<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:
        # def recursion(i,j):
        #     if j < 0 or i < 0 or j >= cols: #dont take this path
        #         return float("inf")
        #     if i == 0: #destination
        #         return matrix[i][j]
        #     if (i,j) in dp:
        #         return dp[(i,j)]
        #     up = matrix[i][j] + recursion(i-1, j)
        #     left_diagnol = matrix[i][j] + recursion(i-1, j-1)
        #     right_diagnol = matrix[i][j] + recursion(i-1, j+1)
        #     dp[(i,j)] = min(up, min(left_diagnol, right_diagnol))
        #     return dp[(i,j)]

        

        #write memoization code first in notebook.
        ##TABULATION:

        rows = len(matrix)
        cols = len(matrix[0])
        dp = {} 
        for i in range(rows):
            for j in range(cols):
                if i==0:
                    dp[(i,j)] = matrix[i][j]
                else:
                    up = matrix[i][j]
                    if i > 0:
                        up += dp[(i-1, j)]
                    else:
                        up += float("inf")
                    
                    left_diagnol = matrix[i][j]
                    if i > 0 and j > 0:
                        left_diagnol += dp[(i-1, j-1)]
                    else:
                        left_diagnol = float("inf")

                    right_diagnol = matrix[i][j]
                    if i > 0 and j+1 < cols:
                        right_diagnol += dp[(i-1, j+1)]
                    else:
                        right_diagnol = float("inf")

                    dp[(i,j)] = min(up, min(left_diagnol, right_diagnol))
        print(dp)
        ans = float("inf")
        for j in range(cols):
            ans =  min(ans, dp[(rows-1, j)])

        return ans




18900