<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 Trie:
    def __init__(self):
        self.children = [None] * 2
        self.dig = -1

    def insert(self, word):
        # Implement the 'insert' method here.
        node = self
        for i in range(len(word)):
            o = int(word[i])
            if not node.children[o]:
                node.children[o] = Trie()
                node.children[o].dig=o
            node = node.children[o]

def binarize(n):
    if n<2:
        return str(n)
    return binarize(n//2) + str(n%2)

def decimalize(n,e):
    if n==0:
        return n
    else:
        return decimalize(n//10,e<<1) + e*(n%10)

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()
        bins = list(map(binarize,nums))
        qbins = [list(map(binarize,q)) for q in queries]
        qbins = [(e,q) for e,q in enumerate(qbins)]
        maxpow = max([len(b) for b in bins])
        maxpow = max(maxpow,max([max(len(q[1][0]),len(q[1][1])) for q in qbins]))
        bins = ['0'*(maxpow-len(b))+b for b in bins]
        qbins = [[q[0],['0'*(maxpow-len(b))+b for b in q[1]]] for q in qbins]
        qbins.sort(key=lambda x:x[1][1])
        res = [-1]*len(queries)
        k=0
        trie = Trie()
        for e,q in qbins:
            if q[1]>=bins[0]:
                for i in range(k,len(nums)):
                    if q[1]>=bins[i]:
                        trie.insert(bins[i])
                        k+=1
                    else:
                        break
                xor = self.findMaxXor(q[0],trie)
                if xor:
                    res[e]=decimalize(int(xor),1)
        return res

    def findMaxXor(self,x,trie):
        if trie.children[0] and int(x[0])^0==1:
            return '1'+ self.findMaxXor(x[1:],trie.children[0])
        elif trie.children[1] and int(x[0])^1==1:
            return '1'+ self.findMaxXor(x[1:],trie.children[1])
        else:
            if trie.children[0] is None and trie.children[1] is None:
                return ''
            elif trie.children[0]:
                return '0'+self.findMaxXor(x[1:],trie.children[0])
            else:
                return '0'+self.findMaxXor(x[1:],trie.children[1])
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