Introduction

In this blog post, we will explore how to solve the LeetCode problem "2617" 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 integer matrix grid. Your initial position is at the top-left cell (0, 0). Starting from the cell (i, j), you can move to one of the following cells: Cells (i, k) with j < k <= grid[i][j] + j (rightward movement), or Cells (k, j) with i < k <= grid[i][j] + i (downward movement). Return the minimum number of cells you need to visit to reach the bottom-right cell (m - 1, n - 1). If there is no valid path, return -1. Example 1: Input: grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]] Output: 4 Explanation: The image above shows one of the paths that visits exactly 4 cells. Example 2: Input: grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]] Output: 3 Explanation: The image above shows one of the paths that visits exactly 3 cells. Example 3: Input: grid = [[2,1,0],[1,0,0]] Output: -1 Explanation: It can be proven that no path exists. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 105 1 <= m n <= 105 0 <= grid[i][j] < m n grid[m - 1][n - 1] == 0

Explanation

Here's a solution to the problem, incorporating efficiency and clarity:

• Breadth-First Search (BFS): Explore the grid level by level, starting from the top-left cell. This guarantees finding the shortest path if one exists.
• Visited Set: Maintain a set of visited cells to avoid cycles and redundant exploration, drastically improving efficiency.

• Early Exit: If the target cell is reached, immediately return the path length.

• Runtime Complexity: O(m*n), where m is the number of rows and n is the number of columns.

• Storage Complexity: O(m*n) in the worst case, for the queue and the visited set.

Code

    from collections import deque

def min_cells_to_reach_bottom_right(grid):
    """
    Finds the minimum number of cells to visit to reach the bottom-right cell in a grid.

    Args:
        grid: A 0-indexed m x n integer matrix.

    Returns:
        The minimum number of cells needed to reach the bottom-right cell, or -1 if no path exists.
    """

    m = len(grid)
    n = len(grid[0])

    if m == 1 and n == 1:
        return 1

    q = deque([(0, 0, 1)])  # (row, col, path_length)
    visited = {(0, 0)}

    while q:
        row, col, path_length = q.popleft()

        if row == m - 1 and col == n - 1:
            return path_length

        # Rightward movement
        for k in range(col + 1, min(col + grid[row][col] + 1, n)):
            if (row, k) not in visited:
                q.append((row, k, path_length + 1))
                visited.add((row, k))

        # Downward movement
        for k in range(row + 1, min(row + grid[row][col] + 1, m)):
            if (k, col) not in visited:
                q.append((k, col, path_length + 1))
                visited.add((k, col))

    return -1