# Max XOR of Two Numbers in an Array

### Introduction

In computer science, **the maximum XOR of two integers in an array is a compute-intensive operation that is typically used in data mining and network traffic analysis. **The problem is to find the maximum value of the XOR operation applied to all pairs of integers in a given array.

The XOR operation is a binary operation that takes two operands and returns ‘**1**’ if the operands are different, and ‘**0’** if they are the same. For example, the XOR of 5 and 3 is 6 (5 XOR 3 = 0101 XOR 0011 = 0110 = 6).

The maximum XOR of two integers in an array **is a well-knownNP-hard problem**. However, there are efficient algorithms for solving it, such as the **bit-parallel algorithm.**

### What is the XOR Operator?

It is a binary operator that takes two operands and returns ‘1’ if the operands are different, and ‘0’ if they are the same. **The XOR operator is a useful tool as it can be used to encode data efficiently,** as well as to quickly calculate various arithmetic functions. XOR is associative, meaning that the order of the two operands does not matter when it comes to the result. This means that the XOR of two numbers is the same regardless of their order.

**The problem of finding the maximum XOR of two numbers in an array can be solved using a bitwise trie data structure. **The approach involves iterating over the bits of each number in the array, and for each bit, adding it to the trie.

To explain this approach, let's consider an example array of integers: [3, 10, 5, 25, 2, 8]. We can represent each integer in binary form and visualize them as a trie, where the root node is empty and each level represents a bit in the binary representation of the number

For each number in the array, we start at the root node and iterate over its bits, adding a new node to the trie for each bit. If the current bit is 0, we follow the left child, and if it is 1, we follow the right child.

After adding all the numbers to the trie, we can find the maximum XOR by iterating over the numbers again and for each number, finding the complementary number in the trie that will result in the maximum XOR. To do this, we start at the root node and iterate over the bits of the current number. At each bit, we check if the complementary bit exists in the trie. If it does, we follow that path, otherwise, we follow the opposite path.

** For example**: if the current number is 5 (binary representation: 101), we start at the root node and follow the path: 1->0->1. At each bit, we check if the complementary bit exists in the trie. If it does, we follow that path, otherwise, we follow the opposite path. In this case, we have complementary bits at the first and last level of the trie, so we follow the path 0->1->0, resulting in the complementary number 10 (binary representation: 1010).

We repeat this process for all the numbers in the array, keeping track of the maximum XOR we find. In this case, the maximum XOR is between 5 and 25 (101 XOR 11001 = 11000, or decimal 24).

### What is the maximum XOR of two numbers in an Array?

The maximum XOR of two numbers in an array is the maximum value of the XOR operation which is applied to all pairs of integers in that array. For example, the XOR of 5 and 3 is 6 (5 XOR 3 = 0101 XOR 0011 = 0110 = 6). The maximum XOR of two numbers in an array is useful for many tasks, such as *pattern matching, network traffic analysis, and data mining*. **It can also be used to find the maximum distance between any two points on a map.** The problem is somewhat difficult to solve, however, because it involves considering all possible combinations of two numbers from the array.** By XORing two numbers, we can reduce the size of the numbers and increase its data compression efficiency. **We can also use XOR for encryption, as it can hide data contained within a message. Finally, XOR can be used to improve the efficiency of sorting algorithms.

### How can we use an array to XOR two numbers?

It is possible to XOR two numbers by making use of an array of bits.

- Begin by initializing the array.
- Set all elements in the array to false (0).
- Then, replace each index in the array with the XOR of its corresponding bit in the two numbers. This can be done by looping through the two numbers, row by row. For example, if we have the numbers 11 and 12, then we can find the XOR of these two numbers by looping through their binary representation and replacing each position in the array accordingly.
- This process is being illustrated in the example below.

11 = 00001011 12 = 00001100 XOR = 00000111.

The **bit-parallel algorithm** is the most effective way to calculate the maximum XOR between two numbers from an array.

- The algorithm begins by breaking up the array into two halves, then computing the maximum XOR of two elements in each half.
- Finally, these two results are combined for a result that represents the maximum XOR for two of any number in the given array. This approach runs in
**O(n log n)**time.

**Algorithm and Code for Max XOR of Two Numbers in an Array**

Here is the algorithm and code implementation in C++ for finding the maximum XOR of two numbers in an array using the trie data structure.

**Algorithm**:

- Define a trie node
**struct**to represent each node of the trie. - Define a Trie class to represent the trie data structure.
- Create an instance of the Trie class.
- For each number in an array, insert its binary representation into the trie.
- For each number in an array, find its XOR value with all other numbers in the array by traversing the trie.
- Keep track of the maximum XOR value found so far.
- Return the maximum XOR value.

The** code implementation:**

```
#include <iostream>
#include <vector>
using namespace std;
struct TrieNode {
TrieNode* children[2];
TrieNode() {
children[0] = nullptr;
children[1] = nullptr;
}
};
class Trie {
TrieNode* root;
public:
Trie() {
root = new TrieNode();
}
void insert(int num) {
TrieNode* node = root;
for(int i=31; i>=0; i--) {
int bit = (num >> i) & 1;
if(node->children[bit] == nullptr) {
node->children[bit] = new TrieNode();
}
node = node->children[bit];
}
}
int findMaxXOR(int num) {
TrieNode* node = root;
int xor_val = 0;
for(int i=31; i>=0; i--) {
int bit = (num >> i) & 1;
if(node->children[1-bit] != nullptr) {
xor_val |= (1 << i);
node = node->children[1-bit];
}
else {
node = node->children[bit];
}
}
return xor_val;
}
};
int findMaxXOR(vector<int>& arr) {
Trie trie;
for(int num : arr) {
trie.insert(num);
}
int max_xor = 0;
for(int num : arr) {
max_xor = max(max_xor, trie.findMaxXOR(num));
}
return max_xor;
}
int main() {
vector<int> arr = {3, 10, 5, 25, 2, 8};
int max_xor = findMaxXOR(arr);
cout << "Maximum XOR value of two numbers in the array: " << max_xor << endl;
return 0;
}
```

**Output**

The code given above uses a vector to store the array of integers. The Trie Node struct has two children pointers to represent the two possible values of a binary digit (0 and 1). The Trie class has methods for inserting a number into the trie and finding the maximum XOR value of a number by traversing the trie. The main() function creates an instance of the Trie class, inserts the numbers from the array into the trie, and finds the maximum XOR value of two numbers in the array. And finally the code prints the max XOR value.

## Conclusion

The maximum XOR of two numbers in an array is an important compute-intensive operation that is typically used for data mining, pattern matching, and network traffic analysis. The problem is somewhat difficult to solve, but there are efficient algorithms available, such as the bit-parallel algorith**m. The algorithm works by recursively partitioning the array into two halves and computing the maximum XOR of two elements in each half. **Then, the results of the two halves are combined and the result is the maximum XOR of two numbers in the array. After finding the maximum XOR, we can use a variety of data structures to store the result. In conclusion, the maximum XOR of two numbers in an array is an important problem that can be solved efficiently with the help of algorithms such as the bit-parallel algorithm.