<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 BitTrie:
    def __init__(self):
        self.root = {}
        # self.BITS = BITS
    
    def insert(self, num):
        # HIS CODE
        p = self.root
        for i in range(31, -1, -1):
            cur = (num >> i) & 1
            if cur not in p:
                p[cur] = {}
            p = p[cur]
            
        # MY CODE
        # curr = self.root
        # for bitOffset in range(self.BITS - 1, -1, -1):
        #     bit = (num >> bitOffset) & 1
        #     if bit in curr:
        #         curr = curr[bit]
        #     else:
        #         newChild = {}
        #         curr[bit] = newChild
        #         curr = newChild
    
    def queryMaxXor(self, numToBeXord):
        # HIS CODE
        if not self.root: 
            return -1
        p, ans = self.root, 0
        for i in range(31, -1, -1):
            cur = (numToBeXord >> i) & 1
            if 1 - cur in p:
                p = p[1 - cur]
                ans |= (1 << i)
            else:
                p = p[cur]
        return ans


        # MY CODE
        # res = 0
        # curr = self.root
        # if not curr: # TODO
        #     return False
        # for bitOffset in range(self.BITS - 1, -1, -1):
        #     numBit = (numToBeXord >> bitOffset) & 1
        #     desiredBit = numBit ^ 1

        #     if desiredBit in curr:
        #         res |= (1 << bitOffset)
        #         curr = curr[desiredBit]
        #     else:
        #         curr = curr[numBit]
        # return res



    # MY MIN CODE
    # def queryMinXor(self, numToBeXord):
    #     res = 0
    #     curr = self.root
    #     for bitOffset in range(self.BITS - 1, -1, -1):
    #         numBit = (numToBeXord >> bitOffset) & 1

    #         if numBit in curr:
    #             curr = curr[numBit]
    #         else:
    #             res |= (1 << bitOffset)
    #             curr = curr[numBit ^ 1]
                
    #     return res


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        nums.sort()
        # BITS = math.ceil(math.log2(max(nums))) + 1
        # BITS = 32
        # print(f'bits: {BITS}')
        # t = BitTrie(BITS)

        q = sorted(enumerate(queries),key=lambda x: x[1][1])
        t = BitTrie()
        ans = [-1] * len(queries)
        j = 0
        for i, (x, m) in q:
            while j < len(nums) and nums[j] <= m:
                t.insert(nums[j])
                j += 1
            ans[i] = t.queryMaxXor(x)

            # ans[i] = t.queryMaxXor(x)
        return ans

        # newQueries = queries[::]
        # newQueries.sort(key=lambda tup: (tup[1]))
        # queriesToAnswer = {} # maps tup to answer
        # print(f'sorted queries: {queries}')

        # r = 0 # the right edge, we will add elements from nums while we don't exceed the threshold
        # for i in range(len(newQueries)):
        #     num, allowedUpTo = newQueries[i]
        #     while r < len(nums) and nums[r] <= allowedUpTo:
        #         t.insert(nums[r])
        #         r += 1
        #     biggestXor = t.queryMaxXor(num)
        #     queriesToAnswer[(num, allowedUpTo)] = biggestXor if biggestXor != False else -1
        
        # res = []
        # for num, biggestAllowedUpTo in queries:
        #     res.append(queriesToAnswer[(num), biggestAllowedUpTo])
        # return res














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