<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:
        m, n = len(matrix), len(matrix[0])
        
        memo = [[-sys.maxsize] * n for _ in range(m)]
        
        def dfs(row, col):
            if m - 1 == row:
                return matrix[row][col]
            
            if -sys.maxsize != memo[row][col]:
                return memo[row][col]
            
            dirs = ((row + 1, col - 1), (row + 1, col), (row + 1, col + 1))

            min_sum = sys.maxsize

            for new_r, new_c in dirs:
                if new_r not in range(m) or new_c not in range(n):
                    continue
                min_sum = min(min_sum, dfs(new_r, new_c))
            
            memo[row][col] = matrix[row][col] + min_sum 
            return matrix[row][col] + min_sum
        

        # return min(dfs(0, col) for col in range(n))
        prev= copy.copy(matrix[-1])
        dp = [sys.maxsize] * n

        for i in range(m - 2, -1, -1):
            for j in range(n):
                dp[j] = min(dp[j], prev[j - 1]) if j > 0 else dp[j]
                dp[j] = min(dp[j], prev[j])
                dp[j] = min(dp[j], prev[j + 1]) if j + 1 < n else dp[j]
                dp[j] += matrix[i][j]
            
            dp, prev = prev, dp
            dp = [sys.maxsize] * n
        
        return min(prev[col] for col in range(n))

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