<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:
  
    int findMaximumXOR(vector<int>& nums) {
        
        int candidate = 0;
        int mask = 0;
        int max_xor = 0;

        unordered_set<int> set;

        for(int i = 31; i >=0; i--){

            mask = mask | (1<<i);

            for(int i = 0; i < nums.size(); i++){
                set.insert(nums[i] & mask);
            }

            candidate = max_xor | (1 << i);

            for(auto prefix : set){

                if(set.count(prefix^candidate)){
                    max_xor = candidate;
                    break;
                }
            }

            set.clear();
         
        }

        return max_xor;
   
    }
};
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