#3568

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Moves to Clean the Classroom

Solve the "Minimum Moves to Clean the Classroom" problem using array scanning and hash lookup for an optimized solution.

array hash table bit manipulation breadth first search matrix

Problem Statement

In this problem, you are given an m x n grid representing a classroom. Each cell in the grid contains one of the following characters: 'S' for the starting position, 'L' for litter to be collected, 'R' for reset areas that restore energy, 'X' for obstacles, or '.' for empty spaces. A student volunteer must collect all the litter ('L') while managing their energy. Starting at the position 'S', the student has a maximum energy capacity. Each move to an adjacent cell costs 1 unit of energy.

If the energy reaches 0, the student can only continue if they are on a reset area 'R', which restores their energy to its maximum. Your task is to determine the minimum number of moves needed to collect all 'L' cells and return to the starting position. If it is impossible to collect all 'L' or if the energy runs out without reaching all 'L', return -1.

Examples

Example 1

Input: classroom = ["S.", "XL"], energy = 2

Output: 2

Example 2

Input: classroom = ["LS", "RL"], energy = 4

Output: 3

Example 3

Input: classroom = ["L.S", "RXL"], energy = 3

Output: -1

No valid path collects all 'L' .

Constraints

1 <= m == classroom.length <= 20
1 <= n == classroom[i].length <= 20
classroom[i][j] is one of 'S', 'L', 'R', 'X', or '.'
1 <= energy <= 50
There is exactly one 'S' in the grid.
There are at most 10 'L' cells in the grid.

Solution Approach

BFS with States (x, y, mask, energy, steps)

The problem can be solved efficiently using BFS. For each state, track the student's position (x, y), energy level, the collected litter mask (mask), and the number of steps taken. Starting from the 'S' position, the algorithm explores all possible moves, updating energy and the collected litter mask. A state is represented as (x, y, mask, energy, steps), and BFS explores all possible transitions while ensuring energy limits are respected.

Array Scanning with Hash Lookup

Array scanning helps identify key elements ('S', 'L', 'R', 'X', '.') in the grid, and hash lookup provides an efficient way to track visited states. By scanning the grid and recording the positions of the 'L' cells, the algorithm can efficiently check if all litter has been collected by examining the state mask. This allows for quick identification of which litter has been collected and which still needs to be gathered.

Energy Management with Reset Areas

Energy management is crucial in this problem, especially when energy reaches 0. The student can only continue moving if they land on a reset area 'R', which restores their energy. By modeling energy consumption and reset areas within the BFS states, the algorithm ensures that energy is optimally used and that the student can continue collecting litter as long as energy is available or reset areas are encountered.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depends on the approach used. In the worst case, BFS explores all possible grid states, including all positions, energy levels, and litter collection masks. Given that there are at most 10 litter cells and the energy is capped at 50, the overall complexity is influenced by the number of possible states and the transitions between them, making it O(m * n * 2^L * energy), where L is the number of litter cells and energy is the energy capacity.

What Interviewers Usually Probe

The candidate demonstrates understanding of BFS and state space traversal.
The candidate emphasizes efficient energy management and use of reset areas.
The candidate shows how array scanning and hash lookup can optimize the solution.

Common Pitfalls or Variants

Common pitfalls

Failing to account for energy depletion and the use of reset areas.
Incorrectly tracking visited states, which can lead to unnecessary recalculations.
Not handling cases where it is impossible to collect all litter due to energy constraints.

Follow-up variants

Increasing grid size beyond the 20x20 limit to test scalability.
Introducing more complex obstacles ('X') or additional reset areas ('R').
Adding a time constraint for energy management to increase difficulty.

FAQ

How does BFS help solve the Minimum Moves to Clean the Classroom problem?

BFS is ideal for this problem because it explores all possible states, ensuring that the shortest path to collect all litter is found while managing energy constraints.

What role do reset areas ('R') play in this problem?

Reset areas restore energy when the student runs out, allowing them to continue collecting litter. They are crucial for extending the exploration path in energy-constrained scenarios.

What happens if there is no valid path to collect all litter?

If no valid path exists to collect all 'L' due to energy constraints or obstacles, the solution should return -1.

How can array scanning optimize this problem?

Array scanning helps identify and track important positions ('S', 'L', 'R', 'X'), and hash lookup efficiently manages the state transitions during BFS traversal.

How do you handle energy limits in the solution?

Energy limits are managed by tracking the energy at each BFS state and using reset areas ('R') to restore energy when it reaches 0, ensuring continuous movement.

terminal

Solution

Solution 1: BFS

We can use Breadth-First Search (BFS) to solve this problem. First, we need to find the student's starting position and record the locations of all garbage. Then, we can use BFS to explore all possible paths starting from the initial position, while tracking the current energy and the collected garbage.

class Solution:
    def minMoves(self, classroom: List[str], energy: int) -> int:
        m, n = len(classroom), len(classroom[0])
        d = [[0] * n for _ in range(m)]
        x = y = cnt = 0
        for i, row in enumerate(classroom):
            for j, c in enumerate(row):
                if c == "S":
                    x, y = i, j
                elif c == "L":
                    d[i][j] = cnt
                    cnt += 1
        if cnt == 0:
            return 0
        vis = [
            [[[False] * (1 << cnt) for _ in range(energy + 1)] for _ in range(n)]
            for _ in range(m)
        ]
        q = [(x, y, energy, (1 << cnt) - 1)]
        vis[x][y][energy][(1 << cnt) - 1] = True
        dirs = (-1, 0, 1, 0, -1)
        ans = 0
        while q:
            t = q
            q = []
            for i, j, cur_energy, mask in t:
                if mask == 0:
                    return ans
                if cur_energy <= 0:
                    continue
                for k in range(4):
                    x, y = i + dirs[k], j + dirs[k + 1]
                    if 0 <= x < m and 0 <= y < n and classroom[x][y] != "X":
                        nxt_energy = (
                            energy if classroom[x][y] == "R" else cur_energy - 1
                        )
                        nxt_mask = mask
                        if classroom[x][y] == "L":
                            nxt_mask &= ~(1 << d[x][y])
                        if not vis[x][y][nxt_energy][nxt_mask]:
                            vis[x][y][nxt_energy][nxt_mask] = True
                            q.append((x, y, nxt_energy, nxt_mask))
            ans += 1
        return -1

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