<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>>& grid) {
        int n = grid.size();
        vector<vector<int>> dp(n, vector<int>(n));
        for(int i=0; i<n; i++) dp[0][i] = grid[0][i];
        for(int i=1; i<n; i++) {
            for(int j=0; j<n; j++) {
                int left = (j-1 >= 0) ? dp[i-1][j-1] : INT_MAX;
                int right = (j+1 < n) ? dp[i-1][j+1] : INT_MAX;
                int straight = dp[i-1][j];
                dp[i][j] = grid[i][j] + min({left,right,straight});
            }
        }
        int ans=INT_MAX;
        for(int i=0; i<n; i++) ans = min(ans, dp[n-1][i]);
        return ans;
    }
};
13700