<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 Node{
    public:
    Node *links[2];

    bool containsKey(int bit){
        return (links[bit]!=NULL);
    }

    void put(int bit, Node* node){
        links[bit] = new Node();
    }

    Node* get(int bit){
        return links[bit];
    }
};

class Trie{
    private:
    Node* root;

    public:
    Trie(){
        root = new Node();
    }

    void insert(int num){
        Node *node = root;

        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(!node->containsKey(bit)){
                node->put(bit, new Node());
            }
            node = node->get(bit);
        }
    }

    int getMax(int num){
        Node *node = root;

        int maxNum=0;
        for(int i=31; i>=0; i--){
            int bit = (num>>i) & 1;
            if(node->containsKey(1 - bit)){
                maxNum |= (1<<i);
                node = node->get(1- bit);
            }
            else{
                node = node->get(bit);
            }
        }
        return maxNum;
    }

};

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        Trie trie;

        for(auto &it : nums){
            trie.insert(it);
        }

        int maxi=0;

        for(auto &it : nums){
            maxi = max(maxi, trie.getMax(it));
        }
        return maxi;
    }
};
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