<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]:
        n = len(heights)
        block_size = int(n ** 0.5) + 1
        blocks = [0] * block_size

        # Preprocess queries to keep the original index
        queries = [sorted(query) + [index] for index, query in enumerate(queries)]

        # Process heights and update blocks
        for index, height in enumerate(heights):
            blocks[index // block_size] = max(height, blocks[index // block_size])

        # Sort queries based on block index
        queries.sort(key=lambda x: x[0] // block_size)

        result = [-1] * len(queries)

        # Iterate through the sorted queries
        for query in queries:
            left, right, index = query

            # Directly set the result if right building is greater than left building
            if heights[right] > heights[left]:
                result[index] = right
                continue

            # Set the result if the left and right indices are the same
            elif right == left:
                result[index] = right
                continue

            right_block = right // block_size
            right_block_val = blocks[right_block]
            flag = True

            # Check if the right block is higher than both left and right buildings
            if right_block_val > heights[left] and right_block_val > heights[right]:
                # Iterate through the buildings in the block to find the leftmost building
                for i, value in enumerate(heights[right_block * block_size : block_size * (right_block + 1) + 1]):
                    if value > heights[left] and value > heights[right] and right_block * block_size + i > right:
                        result[index] = right_block * block_size + i
                        flag = False
                        break
                else:
                    right_block += 1

            else:
                right_block += 1
                flag = True

            # Continue iterating to find the leftmost building
            if flag:
                while right_block < len(blocks) and (blocks[right_block] < heights[left] or blocks[right_block] < heights[right]):
                    right_block += 1
                if right_block >= len(blocks):
                    continue

                # Iterate through the buildings in the block to find the leftmost building
                for i, value in enumerate(heights[right_block * block_size : block_size * (right_block + 1) + 1]):
                    if value > heights[left] and value > heights[right]:
                        result[index] = right_block * block_size + i
                        break

        return result
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