Maximum XOR of Two Numbers in an Array
<p>Given an integer array <code>nums</code>, return <em>the maximum result of </em><code>nums[i] XOR nums[j]</code>, where <code>0 &lt;= i &lt;= j &lt; n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,10,5,25,2,8]
<strong>Output:</strong> 28
<strong>Explanation:</strong> The maximum result is 5 XOR 25 = 28.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [14,70,53,83,49,91,36,80,92,51,66,70]
<strong>Output:</strong> 127
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
Bit Manipulation; Trie; Array; Hash Table
struct Trie {
    children: [Option<Box<Trie>>; 2],
}

impl Trie {
    fn new() -> Trie {
        Trie {
            children: [None, None],
        }
    }

    fn insert(&mut self, x: i32) {
        let mut node = self;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if node.children[v].is_none() {
                node.children[v] = Some(Box::new(Trie::new()));
            }
            node = node.children[v].as_mut().unwrap();
        }
    }

    fn search(&self, x: i32) -> i32 {
        let mut node = self;
        let mut ans = 0;
        for i in (0..=30).rev() {
            let v = ((x >> i) & 1) as usize;
            if let Some(child) = &node.children[v ^ 1] {
                ans |= 1 << i;
                node = child.as_ref();
            } else {
                node = node.children[v].as_ref().unwrap();
            }
        }
        ans
    }
}

impl Solution {
    pub fn find_maximum_xor(nums: Vec<i32>) -> i32 {
        let mut trie = Trie::new();
        let mut ans = 0;
        for &x in nums.iter() {
            trie.insert(x);
            ans = ans.max(trie.search(x));
        }
        ans
    }
}
