Introduction

In this blog post, we will explore how to solve the LeetCode problem "931" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix. A falling path 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 (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1). Example 1: Input: matrix = [[2,1,3],[6,5,4],[7,8,9]] Output: 13 Explanation: There are two falling paths with a minimum sum as shown. Example 2: Input: matrix = [[-19,57],[-40,-5]] Output: -59 Explanation: The falling path with a minimum sum is shown. Constraints: n == matrix.length == matrix[i].length 1 <= n <= 100 -100 <= matrix[i][j] <= 100

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: We use dynamic programming to store the minimum falling path sum ending at each cell.
• Bottom-Up Approach: We iterate through the matrix from the second row onwards, updating each cell with the minimum sum from the valid cells in the row above.

• In-Place Optimization: We modify the input matrix directly to store the DP table, avoiding the need for extra space.

• Runtime Complexity: O(n²), where n is the size of the matrix. Storage Complexity: O(1) (in-place modification of the input matrix).

Code

    def minFallingPathSum(matrix):
    n = len(matrix)

    for i in range(1, n):
        for j in range(n):
            # Calculate the minimum sum from the previous row
            if j == 0:
                matrix[i][j] += min(matrix[i - 1][j], matrix[i - 1][j + 1])
            elif j == n - 1:
                matrix[i][j] += min(matrix[i - 1][j - 1], matrix[i - 1][j])
            else:
                matrix[i][j] += min(matrix[i - 1][j - 1], matrix[i - 1][j], matrix[i - 1][j + 1])

    # The minimum falling path sum will be the minimum value in the last row
    return min(matrix[n - 1])