Introduction

In this blog post, we will explore how to solve the LeetCode problem "980" 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 array grid where grid[i][j] could be: 1 representing the starting square. There is exactly one starting square. 2 representing the ending square. There is exactly one ending square. 0 representing empty squares we can walk over. -1 representing obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once. Example 1: Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]] Output: 2 Explanation: We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2) Example 2: Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,2]] Output: 4 Explanation: We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3) Example 3: Input: grid = [[0,1],[2,0]] Output: 0 Explanation: There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 20 1 <= m * n <= 20 -1 <= grid[i][j] <= 2 There is exactly one starting cell and one ending cell.

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Depth-First Search (DFS): Explore all possible paths from the starting cell using DFS.
• Backtracking: If a path hits an obstacle, goes out of bounds, revisits a non-obstacle cell, or doesn't visit all non-obstacle cells exactly once before reaching the end, backtrack and try a different path.

• Count Valid Paths: Increment the count only when a path successfully reaches the end cell after visiting all non-obstacle cells exactly once.

• Runtime Complexity: O(4^m*n), where m and n are the dimensions of the grid. In the worst case, we might explore all possible paths. Storage Complexity: O(m*n) due to the recursion depth of the DFS, corresponding to the stack space.

Code

    def uniquePathsIII(grid):
    """
    Finds the number of unique paths from the start to the end in a grid,
    visiting every non-obstacle cell exactly once.

    Args:
      grid: A 2D list of integers representing the grid.

    Returns:
      The number of unique paths.
    """

    rows = len(grid)
    cols = len(grid[0])

    start_row, start_col = -1, -1
    empty_cells = 0

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                start_row, start_col = i, j
            elif grid[i][j] == 0:
                empty_cells += 1

    def dfs(row, col, visited_count):
        """
        Performs Depth-First Search to find unique paths.

        Args:
          row: The current row index.
          col: The current column index.
          visited_count: The number of non-obstacle cells visited so far.

        Returns:
          The number of unique paths found from the current cell.
        """

        if row < 0 or row >= rows or col < 0 or col >= cols or grid[row][col] == -1:
            return 0

        if grid[row][col] == 2:
            if visited_count == empty_cells + 1:  # +1 for the starting cell
                return 1
            else:
                return 0

        original_value = grid[row][col]
        grid[row][col] = -1  # Mark as visited

        paths = 0
        paths += dfs(row + 1, col, visited_count + 1)
        paths += dfs(row - 1, col, visited_count + 1)
        paths += dfs(row, col + 1, visited_count + 1)
        paths += dfs(row, col - 1, visited_count + 1)

        grid[row][col] = original_value  # Backtrack: unmark the cell
        return paths

    return dfs(start_row, start_col, 0)