<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 r, c;
    int row[3] = {1, 1, 1};
    int col[3] = {-1, 0, 1};
    
    bool valid(int i, int j) {
        return i >= 0 && j >= 0 && i < r && j < c;
    }

    int find(int i, int j, vector<vector<int>>& matrix, vector<vector<int>>& dp) {
        if (i == 0) return matrix[0][j];
        if (dp[i][j] != INT_MAX) return dp[i][j];
        int best = find(i - 1, j, matrix, dp);
        if (valid(i - 1, j - 1)) best = min(best, find(i - 1, j - 1, matrix, dp));
        if (valid(i - 1, j + 1)) best = min(best, find(i - 1, j + 1, matrix, dp));
        return dp[i][j] = matrix[i][j] + best;
    }

    int minFallingPathSum(vector<vector<int>>& matrix) {
        r = matrix.size(), c = matrix[0].size();
        vector<vector<int>> dp(r, vector<int>(c, INT_MAX));
        int ans = INT_MAX;
        for (int j = 0; j < c; j++) {
            ans = min(ans, find(r - 1, j, matrix, dp));
        }
        return ans;
    }
};

14000