Introduction

In this blog post, we will explore how to solve the LeetCode problem "1263" 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

A storekeeper is a game in which the player pushes boxes around in a warehouse trying to get them to target locations. The game is represented by an m x n grid of characters grid where each element is a wall, floor, or box. Your task is to move the box 'B' to the target position 'T' under the following rules: The character 'S' represents the player. The player can move up, down, left, right in grid if it is a floor (empty cell). The character '.' represents the floor which means a free cell to walk. The character '#' represents the wall which means an obstacle (impossible to walk there). There is only one box 'B' and one target cell 'T' in the grid. The box can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. This is a push. The player cannot walk through the box. Return the minimum number of pushes to move the box to the target. If there is no way to reach the target, return -1. Example 1: Input: grid = [["#","#","#","#","#","#"], ["#","T","#","#","#","#"], ["#",".",".","B",".","#"], ["#",".","#","#",".","#"], ["#",".",".",".","S","#"], ["#","#","#","#","#","#"]] Output: 3 Explanation: We return only the number of times the box is pushed. Example 2: Input: grid = [["#","#","#","#","#","#"], ["#","T","#","#","#","#"], ["#",".",".","B",".","#"], ["#","#","#","#",".","#"], ["#",".",".",".","S","#"], ["#","#","#","#","#","#"]] Output: -1 Example 3: Input: grid = [["#","#","#","#","#","#"], ["#","T",".",".","#","#"], ["#",".","#","B",".","#"], ["#",".",".",".",".","#"], ["#",".",".",".","S","#"], ["#","#","#","#","#","#"]] Output: 5 Explanation: push the box down, left, left, up and up. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 20 grid contains only characters '.', '#', 'S', 'T', or 'B'. There is only one character 'S', 'B', and 'T' in the grid.

Explanation

Here's the solution to the storekeeper problem:

• High-Level Approach:

  • Use Breadth-First Search (BFS) to explore possible box positions. The state in the BFS consists of the box's position and the player's position.
  • Maintain a visited set to avoid revisiting states, optimizing efficiency.
  • The BFS queue stores tuples of (pushes, box_row, box_col, player_row, player_col).

• Complexity:

  • Runtime Complexity: O(m*n*m*n) where m is the number of rows and n is the number of columns. The BFS explores all possible box positions (m*n), and for each box position, it explores possible player positions (m*n) to check if the player can push the box.
  • Space Complexity: O(m*n*m*n) to store the visited states.

Code

    from collections import deque

def minPushBox(grid):
    m, n = len(grid), len(grid[0])
    player_row, player_col = -1, -1
    box_row, box_col = -1, -1
    target_row, target_col = -1, -1

    for i in range(m):
        for j in range(n):
            if grid[i][j] == 'S':
                player_row, player_col = i, j
            elif grid[i][j] == 'B':
                box_row, box_col = i, j
            elif grid[i][j] == 'T':
                target_row, target_col = i, j

    def is_valid(row, col):
        return 0 <= row < m and 0 <= col < n and grid[row][col] != '#'

    def can_reach(start_row, start_col, target_row, target_col, box_row, box_col):
        visited = set()
        queue = deque([(start_row, start_col)])
        visited.add((start_row, start_col))

        while queue:
            row, col = queue.popleft()
            if row == target_row and col == target_col:
                return True

            directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
            for dr, dc in directions:
                new_row, new_col = row + dr, col + dc
                if is_valid(new_row, new_col) and (new_row, new_col) != (box_row, box_col) and (new_row, new_col) not in visited:
                    queue.append((new_row, new_col))
                    visited.add((new_row, new_col))

        return False

    queue = deque([(0, box_row, box_col, player_row, player_col)])
    visited = set()
    visited.add((box_row, box_col, player_row, player_col))
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while queue:
        pushes, box_row, box_col, player_row, player_col = queue.popleft()

        if box_row == target_row and box_col == target_col:
            return pushes

        for dr, dc in directions:
            new_box_row, new_box_col = box_row + dr, box_col + dc
            new_player_row, new_player_col = box_row - dr, box_col - dc

            if is_valid(new_box_row, new_box_col) and is_valid(new_player_row, new_player_col) and can_reach(player_row, player_col, new_player_row, new_player_col, box_row, box_col):
                if (new_box_row, new_box_col, box_row, box_col) not in visited:
                    queue.append((pushes + 1, new_box_row, new_box_col, box_row, box_col))
                    visited.add((new_box_row, new_box_col, box_row, box_col))

    return -1