You are given an m x n grid classroom where a student volunteer is tasked with cleaning up litter scattered around the room. Each cell in the grid is one of the following:
'S': Starting position of the student'L': Litter that must be collected (once collected, the cell becomes empty)'R': Reset area that restores the student's energy to full capacity, regardless of their current energy level (can be used multiple times)'X': Obstacle the student cannot pass through'.': Empty spaceYou are also given an integer energy, representing the student's maximum energy capacity. The student starts with this energy from the starting position 'S'.
Each move to an adjacent cell (up, down, left, or right) costs 1 unit of energy. If the energy reaches 0, the student can only continue if they are on a reset area 'R', which resets the energy to its maximum capacity energy.
Return the minimum number of moves required to collect all litter items, or -1 if it's impossible.
Example 1:
Input: classroom = ["S.", "XL"], energy = 2
Output: 2
Explanation:
(0, 0) with 2 units of energy.(1, 0) contains an obstacle 'X', the student cannot move directly downward.(0, 0) → (0, 1) with 1 unit of energy and 1 unit remaining.(0, 1) → (1, 1) to collect the litter 'L'.Example 2:
Input: classroom = ["LS", "RL"], energy = 4
Output: 3
Explanation:
(0, 1) with 4 units of energy.(0, 1) → (0, 0) to collect the first litter 'L' with 1 unit of energy used and 3 units remaining.(0, 0) → (1, 0) to 'R' to reset and restore energy back to 4.(1, 0) → (1, 1) to collect the second litter 'L'.Example 3:
Input: classroom = ["L.S", "RXL"], energy = 3
Output: -1
Explanation:
No valid path collects all 'L'.
Constraints:
1 <= m == classroom.length <= 201 <= n == classroom[i].length <= 20classroom[i][j] is one of 'S', 'L', 'R', 'X', or '.'1 <= energy <= 50'S' in the grid.'L' cells in the grid.When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The goal is to find the fewest actions to clean a row of dirty spots using a special tool. The brute force method explores every single possible sequence of cleaning actions to see which one gets the job done with the minimum number of steps.
Here's how the algorithm would work step-by-step:
def min_moves_to_clean_brute_force(classroom):
if not classroom or not classroom[0]:
return 0
number_of_rows = len(classroom)
number_of_cols = len(classroom[0])
memoization_cache = {}
def is_completely_clean(classroom_state_tuple):
for row_tuple in classroom_state_tuple:
if 'D' in row_tuple:
return False
return True
def perform_cleaning_move(classroom_state_tuple, move_row, move_col):
mutable_grid = [list(row_tuple) for row_tuple in classroom_state_tuple]
for row_offset in range(2):
for col_offset in range(2):
target_row = move_row + row_offset
target_col = move_col + col_offset
if 0 <= target_row < number_of_rows and 0 <= target_col < number_of_cols:
mutable_grid[target_row][target_col] = 'C'
return tuple(tuple(row_list) for row_list in mutable_grid)
def explore_all_move_sequences(current_classroom_state):
# To avoid re-solving for the same classroom layout, we cache results for states we see.
if current_classroom_state in memoization_cache:
return memoization_cache[current_classroom_state]
# Base case: if the room is clean, a path is complete and requires 0 more moves.
if is_completely_clean(current_classroom_state):
return 0
minimum_moves_for_this_path = float('inf')
# This loop tries every possible cleaning move from the current state to explore all paths.
for row_index in range(number_of_rows):
for col_index in range(number_of_cols):
next_classroom_state = perform_cleaning_move(current_classroom_state, row_index, col_index)
# A move must change the classroom state, otherwise we risk infinite recursion on a dead-end.
if next_classroom_state != current_classroom_state:
result_from_next_state = explore_all_move_sequences(next_classroom_state)
if result_from_next_state != float('inf'):
minimum_moves_for_this_path = min(minimum_moves_for_this_path, 1 + result_from_next_state)
memoization_cache[current_classroom_state] = minimum_moves_for_this_path
return minimum_moves_for_this_path
initial_state_tuple = tuple(tuple(row) for row in classroom)
final_answer = explore_all_move_sequences(initial_state_tuple)
return final_answer if final_answer != float('inf') else -1The clever trick is to figure out the maximum number of students who can be left in place, rather than tracking individual moves. By identifying the largest group of students already in a correct height order relative to each other, we know everyone else must be moved.
Here's how the algorithm would work step-by-step:
def minimum_moves_to_clean_classroom(messes_per_section):
number_of_sections = len(messes_per_section)
if not number_of_sections:
return 0
total_messes = sum(messes_per_section)
# If the total messes cannot be evenly distributed, a balanced state is impossible to achieve.
if total_messes % number_of_sections != 0:
return -1
# Calculate the target number of messes each section must have for the classroom to be 'clean'.
target_messes_per_section = total_messes // number_of_sections
total_moves_required = 0
# The total moves is the sum of all items that must be relocated from over-filled sections.
for current_section_messes in messes_per_section:
if current_section_messes > target_messes_per_section:
total_moves_required += current_section_messes - target_messes_per_section
return total_moves_required| Case | How to Handle |
|---|---|
| The classroom contains no litter ('L') items to collect. | The solution should return 0 immediately as no moves are required to complete the already-finished task. |
| A litter item is unreachable, for instance, completely surrounded by obstacles ('X'). | The algorithm will find no valid path to the state where that litter is collected, correctly returning -1. |
| The initial energy is insufficient to reach even the closest litter or reset area. | The search will exhaust all possibilities from the start without reaching any litter, thus concluding it's impossible and returning -1. |
| The grid contains no reset areas ('R'), forcing the entire task to be completed on a single energy bar. | The pathfinding logic correctly handles the absence of reset areas by only considering direct paths between objectives. |
| A path to a litter item consumes the student's exact remaining energy, leaving them with zero. | The state correctly reflects zero remaining energy, requiring a trip to a reset area for any subsequent move. |
| The optimal path involves a detour to a reset area even when a direct path to the next litter is possible but leaves too little energy for later. | A shortest path algorithm like Dijkstra explores both direct and reset-area paths, ensuring it finds the true global minimum moves. |
| Two paths to the same litter with the same items collected take equal moves but leave different remaining energy. | The algorithm prioritizes the path leaving more residual energy, as it provides greater flexibility for future moves. |
| All key locations (S, L, R) are in a single row or column. | The underlying BFS for distance calculation and the main search algorithm are robust to constrained grid layouts. |