<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 Node:
    def __init__(self):
        self.children = [None, None]

class TrieNode:
    def __init__(self):
        self.root = Node()
    
    def insert(self, num):
        node = self.root

        # traverse starting from the most significant bit
        for i in range(31, -1, -1):
            # extract the ith bit
            bit = (num >> i) & 1

            # if not present, insert the bit
            if not node.children[bit]:
                node.children[bit] = Node()

            node = node.children[bit]
    
    # find the maximum value achievale by that num from trie
    def findMax(self, num):
        node = self.root
        maxNum = 0

        for i in range(31, -1, -1):
            # extract the ith bit
            bit = (num >> i) & 1

            # if opp bit in trie then choose it to get max number
            if node.children[not bit]:
                # set ith bit in max num
                maxNum |= (1 << i)
                node = node.children[not bit]
            # if no opp bit in trie, move to child with same bit
            else:
                node = node.children[bit]

        return maxNum


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        n = len(queries)
        a = len(nums)
        nums.sort()
        offlineQueries = []

        for i in range(n):
            offlineQueries.append((queries[i][1], (queries[i][0], i)))
        
        # sort queries by end to have incremental additions of nums to trie
        offlineQueries.sort()

        i = 0
        ans = [0]*n
        trie = TrieNode()

        for end, (start, ind) in offlineQueries:
            # insert numbers in trie only which are less than end to ensure max is found with
            # condition satisfied
            while i < a and nums[i] <= end:
                trie.insert(nums[i])
                i += 1
            
            if i != 0:
                ans[ind] = trie.findMax(start)
            else:
                ans[ind] = -1
        
        return ans

        
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