<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 {
public:
    int minFallingPathSum(vector<vector<int>>& matrix) {
        int n = matrix.size();
        if (n == 0) return 0;
        
        // Create a dp array initialized with matrix values as we will overwrite it
        vector<vector<int>> dp(n, vector<int>(n, 0));

        // Initialize the last row of dp, as this is the base case in recursion
        for (int i = 0; i < n; i++) {
            dp[n-1][i] = matrix[n-1][i];
        }

        // Fill the dp table from the second-last row up to the first row
        for (int row = n - 2; row >= 0; row--) {
            for (int col = 0; col < n; col++) {
                int down = dp[row + 1][col]; // Move down
                int dgR = (col + 1 < n) ? dp[row + 1][col + 1] : 1e9; // Move diagonally right
                int dgL = (col - 1 >= 0) ? dp[row + 1][col - 1] : 1e9; // Move diagonally left

                dp[row][col] = matrix[row][col] + min(down, min(dgR, dgL));
            }
        }

        // The answer will be the minimum value in the first row
        int ans = INT_MAX;
        for (int i = 0; i < n; i++) {
            ans = min(ans, dp[0][i]);
        }

        return ans;
    }
};

13800