<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:
    bool isTerminal;
    char ch;
    TrieNode *children[2];

    TrieNode(char ch){
        this->ch=ch;
        isTerminal = false;
        for(int i=0;i<2;i++){
            children[i]=NULL;
        }
    }
};

class Solution {
public:

    string getbinaryString(int n){
        string s = bitset<32>(n).to_string();
        return s;
    }
    int findMaximumXOR(vector<int>& nums) {
        vector<string>bsa;
        for(int i=0;i<nums.size();i++){
            bsa.push_back(getbinaryString(nums[i]));
        }

        TrieNode *root = new TrieNode('\0');

        for(auto bs:bsa){
            TrieNode *currNode = root;
            for(auto c:bs){
                if(currNode->children[c-'0']==NULL){
                    currNode->children[c-'0'] = new TrieNode(c);
                }
                currNode = currNode->children[c-'0'];
            }
            currNode->isTerminal=true;
        }

        int maxXor=0;
        for(auto bs:bsa){
            int res=0;
            int mul = 31;
            TrieNode * currNode = root;
            for(auto c:bs){
                auto k = (c=='0')? '1':'0';
                if(currNode->children[k-'0']!=NULL){
                    currNode = currNode->children[k-'0'];
                    res = res+pow(2,mul);
                }
                else{
                    currNode = currNode->children[!(k-'0')];
                }
                mul--;
            }
            maxXor=max(maxXor,res);
        }
        return maxXor;
    }
};
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