<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>

MAX_VALUE = 2**31-1

from typing import List

class Node:
    def __init__(self,x:int):
        self.isTerminal = True
        self.minValueReachable = x
        self.children = [None,None]

def addNumber(x:int,root:Node)->None:
    crt = root
    mask = 1<<30
    for i in range(31):
        branch = 1 if x & mask > 0 else 0
        crt.isTerminal = False
        crt.minValueReachable = min(crt.minValueReachable,x)
        if crt.children[branch] is None:
            crt.children[branch] = Node(x)
        #going to next node
        crt = crt.children[branch]
        mask = mask>>1


def findRequiredInTrie(root:Node,x:int,m:int,pow=30,intermediary=0)->int:
    """
    use the trie's extra info to determine if choosing the next bit 1 makes it imposible to find numbers <= m
    . Otherwise check if we can choose the bit to be to negated value of the coresponding one from x
    in order to maximise xor
    """
    if root.isTerminal:
        return intermediary
    
    if root.children[1] and root.children[1].minValueReachable>m:
        desiredBitValue = 0
    else:
        desiredBitValue = 1 if (~x)&(1<<pow)>0 else 0
        if root.children[desiredBitValue] is None:
            desiredBitValue = 1^desiredBitValue
    
    intermediary += desiredBitValue * (1<<pow)

    return findRequiredInTrie( root.children[desiredBitValue], x, m, pow-1, intermediary)

class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        root = Node(MAX_VALUE)
        for x in nums:
            addNumber(x,root)

        ans = []
        for q in queries:
            x = q[0]
            m = q[1]
            if(root.minValueReachable > m):
                ans.append(-1)
            else:
                valueFromTrie = findRequiredInTrie(root,x,m) 
                ans.append(valueFromTrie ^ x)
        return ans

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