<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();
        
        // DP array to store the minimum path sum for each column in the previous row
        vector<int> dp = matrix[0];  // Initialize dp with the first row values

        // Iterate over each row, starting from the second row
        for (int row = 1; row < n; ++row) {
            vector<int> current(n, 0);  // Temporary array for the current row's dp values
            for (int col = 0; col < n; ++col) {
                // Find the minimum path sum by considering the three possible previous positions
                int best = dp[col];  // Directly above
                if (col > 0) {
                    best = min(best, dp[col - 1]);  // Diagonally left
                }
                if (col < n - 1) {
                    best = min(best, dp[col + 1]);  // Diagonally right
                }
                
                // Update the current dp value
                current[col] = matrix[row][col] + best;
            }
            // Update the dp array to the current row's values
            dp = current;
        }
        
        // The result is the minimum value in the last row of dp
        return *min_element(dp.begin(), dp.end());
    }
};

13800