<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) {
        std::ios_base::sync_with_stdio(false);
        std::cin.tie(nullptr);
        unordered_set<uint32_t> pr[32];
        for(int x_: nums) {
            uint32_t x = x_;
            for(int l=0; l<=31; l++) {
                pr[l].insert((x>>(31-l)) << (31-l));
            }
        }
        uint32_t ans = 0;
        for(int x_: nums) {
            uint32_t x = x_;
            uint32_t pre = 0;
            for(int i=0; i<=31; i++) {
                uint32_t bi = (uint32_t)1 << (31-i);
                uint32_t prelarger = pre | ( (x&bi) ^ bi);
                if( pr[i].contains(prelarger) ) {
                    pre = prelarger;
                }
                else {
                    pre = pre | (x&bi);
                }
            }
            ans = max(ans,x^pre);
        }
        return ans;
    }
};
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