Maximum Strong Pair XOR II
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. A pair of integers <code>x</code> and <code>y</code> is called a <strong>strong</strong> pair if it satisfies the condition:</p>

<ul>
	<li><code>|x - y| &lt;= min(x, y)</code></li>
</ul>

<p>You need to select two integers from <code>nums</code> such that they form a strong pair and their bitwise <code>XOR</code> is the <strong>maximum</strong> among all strong pairs in the array.</p>

<p>Return <em>the <strong>maximum</strong> </em><code>XOR</code><em> value out of all possible strong pairs in the array</em> <code>nums</code>.</p>

<p><strong>Note</strong> that you can pick the same integer twice to form a pair.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 11 strong pairs in the array <code>nums</code>: (1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) and (5, 5).
The maximum XOR possible from these pairs is 3 XOR 4 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,100]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are 2 strong pairs in the array nums: (10, 10) and (100, 100).
The maximum XOR possible from these pairs is 10 XOR 10 = 0 since the pair (100, 100) also gives 100 XOR 100 = 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [500,520,2500,3000]
<strong>Output:</strong> 1020
<strong>Explanation:</strong> There are 6 strong pairs in the array nums: (500, 500), (500, 520), (520, 520), (2500, 2500), (2500, 3000) and (3000, 3000).
The maximum XOR possible from these pairs is 500 XOR 520 = 1020 since the only other non-zero XOR value is 2500 XOR 3000 = 636.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2<sup>20</sup> - 1</code></li>
</ul>
Bit Manipulation; Trie; Array; Hash Table; Sliding Window
class Trie {
public:
    Trie* children[2];
    int cnt;

    Trie()
        : cnt(0) {
        children[0] = nullptr;
        children[1] = nullptr;
    }

    void insert(int x) {
        Trie* node = this;
        for (int i = 20; ~i; --i) {
            int v = (x >> i) & 1;
            if (node->children[v] == nullptr) {
                node->children[v] = new Trie();
            }
            node = node->children[v];
            ++node->cnt;
        }
    }

    int search(int x) {
        Trie* node = this;
        int ans = 0;
        for (int i = 20; ~i; --i) {
            int v = (x >> i) & 1;
            if (node->children[v ^ 1] != nullptr && node->children[v ^ 1]->cnt > 0) {
                ans |= 1 << i;
                node = node->children[v ^ 1];
            } else {
                node = node->children[v];
            }
        }
        return ans;
    }

    void remove(int x) {
        Trie* node = this;
        for (int i = 20; ~i; --i) {
            int v = (x >> i) & 1;
            node = node->children[v];
            --node->cnt;
        }
    }
};

class Solution {
public:
    int maximumStrongPairXor(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        Trie* tree = new Trie();
        int ans = 0, i = 0;
        for (int y : nums) {
            tree->insert(y);
            while (y > nums[i] * 2) {
                tree->remove(nums[i++]);
            }
            ans = max(ans, tree->search(y));
        }
        return ans;
    }
};