<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>

from typing import List

class TrieNode:
    def __init__(self, val: int) -> None:
        self.val = val
        self.childrens = [None, None]

class Trie:
    def __init__(self) -> None:
        self.root = TrieNode(val=-1)
    
    def add(self, num: int) -> None:
        tmp = self.root
        repr = ""
        for i in range(31, -1, -1):
            bit = 1 if((1<<i)&num)>0 else 0
            if tmp.childrens[bit]==None:
                nn = TrieNode(val=bit)
                tmp.childrens[bit]=nn
            repr += f"{bit}"
            tmp = tmp.childrens[bit]

    def print(self):
        tmp = self.root
        queue = [tmp]
        pos = 31
        while len(queue) != 0:
            n = len(queue)
            while n!= 0:
                item = queue.pop(0)
                leftchild = item.childrens[0]
                rightchild = item.childrens[1]
                print(f"pos: {pos}, lc: {leftchild!=None}, rc: {rightchild!=None}")
                if leftchild != None:
                    queue.append(leftchild)
                if rightchild != None:
                   queue.append(rightchild)
                
                n-=1
            
            pos-=1
    
    def maxXor(self, num: int) -> int:
        value = 0
        tmp = self.root

        for i in range(31, -1, -1):
            bit = 1 if ((num & (1<<i)) > 0 ) else 0
            rev_bit = 1 if bit==0 else 0
            #print(f"bit pos: {i}, bit val: {bit}")

            if tmp.childrens[rev_bit] != None:
                #print(f"rev bit found for bit pos: {i}")
                value += (1<<i)
                tmp = tmp.childrens[rev_bit]
            else:
                tmp = tmp.childrens[bit]
        
        return value


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        res = [None] * len(queries)
        tr = Trie()
        nums.sort()
        
        idx_queries = []
        for i in range(len(queries)):
            xi = queries[i][0]
            mi = queries[i][1]
            idx_queries.append((i, xi, mi))

        sorted_list = sorted(idx_queries, key=lambda qi: qi[2])

        ni = 0
        for i in range(len(sorted_list)):
            xi = sorted_list[i][1]
            mi = sorted_list[i][2]
            max_xor = -1
            while ni < len(nums) and nums[ni] <= mi:
                tr.add(nums[ni])
                ni+=1
            
            if ni != 0:
                max_xor = tr.maxXor(xi)
            
            res[sorted_list[i][0]] = max_xor

        return res
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