How do you solve Minimum Moves to Spread Stones Over Grid on LeetCode?

Minimum Moves to Spread Stones Over Grid (LeetCode #2850) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The problem can be reduced to moving stones from overfilled to empty cells.

What is the brute force solution for Minimum Moves to Spread Stones Over Grid?

The brute force approach for Minimum Moves to Spread Stones Over Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries all possible ways to distribute the stones across the grid. We can generate all possible configurations of stone distributions and calculate the number of moves needed for each configuration. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Moves to Spread Stones Over Grid?

Clearly outline your thought process before jumping into coding. Consider edge cases, such as all stones in one cell or all cells empty. Explain your approach to the interviewer as you code to demonstrate your understanding.

What are common mistakes when solving Minimum Moves to Spread Stones Over Grid?

Failing to consider all possible configurations of stone movements. Not keeping track of the minimum moves correctly during backtracking.

#2850

Minimum Moves to Spread Stones Over Grid

Medium

ArrayDynamic ProgrammingBacktrackingBit ManipulationMatrixBitmaskBacktrackingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a backtracking approach to efficiently explore the distribution of stones. It leverages the small grid size and constraints to minimize unnecessary calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a function to perform backtracking that tries to place stones in empty cells.
2Step 2: For each cell with excess stones, attempt to move stones to adjacent empty cells and recursively call the function.
3Step 3: Keep track of the number of moves and update the minimum moves found.

solution.py23 lines

1def minMoves(grid):
2    def backtrack(excess, empty, moves):
3        if not excess:
4            return moves
5        min_moves = float('inf')
6        x, y, count = excess[0]
7        for ex, ey in empty:
8            if abs(x - ex) + abs(y - ey) <= count:
9                new_excess = excess[1:]
10                new_empty = empty.copy()
11                new_empty.remove((ex, ey))
12                min_moves = min(min_moves, backtrack(new_excess, new_empty, moves + abs(x - ex) + abs(y - ey)))
13        return min_moves
14
15    excess = []
16    empty = []
17    for i in range(3):
18        for j in range(3):
19            if grid[i][j] > 1:
20                excess.append((i, j, grid[i][j] - 1))
21            elif grid[i][j] == 0:
22                empty.append((i, j))
23    return backtrack(excess, empty, 0)

ℹ

Complexity note: The complexity is linear due to the limited number of cells (3x3), allowing efficient exploration of possible moves without excessive branching.

1The problem can be reduced to moving stones from overfilled to empty cells.
2The small grid size (3x3) allows for exhaustive search methods like backtracking.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.