Introduction

In this blog post, we will explore how to solve the LeetCode problem "2328" 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 an m x n integer matrix grid, where you can move from a cell to any adjacent cell in all 4 directions. Return the number of strictly increasing paths in the grid such that you can start from any cell and end at any cell. Since the answer may be very large, return it modulo 109 + 7. Two paths are considered different if they do not have exactly the same sequence of visited cells. Example 1: Input: grid = [[1,1],[3,4]] Output: 8 Explanation: The strictly increasing paths are: - Paths with length 1: [1], [1], [3], [4]. - Paths with length 2: [1 -> 3], [1 -> 4], [3 -> 4]. - Paths with length 3: [1 -> 3 -> 4]. The total number of paths is 4 + 3 + 1 = 8. Example 2: Input: grid = [[1],[2]] Output: 3 Explanation: The strictly increasing paths are: - Paths with length 1: [1], [2]. - Paths with length 2: [1 -> 2]. The total number of paths is 2 + 1 = 3. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 1000 1 <= m * n <= 105 1 <= grid[i][j] <= 105

Explanation

Here's the approach to solve this problem:

• Dynamic Programming with Memoization: Use a DP approach where dp[i][j] stores the number of strictly increasing paths starting from cell (i, j). Memoize the results to avoid redundant calculations.
• Depth-First Search (DFS): Implement a DFS function to explore adjacent cells. The DFS function checks if a neighboring cell has a larger value than the current cell. If it does, recursively call the DFS function on the neighbor and add the result to the current cell's path count.

• Iterate and Sum: Iterate through each cell in the grid and start a DFS from that cell. Sum up the number of paths starting from each cell to get the final result.

• Runtime Complexity: O(m * n), Storage Complexity: O(m * n)

Code

    def count_paths(grid: list[list[int]]) -> int:
    """
    Given an m x n integer matrix grid, return the number of strictly increasing paths in the grid
    such that you can start from any cell and end at any cell.
    Since the answer may be very large, return it modulo 109 + 7.
    Two paths are considered different if they do not have exactly the same sequence of visited cells.
    """
    m = len(grid)
    n = len(grid[0])
    dp = [[0] * n for _ in range(m)]
    MOD = 10**9 + 7

    def dfs(i: int, j: int) -> int:
        if dp[i][j] != 0:
            return dp[i][j]

        dp[i][j] = 1  # Path of length 1 starting from this cell
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for dx, dy in directions:
            x = i + dx
            y = j + dy

            if 0 <= x < m and 0 <= y < n and grid[x][y] > grid[i][j]:
                dp[i][j] += dfs(x, y)
                dp[i][j] %= MOD

        return dp[i][j]

    total_paths = 0
    for i in range(m):
        for j in range(n):
            total_paths += dfs(i, j)
            total_paths %= MOD

    return total_paths