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 {
private:
    Trie* children[2];

public:
    Trie()
        : children{nullptr, nullptr} {}

    void insert(int x) {
        Trie* node = this;
        for (int i = 30; ~i; --i) {
            int v = (x >> i) & 1;
            if (!node->children[v]) {
                node->children[v] = new Trie();
            }
            node = node->children[v];
        }
    }

    int search(int x) {
        Trie* node = this;
        int ans = 0;
        for (int i = 30; ~i; --i) {
            int v = (x >> i) & 1;
            if (node->children[v ^ 1]) {
                ans |= 1 << i;
                node = node->children[v ^ 1];
            } else if (node->children[v]) {
                node = node->children[v];
            } else {
                return -1;
            }
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        sort(nums.begin(), nums.end());
        int n = queries.size();
        vector<int> idx(n);
        iota(idx.begin(), idx.end(), 0);
        sort(idx.begin(), idx.end(), [&](int i, int j) { return queries[i][1] < queries[j][1]; });
        vector<int> ans(n);
        Trie trie;
        int j = 0;
        for (int i : idx) {
            int x = queries[i][0], m = queries[i][1];
            while (j < nums.size() && nums[j] <= m) {
                trie.insert(nums[j++]);
            }
            ans[i] = trie.search(x);
        }
        return ans;
    }
};