<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<pair<int, int>> nodes(2);
        int root = 1;
        int blen = std::snprintf(nullptr, 0, "%b", *max_element(nums.begin(), nums.end()));
        cout << "blen = " << blen << endl;
        auto get_child = [&](int node, bool left) {
            if (left) {
                if (!nodes[node].first) { nodes[node].first = nodes.size(); nodes.push_back({0, 0}); }
                return nodes[node].first;
            }
            if (!nodes[node].second) { nodes[node].second = nodes.size(); nodes.push_back({0, 0}); }
            return nodes[node].second;
        };
        for (int x : nums) {
            int node = root;
            for (int i = blen - 1; i >= 0; --i) {
                node = get_child(node, (x & (1<<i)) == 0);
            }
        }
        return search(nodes, root, root, 0);
    }

    int search(vector<pair<int, int>>& nodes, int left, int right, int mask) {
        if (!left || !right) return 0;
        if (nodes[left].first && nodes[right].second || nodes[left].second && nodes[right].first) {
            return max(search(nodes, nodes[left].first, nodes[right].second, (mask << 1) | 1), 
                       search(nodes, nodes[left].second, nodes[right].first, (mask << 1) | 1)); 
        }
        if (nodes[left].first || nodes[left].second || nodes[right].first || nodes[right].second) {
            return max(search(nodes, nodes[left].first, nodes[right].first, mask << 1), 
                       search(nodes, nodes[left].second, nodes[right].second, mask << 1));
        }
        return mask;
    }
};
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