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
type Trie struct {
	children [2]*Trie
}

func newTrie() *Trie {
	return &Trie{}
}

func (t *Trie) insert(x int) {
	node := t
	for i := 30; i >= 0; i-- {
		v := x >> i & 1
		if node.children[v] == nil {
			node.children[v] = newTrie()
		}
		node = node.children[v]
	}
}

func (t *Trie) search(x int) int {
	node := t
	ans := 0
	for i := 30; i >= 0; i-- {
		v := x >> i & 1
		if node.children[v^1] != nil {
			ans |= 1 << i
			node = node.children[v^1]
		} else {
			node = node.children[v]
		}
	}
	return ans
}

func findMaximumXOR(nums []int) (ans int) {
	trie := newTrie()
	for _, x := range nums {
		trie.insert(x)
		ans = max(ans, trie.search(x))
	}
	return ans
}