LeetCode #2850 — MEDIUM

Minimum Moves to Spread Stones Over Grid

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

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.
Patterns Used
📊Dynamic Programming→🔀Backtracking→⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[[1,1,0],[1,1,1],[1,2,1]]

Example 2

[[1,3,0],[1,0,0],[1,0,3]]

Related Problems

  • Minimum Number of Operations to Move All Balls to Each Box (minimum-number-of-operations-to-move-all-balls-to-each-box)
  • Minimum Number of Operations to Make X and Y Equal (minimum-number-of-operations-to-make-x-and-y-equal)
Step 02

Core Insight

What unlocks the optimal approach

  • There are at most <code>4</code> cells with more than one stone.
  • Let <code>a</code> be the number of cells containing more than one stone, and <code>b</code> be the number of cells containing no stones. <code></code>. <code>b^a ≤ 6561</code>. Use this fact to come up with a bruteforce.
  • For all empty cells, bruteforce over all possible cells from which a stone can come. Note that a stone will always come from a cell containing at least 2 stones.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2850: Minimum Moves to Spread Stones Over Grid
class Solution {
    public int minimumMoves(int[][] grid) {
        Deque<String> q = new ArrayDeque<>();
        q.add(f(grid));
        Set<String> vis = new HashSet<>();
        vis.add(f(grid));
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int ans = 0;; ++ans) {
            for (int k = q.size(); k > 0; --k) {
                String p = q.poll();
                if ("111111111".equals(p)) {
                    return ans;
                }
                int[][] cur = g(p);
                for (int i = 0; i < 3; ++i) {
                    for (int j = 0; j < 3; ++j) {
                        if (cur[i][j] > 1) {
                            for (int d = 0; d < 4; ++d) {
                                int x = i + dirs[d];
                                int y = j + dirs[d + 1];
                                if (x >= 0 && x < 3 && y >= 0 && y < 3 && cur[x][y] < 2) {
                                    int[][] nxt = new int[3][3];
                                    for (int r = 0; r < 3; ++r) {
                                        for (int c = 0; c < 3; ++c) {
                                            nxt[r][c] = cur[r][c];
                                        }
                                    }
                                    nxt[i][j]--;
                                    nxt[x][y]++;
                                    String s = f(nxt);
                                    if (!vis.contains(s)) {
                                        vis.add(s);
                                        q.add(s);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    private String f(int[][] grid) {
        StringBuilder sb = new StringBuilder();
        for (int[] row : grid) {
            for (int x : row) {
                sb.append(x);
            }
        }
        return sb.toString();
    }

    private int[][] g(String s) {
        int[][] grid = new int[3][3];
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                grid[i][j] = s.charAt(i * 3 + j) - '0';
            }
        }
        return grid;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × 2^n)
Space
O(2^n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.

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