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
class Trie {
    private Trie[] children = new Trie[2];

    public Trie() {
    }

    public void insert(int x) {
        Trie node = this;
        for (int i = 30; i >= 0; --i) {
            int v = x >> i & 1;
            if (node.children[v] == null) {
                node.children[v] = new Trie();
            }
            node = node.children[v];
        }
    }

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

class Solution {
    public int findMaximumXOR(int[] nums) {
        Trie trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            trie.insert(x);
            ans = Math.max(ans, trie.search(x));
        }
        return ans;
    }
}