# 421. Maximum XOR of Two Numbers in an Array
## Problem Statement
Given an integer array `nums`, return *the maximum result of* `nums[i] XOR nums[j]`, where `0 <= i <= j < n`.
**Example 1:**
Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.
**Example 2:**
Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127
**Constraints:**
* `1 <= nums.length <= 2 * 105`
* `0 <= nums[i] <= 231 - 1`
## Intuition
```
Approach:
Insert all the elements in Trie,
Then, for each element again ,
check if opposite bits exist and move there
This way you generate answer
9 -> 1001
Insert from start-> Start bits are valuable for maximum answer
```
### Links
### Video Links
### Approach 1:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
struct Node{
Node* links[2];
};
class Trie{
public:
Node * root = new Node();
void insert(int num){
Node *temp = root;
for(int i=31; i>=0; i--){
int bit = (num>>i) & 1;
if(temp->links[bit] == nullptr)
temp->links[bit] = new Node();
temp = temp->links[bit];
}
}
int check(int num){
Node* temp = root;
int ans = 0;
for(int i=31; i>=0; i--){
int bit = (num>>i) & 1;
if(temp->links[1-bit] == nullptr){
temp = temp->links[bit];
}
else{
temp = temp->links[1-bit];
ans = ans | 1<& nums) {
Trie trie;
int ans = 0;
for(auto &it: nums)
trie.insert(it);
for(auto &it: nums)
ans = max(ans, trie.check(it));
return ans;
}
};
```
{% endcode %}
### Approach 2:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
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