<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) {
        
       // Find the length of the maximum number in binary representation
        int maxNum = *max_element(nums.begin(), nums.end());
        int L = 0;
        while (maxNum > 0) {
            L++;
            maxNum >>= 1;
        }

        int maxXor = 0;  // This will store the final result
        unordered_set<int> prefixes;  

        for (int i = L - 1; i >= 0; --i) {
            // Go to the next bit by the left shift
            maxXor <<= 1;

            // Set 1 in the smallest bit
            int currXor = maxXor | 1;
            prefixes.clear();

            // compute all possible prefixes of length (L - i) in binary representation
            for (int num : nums) {
                prefixes.insert(num >> i);
            }

            // Update maxXor, if two of these prefixes could result in currXor.
            // Check if p1^p2 == currXor, i.e. p1 == currXor^p2.
            for (int p : prefixes) {
                if (prefixes.find(currXor ^ p) != prefixes.end()) {
                    maxXor = currXor; 
                    break;
                }
            }
        }

        return maxXor;
    }
};
210403