<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 Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        t = Trie()
        answers = [-1] * len(queries)
        for i in range(len(queries)):
            queries[i].append(i)
        queries.sort(key=lambda x: x[1])
        nums.sort(key=lambda x: x)
        numsIdx = 0

        called_insert = 0
        called_query = 0
        for i in range(len(queries)):
            j = numsIdx
            while j < len(nums):
                #if j:
                #    print("j", j, len(nums), i, len(queries))
                if nums[j] <= queries[i][1]:
                    t.insert(nums[j])
                    called_insert += 1
                    j += 1
                    #print("insert",nums[j])
                else:
                    numsIdx = j
                    break
            else:
                #print("numsIdx ", numsIdx)
                numsIdx = len(nums)
            #print("query", i, numsIdx)
            called_query += 1
            answers[i] = t.getMaxXORltM(queries[i][0], nums, queries[i][1])
        toreturn = [-1] * len(queries)

        #print("!!!!!", called_query, called_insert)
        for i in range(len(queries)):
            toreturn[queries[i][2]] = answers[i]
        return toreturn


class Trie:
    def __init__(self):
        self.mp = {}

    def insert(self, num):
        trav = self.mp
        for i in range(31, -1, -1):
            bit = (num >> i) & 1
            if bit not in trav:
                trav[bit] = {}
            trav = trav[bit]
        trav[-1] = {}

    def getMaxXORltM(self, query, nums, m):
        xorSum = 0

        trav = self.mp
        # print("query", query,"mlimit", m)

        i = 31
        while i >= 0:
            qbit = (query >> i) & 1  # query

            '''print(
                i,
                "q",
                qbit,
                "trav",
                list(trav.keys()),
            )'''

            if qbit == 0 and 1 in trav:
                xorSum += 1 << i
                trav = trav[1]
            elif qbit == 1 and 0 in trav:
                xorSum += 1 << i
                trav = trav[0]
            elif 0 in trav:
                trav = trav[0]
            elif 1 in trav:
                trav = trav[1]
            else:
                return -1

            i -= 1
        return xorSum

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