Find Building Where Alice and Bob Can Meet
<p>You are given a <strong>0-indexed</strong> array <code>heights</code> of positive integers, where <code>heights[i]</code> represents the height of the <code>i<sup>th</sup></code> building.</p>

<p>If a person is in building <code>i</code>, they can move to any other building <code>j</code> if and only if <code>i &lt; j</code> and <code>heights[i] &lt; heights[j]</code>.</p>

<p>You are also given another array <code>queries</code> where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>. On the <code>i<sup>th</sup></code> query, Alice is in building <code>a<sub>i</sub></code> while Bob is in building <code>b<sub>i</sub></code>.</p>

<p>Return <em>an array</em> <code>ans</code> <em>where</em> <code>ans[i]</code> <em>is <strong>the index of the leftmost building</strong> where Alice and Bob can meet on the</em> <code>i<sup>th</sup></code> <em>query</em>. <em>If Alice and Bob cannot move to a common building on query</em> <code>i</code>, <em>set</em> <code>ans[i]</code> <em>to</em> <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> heights = [6,4,8,5,2,7], queries = [[0,1],[0,3],[2,4],[3,4],[2,2]]
<strong>Output:</strong> [2,5,-1,5,2]
<strong>Explanation:</strong> In the first query, Alice and Bob can move to building 2 since heights[0] &lt; heights[2] and heights[1] &lt; heights[2]. 
In the second query, Alice and Bob can move to building 5 since heights[0] &lt; heights[5] and heights[3] &lt; heights[5]. 
In the third query, Alice cannot meet Bob since Alice cannot move to any other building.
In the fourth query, Alice and Bob can move to building 5 since heights[3] &lt; heights[5] and heights[4] &lt; heights[5].
In the fifth query, Alice and Bob are already in the same building.  
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> heights = [5,3,8,2,6,1,4,6], queries = [[0,7],[3,5],[5,2],[3,0],[1,6]]
<strong>Output:</strong> [7,6,-1,4,6]
<strong>Explanation:</strong> In the first query, Alice can directly move to Bob&#39;s building since heights[0] &lt; heights[7].
In the second query, Alice and Bob can move to building 6 since heights[3] &lt; heights[6] and heights[5] &lt; heights[6].
In the third query, Alice cannot meet Bob since Bob cannot move to any other building.
In the fourth query, Alice and Bob can move to building 4 since heights[3] &lt; heights[4] and heights[0] &lt; heights[4].
In the fifth query, Alice can directly move to Bob&#39;s building since heights[1] &lt; heights[6].
For ans[i] != -1, It can be shown that ans[i] is the leftmost building where Alice and Bob can meet.
For ans[i] == -1, It can be shown that there is no building where Alice and Bob can meet.

</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= heights.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= heights[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= heights.length - 1</code></li>
</ul>
Stack; Binary Indexed Tree; Segment Tree; Array; Binary Search; Monotonic Stack; Heap (Priority Queue)
class BinaryIndexedTree:
    __slots__ = ["n", "c"]

    def __init__(self, n: int):
        self.n = n
        self.c = [inf] * (n + 1)

    def update(self, x: int, v: int):
        while x <= self.n:
            self.c[x] = min(self.c[x], v)
            x += x & -x

    def query(self, x: int) -> int:
        mi = inf
        while x:
            mi = min(mi, self.c[x])
            x -= x & -x
        return -1 if mi == inf else mi


class Solution:
    def leftmostBuildingQueries(
        self, heights: List[int], queries: List[List[int]]
    ) -> List[int]:
        n, m = len(heights), len(queries)
        for i in range(m):
            queries[i] = [min(queries[i]), max(queries[i])]
        j = n - 1
        s = sorted(set(heights))
        ans = [-1] * m
        tree = BinaryIndexedTree(n)
        for i in sorted(range(m), key=lambda i: -queries[i][1]):
            l, r = queries[i]
            while j > r:
                k = n - bisect_left(s, heights[j]) + 1
                tree.update(k, j)
                j -= 1
            if l == r or heights[l] < heights[r]:
                ans[i] = r
            else:
                k = n - bisect_left(s, heights[l])
                ans[i] = tree.query(k)
        return ans
