Solution Explanation: Minimum Falling Path Sum

This problem asks to find the minimum sum of a falling path through a given square matrix. A falling path starts at any element in the first row and can move to the directly below element or diagonally left/right in the next row.

The most efficient approach is using dynamic programming. We maintain a 1D array f to store the minimum falling path sum ending at each column in the current row. The algorithm iterates through the matrix row by row. For each cell in a row, it considers the minimum falling path sum from the three possible cells above it (directly above, diagonally left, and diagonally right) and adds the current cell's value.

Algorithm:

1. Initialization: Initialize a 1D array f with the values of the first row of the matrix. This represents the minimum sum to reach each column in the first row.

2. Iteration: Iterate through the remaining rows of the matrix. For each cell matrix[i][j]:

   • Create a temporary array g to store the minimum path sums for the current row.
   • For each column j, calculate g[j] by considering the minimum of f[j-1], f[j], and f[j+1] (handling boundary cases where j-1 or j+1 is out of bounds). This represents the minimum path sum reaching the cell matrix[i][j] from the previous row.
   • Add matrix[i][j] to g[j].
   • After processing all columns in the current row, copy g to f for the next iteration.

3. Result: After processing all rows, the minimum value in f represents the minimum falling path sum from the top row to the bottom row.

Time and Space Complexity Analysis:

• Time Complexity: O(n²), where n is the number of rows (and columns) in the matrix. We iterate through each cell of the matrix once.

• Space Complexity: O(n). We use a 1D array f of size n to store the minimum path sums at each column. The temporary array g also has size n, so the overall space complexity remains O(n).

Code Explanations (Python3 as example, concepts are similar across other languages):

class Solution:
    def minFallingPathSum(self, matrix: List[List[int]]) -> int:
        n = len(matrix)
        f = [0] * n  # Initialize with the first row
        for i in range(n):
            f[i] = matrix[0][i]
 
        for i in range(1, n): # Iterate from the second row
            g = [0] * n  # Temporary array for the current row
            for j in range(n):
                min_above = float('inf') # Initialize with a large value
                # Consider the three possible cells above
                if j > 0:
                    min_above = min(min_above, f[j - 1])
                min_above = min(min_above, f[j])
                if j < n - 1:
                    min_above = min(min_above, f[j + 1])
                g[j] = min_above + matrix[i][j] # update with current cell value
            f = g # update f for next row
 
        return min(f) # return the minimum value in f after iterating all rows

The code follows the algorithm described above. Note the use of float('inf') to initialize min_above to ensure that the minimum is correctly calculated even when j-1 or j+1 are out of bounds. The other language implementations are structurally similar, adapting syntax and standard library functions as needed for each language.