<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>

static const auto Initialize = []{
    ios::sync_with_stdio(false); cin.tie(nullptr);
    return nullptr;
}();

class Solution {
public:
    int findMaximumXOR(vector<int>& nums) {
        int maximum = 0, nowMask = 0;
        unordered_set <int> prefixes;
        for (unsigned i = (1 << 31); i; i >>= 1) {
            nowMask |= i;
            prefixes.clear();
            for (int j : nums)
                prefixes.insert(j & nowMask);
            for (int j : prefixes)
                if (prefixes.count((maximum ^ j) ^ i)) {
                    maximum |= i;
                    break;
                }
        }
        return maximum;
    }
};
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