<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]:
        # sort nums
        # binary search for mi in nums; then restrict next searchs from 0 to mi

        # Ideal number has opposite bits from xi in each position
        # get search range for ideal leading bit; restrict search space if we find one, otherwise just move to next bit
        # O(nlogn + qlogn)
        q = len(queries)
        nums = list(set(nums))
        nums.sort()
        answer = [0]*q
        for i, (x, m) in enumerate(queries):
            #print("NEW QUERY")
            #print(i, x, m)
            hi = bisect.bisect_right(nums, m)
            # handle edge cases now
            if hi == 0:
                answer[i] = -1
                continue
            if hi == 1:
                answer[i] = x ^ nums[0]
                continue
            lo = 0
            goal = ~x
            xlen = x.bit_length() - 1
            bit = nums[hi - 1].bit_length() - 1
            maxVal = 0
            # we maintain the invariant that the elements in [lo, high] have all the same bits before bit
            # so searching for things with the desired bit is just a binary search
            while bit >= 0:
                mask = 1 << bit
                goalbit = mask if bit > xlen else goal & mask
                key = lambda v : v & mask
                bit -= 1
                #print(lo, hi)
                left = bisect.bisect_left(nums, goalbit, lo = lo, hi = hi, key = key)
                # if we found nothing with the bit we want, or everything has bit we want, we cant narrow the search space
                if left == hi:
                    continue
                right = bisect.bisect_right(nums, goalbit, lo = left, hi = hi, key = key)
                if right == lo:
                    continue
                lo = left
                hi = right
                if lo == hi - 1:
                    answer[i] = x ^ nums[lo]
                    break
            # the only case where we didn't break from the loop early is if we never found anything with a bit in common with goal
            # in which case we want answer[i] to be 0, which we already set it to
        return answer

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