<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>

class Solution {
    
    class TreeNode {
    public:
        TreeNode* next[2];
        TreeNode () {next[0] = nullptr; next[1] = nullptr;};
    };
    TreeNode* buildTree(vector<int>& nums) {
        TreeNode* root = new TreeNode(), *cur;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            int num = nums[i];
            cur = root;
            for (int j = 31; j >= 0; j--) {
                int index = ((num >> j) & 1);
                if (cur->next[index] == nullptr)
                    cur->next[index] = new TreeNode();
                cur = cur->next[index];
            }
        }
        return root;
    }
    
    int dfs(TreeNode* root, int x, int limit, int value, int height) {
        if (value > limit) return -1;
        
        if (height == -1) return x^value;
        
        int bit_x = (x >> height) & 1;
        
        if (root->next[1-bit_x] != nullptr) {
            int v = dfs(root->next[1-bit_x], x, limit, (value | ((1-bit_x) << height)), height-1);
            if (v >= 0) return v;
        }
        if (root->next[bit_x] != nullptr) {
            int v = dfs(root->next[bit_x], x, limit, (value | (bit_x << height)), height-1);
            if (v >= 0) return v;
        }
        
        return -1;
    }
    
public:
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        vector<int> ans;
        TreeNode* root = buildTree(nums);
        
        for (const vector<int>& query : queries) {            
            int tmp = dfs(root, query[0], query[1], 0, 31);
            ans.push_back(tmp);
        }
        
        return ans;
    }
};
214281