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 {
    children: (Trie | null)[];
    cnt: number;

    constructor() {
        this.children = [null, null];
        this.cnt = 0;
    }

    insert(x: number): void {
        let node: Trie | null = this;
        for (let i = 20; i >= 0; i--) {
            const v = (x >> i) & 1;
            if (node.children[v] === null) {
                node.children[v] = new Trie();
            }
            node = node.children[v] as Trie;
            node.cnt++;
        }
    }

    search(x: number): number {
        let node: Trie | null = this;
        let ans = 0;
        for (let i = 20; i >= 0; i--) {
            const v = (x >> i) & 1;
            if (node.children[v ^ 1] !== null && (node.children[v ^ 1] as Trie).cnt > 0) {
                ans |= 1 << i;
                node = node.children[v ^ 1] as Trie;
            } else {
                node = node.children[v] as Trie;
            }
        }
        return ans;
    }

    remove(x: number): void {
        let node: Trie | null = this;
        for (let i = 20; i >= 0; i--) {
            const v = (x >> i) & 1;
            node = node.children[v] as Trie;
            node.cnt--;
        }
    }
}

function maximumStrongPairXor(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const tree = new Trie();
    let ans = 0;
    let i = 0;

    for (const y of nums) {
        tree.insert(y);

        while (y > nums[i] * 2) {
            tree.remove(nums[i++]);
        }

        ans = Math.max(ans, tree.search(y));
    }

    return ans;
}
