<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 findMaximumXOR(vector<int>& nums) {
        //int N = 2e5 + 10, M = 31 * N;
        int son[31*(int)(2e5 + 10)][2] = {0}, idx = 0;
        for(int j = 0; j < nums.size(); j ++)
        {
            int p = 0;
            for(int i = 30; i >= 0; i --)
            {
                int u = nums[j] >> i & 1;
                if(!son[p][u])
                    son[p][u] = ++ idx;
                p = son[p][u];
            }
        }
        int res = 0;
        for(int i = 0; i < nums.size(); i ++)
        {
            int p = 0, t = 0;
            for(int j = 30; j >= 0; j --)
            {
                int u = nums[i] >> j & 1;
                if(son[p][!u])
                {
                    t = t * 2 + !u;
                    p = son[p][!u];
                }
                else
                {
                    t = t * 2 + u;
                    p = son[p][u];
                }
            }
            res = max(res, t ^ nums[i]);
        }
        return res;
    }
};
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