<p>Given an <code>n x n</code> array of integers <code>matrix</code>, return <em>the <strong>minimum sum</strong> of any <strong>falling path</strong> through</em> <code>matrix</code>.</p>

<p>A <strong>falling path</strong> starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position <code>(row, col)</code> will be <code>(row + 1, col - 1)</code>, <code>(row + 1, col)</code>, or <code>(row + 1, col + 1)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/03/failing1-grid.jpg" style="width: 499px; height: 500px;" />
<pre>
<strong>Input:</strong> matrix = [[2,1,3],[6,5,4],[7,8,9]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> There are two falling paths with a minimum sum as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/03/failing2-grid.jpg" style="width: 164px; height: 365px;" />
<pre>
<strong>Input:</strong> matrix = [[-19,57],[-40,-5]]
<strong>Output:</strong> -59
<strong>Explanation:</strong> The falling path with a minimum sum is shown.
</pre>

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

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
</ul>

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        R, C = len(matrix), len(matrix[0])
        INF = float('inf')

        graph = {}
        for r in range(R - 1):
            for c in range(C):
                graph[(r, c)] = []
                graph[(r, c)].append((matrix[r + 1][c], r + 1, c))
                if c != 0:
                    graph[(r, c)].append((matrix[r + 1][c - 1], r + 1, c - 1))
                if c != C - 1:
                    graph[(r, c)].append((matrix[r + 1][c + 1], r + 1, c + 1))
        
        pq = []
        heapify(pq)
        distances = [[INF] * C for _ in range(R)]
        for c in range(C):
            distances[0][c] = matrix[0][c]
            heappush(pq, (matrix[0][c], 0, c))
    
        while pq:
            dist, r, c = heappop(pq)
            for weight, nR, nC in graph.get((r, c), []):
                if weight + dist < distances[nR][nC]:
                    distances[nR][nC] = min(distances[nR][nC], weight + dist)
                    heappush(pq, (weight + dist, nR, nC))   

        return min(distances[-1]) 
21900