Introduction

In this blog post, we will explore how to solve the LeetCode problem "1289" 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 integer matrix grid, return the minimum sum of a falling path with non-zero shifts. A falling path with non-zero shifts is a choice of exactly one element from each row of grid such that no two elements chosen in adjacent rows are in the same column. Example 1: Input: grid = [[1,2,3],[4,5,6],[7,8,9]] Output: 13 Explanation: The possible falling paths are: [1,5,9], [1,5,7], [1,6,7], [1,6,8], [2,4,8], [2,4,9], [2,6,7], [2,6,8], [3,4,8], [3,4,9], [3,5,7], [3,5,9] The falling path with the smallest sum is [1,5,7], so the answer is 13. Example 2: Input: grid = [[7]] Output: 7 Constraints: n == grid.length == grid[i].length 1 <= n <= 200 -99 <= grid[i][j] <= 99

Explanation

Here's the solution:

• Dynamic Programming: The core idea is to use dynamic programming to store the minimum falling path sums ending at each cell in the grid. We'll build up the solution row by row.
• Maintaining Two Minimums: For each row, instead of storing the entire DP table of the previous row, we only need to keep track of the two smallest path sums from the previous row along with their corresponding column indices. This significantly optimizes the calculation of the current row's minimum path sums.

• Non-Zero Shifts: When calculating the minimum path sum for a cell in the current row, we ensure that we don't use the cell in the previous row that's in the same column. If the smallest path sum from the previous row comes from the same column, we use the second smallest.

• Runtime Complexity: O(n²), Storage Complexity: O(n)

Code

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

    if n == 1:
        return grid[0][0]

    dp = [0] * n

    # Initialize dp with the first row of the grid
    for j in range(n):
        dp[j] = grid[0][j]

    for i in range(1, n):
        # Find the two smallest values in the previous dp array
        first_min = float('inf')
        second_min = float('inf')
        first_min_index = -1

        for j in range(n):
            if dp[j] < first_min:
                second_min = first_min
                first_min = dp[j]
                first_min_index = j
            elif dp[j] < second_min:
                second_min = dp[j]

        new_dp = [0] * n  # temporary dp array for current row

        for j in range(n):
            if j != first_min_index:
                new_dp[j] = grid[i][j] + first_min
            else:
                new_dp[j] = grid[i][j] + second_min

        dp = new_dp  # Update dp for the next row

    return min(dp)