Minimum Falling Path Sum II
<p>Given an <code>n x n</code> integer matrix <code>grid</code>, return <em>the minimum sum of a <strong>falling path with non-zero shifts</strong></em>.</p>

<p>A <strong>falling path with non-zero shifts</strong> is a choice of exactly one element from each row of <code>grid</code> such that no two elements chosen in adjacent rows are in the same column.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1289.Minimum%20Falling%20Path%20Sum%20II/images/falling-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> 
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is&nbsp;[1,5,7], so the answer is&nbsp;13.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[7]]
<strong>Output:</strong> 7
</pre>

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

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>-99 &lt;= grid[i][j] &lt;= 99</code></li>
</ul>
Array; Dynamic Programming; Matrix
class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        n = len(grid)
        f = [0] * n
        for row in grid:
            g = row[:]
            for i in range(n):
                g[i] += min((f[j] for j in range(n) if j != i), default=0)
            f = g
        return min(f)
