Minimum Moves to Spread Stones Over Grid

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Guidewire

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ArraysGreedy Algorithms

You are given a 0-indexed 2D integer matrix grid of size 3 * 3, representing the number of stones in each cell. The grid contains exactly 9 stones, and there can be multiple stones in a single cell.

In one move, you can move a single stone from its current cell to any other cell if the two cells share a side.

Return the minimum number of moves required to place one stone in each cell.

Example 1:

Input: grid = [[1,1,0],[1,1,1],[1,2,1]]
Output: 3
Explanation: One possible sequence of moves to place one stone in each cell is: 
1- Move one stone from cell (2,1) to cell (2,2).
2- Move one stone from cell (2,2) to cell (1,2).
3- Move one stone from cell (1,2) to cell (0,2).
In total, it takes 3 moves to place one stone in each cell of the grid.
It can be shown that 3 is the minimum number of moves required to place one stone in each cell.

Example 2:

Input: grid = [[1,3,0],[1,0,0],[1,0,3]]
Output: 4
Explanation: One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (0,1) to cell (0,2).
2- Move one stone from cell (0,1) to cell (1,1).
3- Move one stone from cell (2,2) to cell (1,2).
4- Move one stone from cell (2,2) to cell (2,1).
In total, it takes 4 moves to place one stone in each cell of the grid.
It can be shown that 4 is the minimum number of moves required to place one stone in each cell.

Constraints:

grid.length == grid[i].length == 3
0 <= grid[i][j] <= 9
Sum of grid is equal to 9.

Try coding it in LeetCode

Minimum Moves to Spread Stones Over Grid

Medium

Guidewire

0 views

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed 2D integer matrix grid of size 3 * 3, representing the number of stones in each cell. The grid contains exactly 9 stones, and there can be multiple stones in a single cell.

In one move, you can move a single stone from its current cell to any other cell if the two cells share a side.

Return the minimum number of moves required to place one stone in each cell.

Example 1:

Input: grid = [[1,1,0],[1,1,1],[1,2,1]]
Output: 3
Explanation: One possible sequence of moves to place one stone in each cell is: 
1- Move one stone from cell (2,1) to cell (2,2).
2- Move one stone from cell (2,2) to cell (1,2).
3- Move one stone from cell (1,2) to cell (0,2).
In total, it takes 3 moves to place one stone in each cell of the grid.
It can be shown that 3 is the minimum number of moves required to place one stone in each cell.

Example 2:

Input: grid = [[1,3,0],[1,0,0],[1,0,3]]
Output: 4
Explanation: One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (0,1) to cell (0,2).
2- Move one stone from cell (0,1) to cell (1,1).
3- Move one stone from cell (2,2) to cell (1,2).
4- Move one stone from cell (2,2) to cell (2,1).
In total, it takes 4 moves to place one stone in each cell of the grid.
It can be shown that 4 is the minimum number of moves required to place one stone in each cell.

Constraints:

grid.length == grid[i].length == 3
0 <= grid[i][j] <= 9
Sum of grid is equal to 9.

Try coding it in LeetCode

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:

What are the dimensions of the grid (number of rows and columns), and what are the possible ranges for these dimensions?
What are the possible values for the number of stones in each cell, and can a cell have a negative number of stones (representing a shortage)?
If it's impossible to spread the stones such that each cell has exactly one stone, what should I return (e.g., -1, Integer.MAX_VALUE, throw an exception)?
Are the starting and ending cells for each move (moving one stone from one cell to another) adjacent either horizontally or vertically, or can stones be moved diagonally, or across multiple cells in a single move?
Is the grid represented as a 2D array of integers, and are we guaranteed that the total number of stones across the entire grid will always equal the number of cells in the grid?

Brute Force Solution

Approach

Imagine you have a grid where some squares have extra stones and others are empty. The brute force approach tries every single possible way to move stones from squares with extras to squares that need them. We explore all combinations to find the fewest moves.

