< i);` `          ``}` `          ``else` `{` `            ``temp = temp.one;` `          ``}` `        ``}` `      ``}` `      ``return` `current_ans;` `    ``}`   `    ``public` `int` `max_xor(``int``[] arr, ``int` `n)` `    ``{` `      ``int` `max_val = ``0``;` `      ``insert(arr[``0``]);` `      ``for` `(``int` `i = ``1``; i < n; i++) {` `        ``max_val = Math.max(max_xor_helper(arr[i]),` `                           ``max_val);` `        ``insert(arr[i]);` `      ``}` `      ``return` `max_val;` `    ``}` `  ``}`   `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[] input = { ``25``, ``10``, ``2``, ``8``, ``5``, ``3` `};` `    ``int` `n = input.length;` `    ``trie t = ``new` `trie();` `    ``System.out.print(t.max_xor(input, n));` `  ``}` `}`   `// This code is contributed by lokeshmvs21.`

## Python3

 `class` `TrieNode :` `    ``def` `__init__(``self``):` `        ``self``.children ``=` `{}`   `class` `Trie :` `    ``def` `__init__(``self``) :` `        ``self``.root ``=` `TrieNode()`   `    ``def` `insert(``self``, n) :` `        ``temp ``=` `self``.root` `        ``i ``=` `31` `        ``while` `i >``=` `0` `:` `            ``bit ``=` `(n >> i) & ``1` `            ``if` `bit ``not` `in` `temp.children: ` `                ``temp.children[bit] ``=` `TrieNode()` `            ``temp ``=` `temp.children[bit]` `            ``i ``-``=` `1`   `    ``def` `max_xor_helper(``self``, value) :` `        ``temp ``=` `self``.root` `        ``current_ans ``=` `0` `        ``i ``=` `31` `        ``while` `i >``=` `0``:` `            ``bit ``=` `(value >> i) & ``1` `            ``if` `bit^``1` `in` `temp.children:` `                ``temp ``=` `temp.children[bit^``1``]` `                ``current_ans ``+``=` `(``1` `<< i)` `            ``else``:` `                ``temp ``=` `temp.children[bit]` `            ``i ``-``=` `1` `        ``return` `current_ans`   `class` `Solution:` `    ``def` `solve(``self``, A):` `        ``trie ``=` `Trie()` `        ``max_val ``=` `0` `        ``trie.insert(A[``0``])` `        ``for` `n ``in` `A[``1``:]:` `            ``max_val ``=` `max``(trie.max_xor_helper(n),max_val)` `            ``trie.insert(n)` `        ``return` `max_val`   `if` `__name__``=``=``"__main__"``:` `  ``A ``=` `[``25``, ``10``, ``2``, ``8``, ``5``, ``3``]` `  ``print``(Solution().solve(A))`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG {`   `  ``public` `class` `Node {` `    ``public` `Node one;` `    ``public` `Node zero;` `  ``};`   `  ``public` `class` `trie {` `    ``public` `Node root;` `    ``public` `trie() { root = ``new` `Node(); }`   `    ``public` `void` `insert(``int` `n)` `    ``{` `      ``Node temp = root;` `      ``for` `(``int` `i = 31; i >= 0; i--) {` `        ``int` `bit = (n >> i) & 1;` `        ``if` `(bit == 0) {` `          ``if` `(temp.zero == ``null``) {` `            ``temp.zero = ``new` `Node();` `          ``}` `          ``temp = temp.zero;` `        ``}` `        ``else` `{` `          ``if` `(temp.one == ``null``) {` `            ``temp.one = ``new` `Node();` `          ``}` `          ``temp = temp.one;` `        ``}` `      ``}` `    ``}`   `    ``public` `int` `max_xor_helper(``int` `value)` `    ``{` `      ``Node temp = root;` `      ``int` `current_ans = 0;`   `      ``for` `(``int` `i = 31; i >= 0; i--) {` `        ``int` `bit = (value >> i) & 1;` `        ``if` `(bit == 0) {` `          ``if` `(temp.one != ``null``) {` `            ``temp = temp.one;` `            ``current_ans += (1 << i);` `          ``}` `          ``else` `{` `            ``temp = temp.zero;` `          ``}` `        ``}` `        ``else` `{` `          ``if` `(temp.zero != ``null``) {` `            ``temp = temp.zero;` `            ``current_ans += (1 << i);` `          ``}` `          ``else` `{` `            ``temp = temp.one;` `          ``}` `        ``}` `      ``}` `      ``return` `current_ans;` `    ``}`   `    ``public` `int` `max_xor(``int``[] arr, ``int` `n)` `    ``{` `      ``int` `max_val = 0;` `      ``insert(arr[0]);` `      ``for` `(``int` `i = 1; i < n; i++) {` `        ``max_val = Math.Max(max_xor_helper(arr[i]),` `                           ``max_val);` `        ``insert(arr[i]);` `      ``}` `      ``return` `max_val;` `    ``}` `  ``};`   `  ``static` `public` `void` `Main()` `  ``{` `    ``int``[] input = { 25, 10, 2, 8, 5, 3 };` `    ``int` `n = input.Length;` `    ``trie t = ``new` `trie();` `    ``Console.WriteLine(t.max_xor(input, n));` `  ``}` `}`   `// This code is contributed by akashish__`

## Javascript

 `class TrieNode{` `    ``constructor(){` `        ``this``.children ={}` `    ``}` `}`   `class Trie{` `    ``constructor(){` `        ``this``.root = ``new` `TrieNode();` `    ``}` `    ``insert(n){` `        ``let temp = ``this``.root;` `        ``let i = 31;` `        ``while``(i >= 0){` `            ``const bit = (n >> i) & 1;` `            ``if` `(!temp.children[bit]) temp.children[bit] = ``new` `TrieNode()` `            ``temp = temp.children[bit]` `            ``i -= 1` `        ``}` `    ``}` `    ``max_xor_helper(value){` `        ``let temp = ``this``.root;` `        ``let current_ans = 0;` `        ``let i = 31;` `        ``while` `(i >= 0){` `            ``const bit = (value >> i) & 1;` `            ``if` `(temp.children[bit^1]){` `                ``temp = temp.children[bit^1]` `                ``current_ans += (1 << i)` `            ``}` `            ``else` `temp = temp.children[bit]` `            ``i -= 1` `        ``}` `        ``return` `current_ans` `    ``}` `}`   `function` `solve(A){` `    ``let trie = ``new` `Trie();` `    ``let max_val = 0;` `    ``trie.insert(A[0])` `    ``let i = 1;` `    ``const n = A.length;` `    ``while``(i < n){` `        ``max_val = Math.max(trie.max_xor_helper(A[i]),max_val);` `        ``trie.insert(A[i]);` `        ``i += 1;` `    ``}` `    ``return` `max_val` `}`   `const A = [25, 10, 2, 8, 5, 3]` `console.log(solve(A))`

Output

`28`

Complexity analysis:

The time complexity of inserting a single element into the trie is O( log(maximum element in array) ). Thus, for n elements, the time complexity of building the trie is O( n log(maximum element in array) ).

The time complexity of finding the maximum XOR for a single element in the trie is also O( log(maximum element in array) ). Thus, for n elements, the time complexity of finding the maximum XOR is O( n log(maximum element in array) ).

The space complexity of the trie is O( n log(maximum element in array) ). This is because in the worst case scenario, every bit of every element in the array needs to be stored in the trie. Since the maximum number of bits in an integer is log(maximum element in array), the space complexity is O( n log(maximum element in array) ).

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