The problem you're asking about is a classic dynamic programming problem. Here's a step-by-step solution:

1. First, we need to understand the problem. We're given a square matrix and we need to find the minimum sum of a falling path. A falling path starts at any element in the first row and can go to the next row by either going straight down or diagonally down to the left or right.

2. We'll solve this problem using dynamic programming. We'll start by creating a DP table with the same dimensions as the input matrix. This DP table will store the minimum sum of a falling path ending at each cell.

3. We'll initialize the first row of the DP table to be the same as the first row of the input matrix, because the minimum sum of a falling path ending at the first row is just the value of the cell itself.

4. Then, we'll fill up the rest of the DP table row by row. For each cell, we'll look at the three cells above it (directly above, diagonally left, and diagonally right). The minimum sum of a falling path ending at the current cell is the value of the current cell plus the minimum of the values in the three cells above it.

5. We need to be careful about the cells in the first and last column because they don't have three cells above them. For these cells, we'll only consider the two cells that exist.

6. After we've filled up the DP table, the minimum sum of a falling path in the matrix is the minimum value in the last row of the DP table.

Here's some Python code that implements this solution:

```python
def minFallingPathSum(matrix):
n = len(matrix)
dp = [[0]*n for _ in range(n)]
dp[0] = matrix[0]
for i in range(1, n):
for j in range(n):
if j == 0:
dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j+1])
elif j == n-1:
dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j-1])
else:
dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j-1], dp[i-1][j+1])
return min(dp[-1])
```

This code first initializes the DP table and then fills it up row by row. Finally, it returns the minimum value in the last row of the DP table.

Question

The problem you're asking about is a classic dynamic programming problem. Here's a step-by-step solution:

1. First, we need to understand the problem. We're given a square matrix and we need to find the minimum sum of a falling path. A falling path starts at any element in the first row and can go to the next row by either going straight down or diagonally down to the left or right.

2. We'll solve this problem using dynamic programming. We'll start by creating a DP table with the same dimensions as the input matrix. This DP table will store the minimum sum of a falling path ending at each cell.

3. We'll initialize the first row of the DP table to be the same as the first row of the input matrix, because the minimum sum of a falling path ending at the first row is just the value of the cell itself.

4. Then, we'll fill up the rest of the DP table row by row. For each cell, we'll look at the three cells above it (directly above, diagonally left, and diagonally right). The minimum sum of a falling path ending at the current cell is the value of the current cell plus the minimum of the values in the three cells above it.

5. We need to be careful about the cells in the first and last column because they don't have three cells above them. For these cells, we'll only consider the two cells that exist.

6. After we've filled up the DP table, the minimum sum of a falling path in the matrix is the minimum value in the last row of the DP table.

Here's some Python code that implements this solution:

```python
def minFallingPathSum(matrix):
    n = len(matrix)
    dp = [[0]*n for _ in range(n)]
    dp[0] = matrix[0]
    for i in range(1, n):
        for j in range(n):
            if j == 0:
                dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j+1])
            elif j == n-1:
                dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j-1])
            else:
                dp[i][j] = matrix[i][j] + min(dp[i-1][j], dp[i-1][j-1], dp[i-1][j+1])
    return min(dp[-1])
```

This code first initializes the DP table and then fills it up row by row. Finally, it returns the minimum value in the last row of the DP table.

Knowee AI · Accepted Answer