Here's how the algorithm would work step-by-step:

Start by looking at each square that has too many stones.
For each of these squares, consider every other square that needs stones.
Try moving one stone from the overfull square to the needy square.
Now, see if this makes things better or worse, keeping track of how many moves you've made.
Next, try moving two stones, three stones, and so on, until the overfull square has just the right amount or the needy square is full.
Repeat this process for every possible pair of overfull and needy squares.
Keep doing this until every square has the right number of stones.
Since you've tried many different moves, keep track of how many total moves each combination takes.
Finally, choose the combination of moves that uses the fewest moves overall.

Code Implementation

def minimum_moves_to_spread_stones_brute_force(grid):
    rows = len(grid)
    cols = len(grid[0])
    min_moves = float('inf')

    def calculate_moves(current_grid):
        moves = 0
        for row in range(rows):
            for col in range(cols):
                moves += abs(current_grid[row][col] - 1)
        return moves // 2

    def is_balanced(current_grid):
        for row in range(rows):
            for col in range(cols):
                if current_grid[row][col] != 1:
                    return False
        return True

    def solve(current_grid, moves_taken):
        nonlocal min_moves

        if is_balanced(current_grid):
            min_moves = min(min_moves, moves_taken)
            return

        # Find a cell with stones to move
        for source_row in range(rows):
            for source_col in range(cols):
                if current_grid[source_row][source_col] > 1:

                    # Find a cell that needs stones
                    for dest_row in range(rows):
                        for dest_col in range(cols):
                            if current_grid[dest_row][dest_col] < 1:

                                # Try moving stones from the source to the destination
                                stones_to_move = min(current_grid[source_row][source_col] - 1, 1 - current_grid[dest_row][dest_col])
                                if stones_to_move > 0:

                                    # Create a new grid to try out the move
                                    new_grid = [row[:] for row in current_grid]
                                    new_grid[source_row][source_col] -= stones_to_move
                                    new_grid[dest_row][dest_col] += stones_to_move

                                    # Recursively call solve with the updated grid and moves
                                    solve(new_grid, moves_taken + stones_to_move)

    # Make a deep copy of the input grid to avoid modifying the original
    initial_grid = [row[:] for row in grid]
    solve(initial_grid, 0)

    if min_moves == float('inf'):
        return -1
    return min_moves

Big(O) Analysis

Time Complexity

O(k^m) – The brute force approach explores all possible combinations of moving stones between overfull and needy cells. Let k be the maximum number of stones that could be moved from a single overfull cell to a needy cell. Let m be the number of pairs of overfull and needy cells. For each pair, we are iterating through k possible moves (1 to k stones). Since we iterate through every combination of these moves across the m pairs, the time complexity grows exponentially with the number of stone movements (k) raised to the power of the number of pairs (m). Therefore, the runtime complexity can be expressed as O(k^m).

Space Complexity

O(N!) – The described brute force approach implicitly explores all possible combinations of stone movements between overfull and needy squares. In the worst-case scenario, where every square needs to be considered with every other square, the number of possible combinations grows factorially with the number of squares (N). Keeping track of each move combination, the number of moves made, and determining if it makes the situation 'better or worse' will involve storing multiple combinations that could theoretically approach N! In reality, we are building a call stack to explore each possibility. Therefore, the space complexity can be estimated as O(N!).

Optimal Solution

Approach

The core idea is to think of this as a transportation problem. We want to move 'stones' (excess amounts) from cells with too many stones to cells that need more stones, minimizing the total moves.

Here's how the algorithm would work step-by-step:

First, find all the cells that have more stones than they should (sources) and all the cells that need more stones (destinations).
Think of each source cell wanting to send stones to the destination cells with the smallest possible number of moves.
For each source, find the closest destination and move as many stones as you can to it without exceeding the destination's need or the source's supply.
Update the stone counts for both the source and destination cells after the move.
Repeat this process, finding the next closest destination for the current source or another source until all the cells are satisfied with exactly one stone each.
Sum up all the moves you made during this process. The total will be the minimum number of moves needed.

