<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>

class Solution {
public:
    int cur_node = 0;
    struct Node {
        int child[2] = {0};
    } Trie[200005 * 32];

    void add(int x) {
        int cur = 0;
        for (int i = 30; i >= 0; i--) {
            int bit = (x >> i) & 1;
            if (!Trie[cur].child[bit]) {
                Trie[cur].child[bit] = ++cur_node;
            }
            cur = Trie[cur].child[bit];
        }
    }

    int find_max(int x) {
        int cur = 0;
        int ans = 0;
        for (int i = 30; i >= 0; i--) {
            int bit = (x >> i) & 1;
            if (Trie[cur].child[1 - bit]) {
                ans += (1 << i);
                cur = Trie[cur].child[1 - bit];
            } else {
                cur = Trie[cur].child[bit];
            }
        }
        return ans;
    }

    int findMaximumXOR(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            add(nums[i]);
            ans = max(ans, find_max(nums[i]));
        }
        return ans;
    }
};
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