<p>There is a rectangular brick wall in front of you with <code>n</code> rows of bricks. The <code>i<sup>th</sup></code> row has some number of bricks each of the same height (i.e., one unit) but they can be of different widths. The total width of each row is the same.</p>

<p>Draw a vertical line from the top to the bottom and cross the least bricks. If your line goes through the edge of a brick, then the brick is not considered as crossed. You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.</p>

<p>Given the 2D array <code>wall</code> that contains the information about the wall, return <em>the minimum number of crossed bricks after drawing such a vertical line</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/24/cutwall-grid.jpg" style="width: 493px; height: 577px;" />
<pre>
<strong>Input:</strong> wall = [[1,2,2,1],[3,1,2],[1,3,2],[2,4],[3,1,2],[1,3,1,1]]
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> wall = [[1],[1],[1]]
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li><code>n == wall.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= wall[i].length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= sum(wall[i].length) &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>sum(wall[i])</code> is the same for each row <code>i</code>.</li>
	<li><code>1 &lt;= wall[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

from typing import List
import heapq

class Solution:
    def leastBricks(self, wall: List[List[int]]) -> int:
        total_bricks = 0
        for wI in range(len(wall)):
            w = wall[wI]
            total_bricks += len(w) - 1
            for bI in range(1, len(w)):
                wall[wI][bI] += wall[wI][bI - 1]

        # Trim the last element from each sublist in wall
        for wI in range(len(wall)):
            wall[wI] = wall[wI][:-1]
            heapq.heapify(wall[wI])

        priority_queue = wall[:]
        heapq.heapify(priority_queue)

        current = -1
        at_current = 0
        res = 0
        while total_bricks > 0:
            lowestRow = heapq.heappop(priority_queue)
            if not lowestRow:
                continue
            lowest = heapq.heappop(lowestRow)  # get the smallest element from the row
            total_bricks -= 1

            if current == lowest:
                at_current += 1
            else:
                current = lowest
                at_current = 1
            res = max(res, at_current)

            if lowestRow:
                heapq.heappush(priority_queue, lowestRow)

        return len(wall) - res

# Example usage
# solution = Solution()
# wall = [[1, 2, 2, 1], [3, 1, 2], [1, 3, 2], [2, 4], [3, 1, 2], [1, 3, 1, 1]]
# result = solutionastBricks(wall)
# print(result)

21000