Code Implementation

def minimum_moves_to_spread_stones(grid):
    rows = len(grid)
    cols = len(grid[0])

    sources = []
    destinations = []

    for row in range(rows):
        for col in range(cols):
            if grid[row][col] > 1:
                sources.append((row, col))
            elif grid[row][col] < 1:
                destinations.append((row, col))

    total_moves = 0

    while sources and destinations:
        # Iterate through sources until empty
        source_row, source_col = sources[0]
        if grid[source_row][source_col] <= 1:
            sources.pop(0)
            continue

        min_distance = float('inf')
        closest_destination = None

        for dest_row, dest_col in destinations:
            distance = abs(source_row - dest_row) + abs(source_col - dest_col)
            if distance < min_distance:
                min_distance = distance
                closest_destination = (dest_row, dest_col)

        if closest_destination:
            dest_row, dest_col = closest_destination

            # Calculate the amount to move
            move_amount = min(grid[source_row][source_col] - 1, 1 - grid[dest_row][dest_col])

            # Update counts after the move
            grid[source_row][source_col] -= move_amount
            grid[dest_row][dest_col] += move_amount

            total_moves += move_amount * min_distance

            # Update sources and destinations lists
            if grid[source_row][source_col] <= 1:
                sources.pop(0)
            if grid[dest_row][dest_col] >= 1:
                destinations.remove((dest_row, dest_col))

    return total_moves

Big(O) Analysis

Time Complexity

O(n^3) – Let n be the number of cells in the grid. The algorithm identifies source and destination cells which is O(n). For each source cell, it searches for the closest destination cell, potentially requiring checking the distance to all other cells, resulting in O(n) for each source. Since we iterate potentially until each source is empty, in the worst case we make multiple passes to fulfill the destination needs. Thus, this inner loop will have O(n) complexity. Because we are searching the grid multiple times to perform the stone transfers, the number of iterations for finding the min distance is repeated as we adjust the stone counts. Therefore, this can be summarized as O(n)*O(n)*O(n), which simplifies to O(n^3).

Space Complexity

O(N) – The algorithm identifies source and destination cells, requiring lists to store their coordinates. In the worst-case scenario, where almost all cells are either sources or destinations, these lists could grow to a size proportional to the total number of cells, N (where N is the total number of cells in the grid). The algorithm also involves updating the stone counts in place, which doesn't contribute to auxiliary space. Therefore, the dominant space usage comes from storing the source and destination cell coordinates, leading to O(N) auxiliary space.

Edge Cases

Case	How to Handle
Empty grid (null or zero dimensions)	Return 0 if the grid is empty, as no moves are needed since there are no cells.
Grid with only one cell	If only one cell exists, return 0 as no moves are possible from a single cell.
All cells have the same number of stones (either all excess or all deficiency)	Return 0 as stones are already spread evenly, so no moves are necessary.
The total number of stones is not divisible by the number of cells	Return -1 or throw an error indicating that a balanced distribution is impossible.
Large grid dimensions (potential memory issues)	Optimize memory usage by considering in-place calculations or using generators if possible, and analyze the space complexity.
Extreme stone counts in cells (potential integer overflow)	Use a data type that can handle large integers (e.g., long) to prevent overflow during calculations of excess/deficiency.
No possible solution (e.g., stones are clustered in disconnected regions far apart)	The algorithm should explore all possible paths but may not find a valid solution; return -1 to indicate that no solution exists within a reasonable number of moves/iterations.
Negative stone counts in cells (invalid input based on problem definition)	Handle negative stone counts either by throwing an exception or by converting them to zero and adjusting the overall stone count to compensate for the difference and maintaining total stone count consistency with the grid size.