<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 Trie:
    def __init__(self):
        self.children = [None, None]
    
    def insert(self, n):
        root = self

        for i in range(31, -1, -1):
            bit = (n>>i) & 1
            if not root.children[bit]:
                root.children[bit] = Trie()
            
            root = root.children[bit]
        
    def findMaxXor(self, n):
        root = self
        maxXor = 0
        for i in range(31, -1, -1):
            bit = (n>>i) & 1
            if root.children[1-bit]:
                maxXor |= (1<<i)
                root = root.children[1-bit]
            elif root.children[bit]:
                root = root.children[bit]
            else: break
        return maxXor

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()

        queries = [query + [i] for i, query in enumerate(queries)]
        queries = sorted(queries, key = lambda x:x[1])
        ansArr = [-1 for _ in range(len(queries))]

        itrNums = 0
        n = len(nums)
        trie = Trie()

        for query in queries:
            [x, m, itrQuery] = query
            print("-"*50)
            print(f"{x=}, {m=}, {itrQuery=}")
            while itrNums < n and nums[itrNums] <= m:
                # print(f"inserted {nums[itrNums]}")
                trie.insert(nums[itrNums])
                itrNums += 1
            
            if itrNums > 0:
                maxXor = trie.findMaxXor(x)
                ansArr[itrQuery] = maxXor
        
        return ansArr
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