<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 Solution:
    def leftmostBuildingQueries(self, heights: List[int], queries: List[List[int]]) -> List[int]:
        # Monotonic stack + binary search
        #
        # Process queries from right to left.
        # Maintain a monotonic stack of all heights seen so far st
        #   we prefer going to a heigher building as far left as possible
        queries = [(ai, bi, i, max(ai, bi)) for i, (ai, bi) in enumerate(queries)]
        queries.sort(key=lambda x: x[-1])

        def bisect_left(rev, x):
            # Search for x in rev 
            # NOTE: rev is sorted decreasingly wrt heights
            lo, hi = 0, len(rev)
            while lo < hi:
                mid = (lo + hi) // 2
                if heights[rev[mid]] > x:
                    lo = mid + 1
                else:
                    hi = mid
            return lo

        n = len(queries)
        ans = n * [-1]
        # Monotonic stack decreasing wrt heights
        stack = []
        for i in range(len(heights)-1, -1, -1):
            # Faster to do first as it prunes the stack.
            while stack and heights[i] >= heights[stack[-1]]:
                stack.pop()
            stack.append(i)

            while queries and queries[-1][-1] == i:
                ai, bi, qi, mi = queries.pop()
                if ai == bi:
                    ans[qi] = ai
                elif ai == mi and heights[ai] > heights[bi]:
                    ans[qi] = ai
                elif bi == mi and heights[bi] > heights[ai]:
                    ans[qi] = bi
                else:
                    # We have to try and move to another building
                    mh = max(heights[ai], heights[bi])
                    # O(n) implementation 
                    # k = 0
                    # while k < len(stack) and heights[stack[k]] > mh:
                    #     k += 1
                    # k -= 1
                    k = bisect_left(stack, mh) - 1
                    if k >= 0:
                        ans[qi] = stack[k]

            if not queries:
                break

        return ans

        
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