<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://assets.leetcode.com/uploads/2021/08/10/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>

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:

        length = len(grid)
        dp = [[0 for _ in range(length)] for _ in range(length)]
        

        heap = []
        for i in range(length -1, -1, -1):
            prev_heap = heap
            heap = []
            for j in range(length):
                if i == length - 1:
                    dp[i][j] = grid[i][j]
                else:
                    value = prev_heap[0][0]
                    if prev_heap[0][1] == j:
                        save = heapq.heappop(prev_heap)
                        value = prev_heap[0][0]
                        heapq.heappush(prev_heap, save)
                    dp[i][j] = grid[i][j] + value
                    
                heapq.heappush(heap, (dp[i][j], j))
        print(dp)
        print(heap)
        return heap[0][0]
        
20700