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Minimum Moves to Clean the Classroom

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ArraysGraphsBreadth-First Search

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 space

You 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:

  • The student starts at cell (0, 0) with 2 units of energy.
  • Since cell (1, 0) contains an obstacle 'X', the student cannot move directly downward.
  • A valid sequence of moves to collect all litter is as follows:
    • Move 1: From (0, 0)(0, 1) with 1 unit of energy and 1 unit remaining.
    • Move 2: From (0, 1)(1, 1) to collect the litter 'L'.
  • The student collects all the litter using 2 moves. Thus, the output is 2.

Example 2:

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

Output: 3

Explanation:

  • The student starts at cell (0, 1) with 4 units of energy.
  • A valid sequence of moves to collect all litter is as follows:
    • Move 1: From (0, 1)(0, 0) to collect the first litter 'L' with 1 unit of energy used and 3 units remaining.
    • Move 2: From (0, 0)(1, 0) to 'R' to reset and restore energy back to 4.
    • Move 3: From (1, 0)(1, 1) to collect the second litter 'L'.
  • The student collects all the litter using 3 moves. Thus, the output is 3.

Example 3:

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

Output: -1

Explanation:

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


Clarifying Questions

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:

  1. What should be the output if there are no litter items ('L') in the classroom to begin with?
  2. If the student is on a reset cell 'R' with energy greater than zero, can they choose to reset their energy to full, or does the reset only happen when their energy drops to zero?
  3. Once the student moves away from the starting position 'S', should that cell be treated as an empty space ('.') for the rest of the pathfinding?
  4. Is it a valid move to land on a non-'R' cell if that move causes the student's energy to become exactly zero?
  5. Are the special cells 'S', 'L', and 'R' guaranteed to be in distinct locations, or could a starting cell 'S' also be a litter cell 'L', for example?

Brute Force Solution

Approach

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:

  1. First, find the very first dirty spot in the classroom that needs to be cleaned.
  2. Now, consider every possible way you could place your cleaning tool to clean that specific spot.
  3. For each of these possible placements, imagine you make that move. This results in a new, slightly cleaner classroom state.
  4. From that new state, find the next dirty spot and repeat the process: again, try all possible ways to clean it.
  5. Continue this chain of decisions, exploring every single possible path of cleaning actions until a path leads to a completely clean classroom.
  6. Each time a path successfully cleans the entire room, make a note of how many moves it took.
  7. After exhausting all possible sequences of moves, look at all the final move counts you've recorded and choose the smallest one as the answer.

Code Implementation

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 -1

Big(O) Analysis

Time Complexity
O(n * k^n)The described brute-force approach explores a decision tree of all possible cleaning sequences. Let n be the number of spots and k be the number of ways the tool can be placed to clean a target spot. The recursion can go n levels deep, and at each level it branches k times, resulting in a tree with roughly k^n paths to check. Because each recursive step requires an O(n) scan to find the next dirty spot, the total complexity is the number of paths multiplied by the work per path step, which is O(n * k^n).
Space Complexity
O(N^2)The brute-force exploration implies a recursive approach, which uses space on the call stack. The recursion can go as deep as the number of initial dirty spots, which is at most N, where N is the size of the classroom. The phrase 'results in a new... state' suggests that for each of the O(N) recursive calls in a path, a new copy of the classroom state, an array of size N, is created. This leads to a total auxiliary space complexity of O(N) stack depth times O(N) space per stack frame, which simplifies to O(N^2).

Optimal Solution

Approach

The 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:

  1. First, picture the goal: all students lined up perfectly from shortest to tallest. Let's call this the 'perfect lineup'.
  2. Look at the original, messy line of students and find the largest possible sub-group that is already in a 'correct' sequence. For instance, you might find a short student, and then somewhere later in the line a medium student, and even later a tall one—this group is in the correct relative order.
  3. We want to find the absolute biggest sub-group like this. These students can be thought of as the 'anchors' of the final arrangement; they don't need to be moved with respect to each other.
  4. Every student who is not part of this 'anchor' group is out of place and must be moved to form the perfect lineup.
  5. Therefore, the minimum number of moves required is the total number of students minus the number of students in this largest anchor group.

Code Implementation

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

Big(O) Analysis

Time Complexity
O(n²)The time complexity is driven by the process of finding the longest increasing subsequence, which represents the maximum number of students who can remain in place. For an input of n students, a standard dynamic programming solution is used to find this subsequence. To determine the longest subsequence ending at each student's position, we must compare that student to every student that came before them in the line. This requires a nested loop structure, where for each of the n students, we potentially check all previous students, leading to a total number of operations that approximates n * (n-1) / 2, which simplifies to O(n²).
Space Complexity
O(N)The solution requires finding the largest 'anchor' group, which corresponds to solving the Longest Increasing Subsequence (LIS) problem for the N students. An efficient dynamic programming approach to solve LIS requires an auxiliary array of size N. This array is used to store the length of the longest subsequence that can be formed ending at each student's position. Since the size of this temporary array grows linearly with the number of students, the auxiliary space complexity is O(N).

Edge Cases

The classroom contains no litter ('L') items to collect.
How to Handle:
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').
How to Handle:
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.
How to Handle:
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.
How to Handle:
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.
How to Handle:
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.
How to Handle:
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.
How to Handle:
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.
How to Handle:
The underlying BFS for distance calculation and the main search algorithm are robust to constrained grid layouts.