<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>

class SegNode:
    def __init__(self, l, r, val, left, right):
        self.l, self.r, self.val = l, r, val
        self.left, self.right = left, right


class SegTree:
    def __init__(self, values):
        self.values = values
        self.root = self.build(0, len(values) - 1)
        self.cache = {}

    def build(self, l, r):
        if l == r:
            return SegNode(l, r, self.values[l], None, None)
        m = (l + r) // 2
        left, right = self.build(l, m), self.build(m + 1, r)
        return SegNode(l, r, max(left.val, right.val), left, right)
    
    def max_interval(self, l, r, node, is_root):
        if is_root and (l, r) in self.cache:
            return self.cache[(l, r)]
        if l <= node.l and node.r <= r:
            return node.val
        if node.r < l or r < node.l:
            return -inf
        mx = max(
            self.max_interval(l, r, node.left, False),
            self.max_interval(l, r, node.right, False),
        )
        if is_root:
            self.cache[(l, r)] = mx
        return mx

    def binary_search(self, i, req):
        l, r = i, len(self.values) - 1
        while l < r:
            m = (l + r) // 2
            if req < self.max_interval(l, m, self.root, True):
                r = m
            else:
                l = m + 1

        return r if self.values[r] > req else -1


class Solution:
    def leftmostBuildingQueries(
        self, heights: List[int], queries: List[List[int]]
    ) -> List[int]:

        answer = []
        tree = SegTree(heights)

        for l, r in queries:
            l, r = min(l, r), max(l, r)
            if heights[l] < heights[r] or l == r:
                answer.append(r)
            else:
                answer.append(tree.binary_search(r + 1, heights[l]))

        return answer

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