<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 max = 0, mask = 0;
        unordered_set<int> t;
        // search from left to right, find out for each bit is there 
        // two numbers that has different value
        for (int i = 31; i >= 0; i--){
            // mask contains the bits considered so far
            mask |= (1 << i);
            t.clear();
            // store prefix of all number with right i bits discarded
            for (int n: nums){
                t.insert(mask & n);
            }
            
            // now find out if there are two prefix with different i-th bit
            // if there is, the new max should be current max with one 1 bit at i-th position, which is candidate
            // and the two prefix, say A and B, satisfies:
            // A ^ B = candidate
            // so we also have A ^ candidate = B or B ^ candidate = A
            // thus we can use this method to find out if such A and B exists in the set 
            int candidate = max | (1<<i);
            for (int prefix : t){
                if (t.find(prefix ^ candidate) != t.end()){
                    max = candidate;
                    break;
                }
                
            }
        }
        return max;
    }
};
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