<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, int i = 31){
                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);
                }
            }

            void getMaxFromTrie(bitset<32>& br, int& b, int bri = 31){
                if(bri < 0){
                    return;
                }

                int oppBit = !(br[bri]);
                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);
                }
            }
    };

    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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