<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) {
        vector<vector<int>> nxt;
        nxt.push_back(vector<int>(2));
        int N = 0;
        for (auto num: nums) {
            int node = 0;
            for(int i = 30; i >= 0; i--) {
                int digit = ((1ll<<i) & num) != 0;
                if (nxt[node][digit] == 0) {
                    nxt.push_back(vector<int>(2));
                    nxt[node][digit] = ++N;
                } 
                node = nxt[node][digit];
            }
        }

        int ans = 0;
        for (int num: nums) {
            int node = 0, curr = 0;
            for (int i = 30; i >= 0; i--) {
                int digit = ((1ll<<i) & num) != 0;
                if (nxt[node][!digit] != 0) {
                    curr |= (1<<i);
                    node = nxt[node][!digit];
                }
                else node = nxt[node][digit];
            }
            ans = max(ans, curr);
        }

        return ans;
    }
};
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