Maximum XOR With an Element From Array
<p>You are given an array <code>nums</code> consisting of non-negative integers. You are also given a <code>queries</code> array, where <code>queries[i] = [x<sub>i</sub>, m<sub>i</sub>]</code>.</p>

<p>The answer to the <code>i<sup>th</sup></code> query is the maximum bitwise <code>XOR</code> value of <code>x<sub>i</sub></code> and any element of <code>nums</code> that does not exceed <code>m<sub>i</sub></code>. In other words, the answer is <code>max(nums[j] XOR x<sub>i</sub>)</code> for all <code>j</code> such that <code>nums[j] &lt;= m<sub>i</sub></code>. If all elements in <code>nums</code> are larger than <code>m<sub>i</sub></code>, then the answer is <code>-1</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em> and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

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

<pre>
<strong>Input:</strong> nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
<strong>Output:</strong> [3,3,7]
<strong>Explanation:</strong>
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
<strong>Output:</strong> [15,-1,5]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= nums[j], x<sub>i</sub>, m<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>
Bit Manipulation; Trie; Array
class Trie {
    children: (Trie | null)[];

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

    insert(x: number): void {
        let node: Trie | null = this;
        for (let i = 30; ~i; i--) {
            const v = (x >> i) & 1;
            if (node.children[v] === null) {
                node.children[v] = new Trie();
            }
            node = node.children[v] as Trie;
        }
    }

    search(x: number): number {
        let node: Trie | null = this;
        let ans = 0;
        for (let i = 30; ~i; i--) {
            const v = (x >> i) & 1;
            if (node.children[v ^ 1] !== null) {
                ans |= 1 << i;
                node = node.children[v ^ 1] as Trie;
            } else if (node.children[v] !== null) {
                node = node.children[v] as Trie;
            } else {
                return -1;
            }
        }
        return ans;
    }
}

function maximizeXor(nums: number[], queries: number[][]): number[] {
    nums.sort((a, b) => a - b);
    const n = queries.length;
    const idx = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => queries[i][1] - queries[j][1]);
    const ans: number[] = [];
    const trie = new Trie();
    let j = 0;
    for (const i of idx) {
        const x = queries[i][0];
        const m = queries[i][1];
        while (j < nums.length && nums[j] <= m) {
            trie.insert(nums[j++]);
        }
        ans[i] = trie.search(x);
    }
    return ans;
}
