<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:
    class Trie{
        public:
            int value;
            vector<Trie*> children;
            bool isTerminal;

            Trie(int v, bool it = false):children(2, nullptr){
                value = v;      
                isTerminal = it;      
            }

            void insertInTrie(bitset<32>& br){
                // if(i < 0){
                //     isTerminal = true;
                //     return;
                // }

                // if(children[br[i]] != nullptr){
                //     (*children[br[i]]).insertInTrie(br, i-1);
                // }
                // else{
                //     Trie* newTrie = new Trie(br[i]);
                //     children[br[i]] = newTrie;
                //     (*children[br[i]]).insertInTrie(br, i-1);
                // }

                Trie* root = this;
                for(int i = 31; i >= 0; i--){
                    int bit = br[i];
                    if(root->children[bit] != nullptr){
                        root = root->children[bit];
                    }
                    else{
                        root->children[bit] = new Trie(bit);
                        root = root->children[bit];
                    }
                }
            }

            void getMaxFromTrie(bitset<32>& br, int& b){
                // if(bri < 0){
                //     return;
                // }

                
                Trie* root = this;
                for(int i = 31; i >= 0; i--){
                    int oppBit = !(br[i]);
                    if(root->children[oppBit] != nullptr){
                        b += (1 << i);
                        root = root->children[oppBit];
                    }
                    else{
                        root = root->children[!oppBit];
                    }
                }
                // if(children[oppBit] != nullptr){
                //     // br2[bri] = oppBit;
                //     b += (1 << bri);
                //     (*children[oppBit]).getMaxFromTrie(br, b, bri-1);
                //     //todo
                // }
                // else{
                //     // br2[bri] = !oppBit;
                //     (*children[!oppBit]).getMaxFromTrie(br, b, bri-1);
                // }
            }
    };

    // bitset<32> getBr(int num){
    //     bitset<32> br(32, 0);
    //     for(int j = 0; j < 32; j++){
    //         int bit = num&1;
    //         br[31 - j] = bit;
    //         num = num>>1;
    //     }
    //     return br;
    // }

    // int getNum(bitset<32> br){
    //     int num = 0;
    //     for(int j = 0; j < 32; j++){
    //         int exp = pow(2.0, 1.0*(31-j)) + 0.1;
    //         num += (br[j] * exp);
    //     }
    //     return num;
    // }


    int findMaximumXOR(vector<int>& nums) {
        Trie* t = new Trie(-1);
        int N = nums.size();
        for(int i = 0; i < N; i++){
            int num = nums[i];
            bitset<32> br(num);
            t->insertInTrie(br);
        }

        int maxi = INT_MIN;
        for(int i = 0; i < N; i++){
            int num = nums[i];
            bitset<32> br(num);
            bitset<32> br2;
            int b = 0;
            t->getMaxFromTrie(br, b);
            maxi = max(maxi, b);
        }

        return maxi;
    }
};
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