Introduction

In this blog post, we will explore how to solve the LeetCode problem "2684" 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

You are given a 0-indexed m x n matrix grid consisting of positive integers. You can start at any cell in the first column of the matrix, and traverse the grid in the following way: From a cell (row, col), you can move to any of the cells: (row - 1, col + 1), (row, col + 1) and (row + 1, col + 1) such that the value of the cell you move to, should be strictly bigger than the value of the current cell. Return the maximum number of moves that you can perform. Example 1: Input: grid = [[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]] Output: 3 Explanation: We can start at the cell (0, 0) and make the following moves: - (0, 0) -> (0, 1). - (0, 1) -> (1, 2). - (1, 2) -> (2, 3). It can be shown that it is the maximum number of moves that can be made. Example 2: Input: grid = [[3,2,4],[2,1,9],[1,1,7]] Output: 0 Explanation: Starting from any cell in the first column we cannot perform any moves. Constraints: m == grid.length n == grid[i].length 2 <= m, n <= 1000 4 <= m * n <= 105 1 <= grid[i][j] <= 106

Explanation

Here's the approach to solve this problem:

• Dynamic Programming (DP): Use a DP table dp of the same dimensions as the grid to store the maximum number of moves possible from each cell.
• Iterate Backwards: Start from the last column and iterate backwards to the first column. For each cell, check the three possible move directions and update the DP table based on whether a valid move (strictly greater value) can be made.

• Base Case: The base case for the last column is 0 moves since we cannot move further.

• Runtime Complexity: O(m * n)

• Storage Complexity: O(m * n)

Code

    def maxMoves(grid):
    m = len(grid)
    n = len(grid[0])

    dp = [[0] * n for _ in range(m)]

    for j in range(n - 2, -1, -1):
        for i in range(m):
            for dr in [-1, 0, 1]:
                new_row = i + dr
                if 0 <= new_row < m and grid[new_row][j + 1] > grid[i][j]:
                    dp[i][j] = max(dp[i][j], dp[new_row][j + 1] + 1)

    max_moves = 0
    for i in range(m):
        max_moves = max(max_moves, dp[i][0])

    return max_moves