<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 {
private:
    int n;
    int dp[101][101];
    vector<vector<int>> arr;

    void set_dp() {
        for (int i = 0; i <= 100; i++) {
            for (int j = 0; j <= 100; j++) {
                dp[i][j] = 1e9;
            }
        }
    }

    int rec(int i, int j) {
        // pruning
        if (j < 0 || j >= n) {
            return 1e9;
        }
        // basecase
        if (i == n) {
            return 0;
        }

        // cache check
        if (dp[i][j] != 1e9) {
            return dp[i][j];
        }
        // compute
        int ans = 1e9;
        ans = min(ans, arr[i][j] + rec(i + 1, j));     // down
        ans = min(ans, arr[i][j] + rec(i + 1, j - 1)); // diagonally left
        ans = min(ans, arr[i][j] + rec(i + 1, j + 1)); // diagonally right
        // save and return
        return dp[i][j] = ans;
    }

public:
    int minFallingPathSum(vector<vector<int>>& matrix) {
        n = matrix.size();

        arr = matrix;
        set_dp();
        int ans = 1e9;
        for (int i = 0; i < n; i++) {
            ans = min(ans, rec(0, i));
            set_dp();
        }
        return ans;
    }
};
14800