<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 TrieNode {
public:
    TrieNode* children[2];
    TrieNode() {
        children[0] = nullptr;
        children[1] = nullptr;
    }
};

class Solution {
private:
    TrieNode* root;
    
    void insert(int num) {
        TrieNode* node = root;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (!node->children[bit]) {
                node->children[bit] = new TrieNode();
            }
            node = node->children[bit];
        }
    }
    
    int findMaxXOR(int num) {
        TrieNode* node = root;
        int maxXor = 0;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (node->children[1 - bit]) {
                maxXor |= (1 << i);
                node = node->children[1 - bit];
            } else if (node->children[bit]) {
                node = node->children[bit];
            } else {
                break;
            }
        }
        return maxXor;
    }

public:
    Solution(){
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
    }
    int findMaximumXOR(vector<int>& nums) {
        root = new TrieNode();
        int maxXor = 0;
        
        for (int num : nums) {
            insert(num);
            maxXor = max(maxXor, findMaxXOR(num));
        }
        
        return maxXor;
    }
};
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