What is the time complexity of Minimum Moves to Spread Stones Over Grid?

The optimal solution has a time complexity of O(n!).

What data structures are used to solve Minimum Moves to Spread Stones Over Grid?

This problem uses Array, Dynamic Programming, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Moves to Spread Stones Over Grid?

Insight 1: The total number of stones is fixed at 9, which is equal to the number of cells in the 3x3 grid. Therefore, the problem is about redistributing existing stones, not adding or removing any. Insight 2: Cells with more than one stone (excess stones) need to move stones to cells with zero stones. We need to find the optimal pairing between these 'excess' and 'zero' cells.

Which companies ask LeetCode 2850. Minimum Moves to Spread Stones Over Grid in interviews?

This problem is frequently asked at Geico, Guidewire.

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Minimum Moves to Spread Stones Over Grid - LeetCode 2850 Solution

Minimum Moves to Spread Stones Over Grid - Complete Solution Guide

Minimum Moves to Spread Stones Over Grid is LeetCode problem 2850, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 pl

Detailed Explanation

The problem asks us to find the minimum number of moves needed to redistribute stones in a 3x3 grid such that each cell contains exactly one stone. We start with a grid containing a total of 9 stones, but some cells might have more than one while others have zero. A move consists of transferring a single stone from one cell to an adjacent cell (sharing a side, not diagonally). The goal is to determine the fewest moves required to achieve a state where every cell has exactly one stone.

Solution Approach

The provided solution identifies the cells with excess stones (more than 1) and the empty cells (0 stones). Then, it generates all possible permutations of excess stones and pairs them with the empty cells. For each permutation, it calculates the total Manhattan distance (number of moves) required to move stones from each excess cell to a corresponding empty cell. The solution then returns the minimum total Manhattan distance found across all permutations.

Step-by-Step Algorithm

Step 1: Iterate through the grid and identify cells with zero stones (zeros) and cells with more than one stone (extras). For each cell with more than one stone, calculate the number of excess stones (grid[r][c] - 1) and add the cell's coordinates to the `extras` list as many times as there are excess stones.
Step 2: If there are no `extras` (all cells have exactly one stone), return 0 as no moves are needed.
Step 3: Generate all possible permutations of the `extras` list. Each permutation represents a potential assignment of excess stones to empty cells.
Step 4: For each permutation, calculate the total Manhattan distance. This is done by iterating through the `zeros` and the current permutation of `extras`, calculating the Manhattan distance between each corresponding pair of cells, and summing up these distances.
Step 5: Keep track of the minimum total Manhattan distance encountered across all permutations.
Step 6: After evaluating all permutations, return the minimum total Manhattan distance.
Step 7: In Java version, implement a permutation helper function to generate all permutations manually.

Key Insights

Insight 1: The total number of stones is fixed at 9, which is equal to the number of cells in the 3x3 grid. Therefore, the problem is about redistributing existing stones, not adding or removing any.
Insight 2: Cells with more than one stone (excess stones) need to move stones to cells with zero stones. We need to find the optimal pairing between these 'excess' and 'zero' cells.
Insight 3: The Manhattan distance (sum of absolute differences in row and column indices) represents the minimum number of moves required to move a stone between two cells. We need to minimize the sum of Manhattan distances for all stone transfers.
Insight 4: Since the grid size is small (3x3), a brute-force approach involving permutations is feasible despite its high time complexity. This is because the number of excess stones, and thus the number of permutations, will be relatively small.

Complexity Analysis

Time Complexity: O(n!)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Breadth-First Search, Matrix.

Companies

Asked at: Geico, Guidewire.

Minimum Moves to Spread Stones Over Grid | LeetCode 2850

Medium

Array Dynamic Programming Breadth-First Search Matrix

Time: O(n!)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int minimumMoves(int[][] grid) {
        List<int[]> zeros = new ArrayList<>();
        List<int[]> extras = new ArrayList<>();
        for (int r = 0; r < 3; r++) {
            for (int c = 0; c < 3; c++) {
                if (grid[r][c] == 0) {
                    zeros.add(new int[]{r, c});
                } else if (grid[r][c] > 1) {
                    for (int i = 0; i < grid[r][c] - 1; i++) {
                        extras.add(new int[]{r, c});
                    }
                }
            }
        }

        if (extras.isEmpty()) {
            return 0;
        }

        int min_moves = Integer.MAX_VALUE;

        List<List<int[]>> permutations = permute(extras);

        for (List<int[]> perm : permutations) {
            int total_dist = 0;
            for (int i = 0; i < zeros.size(); i++) {
                int[] z = zeros.get(i);
                int[] e = perm.get(i);
                total_dist += Math.abs(z[0] - e[0]) + Math.abs(z[1] - e[1]);
            }
            min_moves = Math.min(min_moves, total_dist);
        }

        return min_moves;
    }

    private List<List<int[]>> permute(List<int[]> nums) {
        List<List<int[]>> result = new ArrayList<>();
        permuteHelper(nums, new ArrayList<>(), result);
        return result;
    }

    private void permuteHelper(List<int[]> nums, List<int[]> current, List<List<int[]>> result) {
        if (current.size() == nums.size()) {
            result.add(new ArrayList<>(current));
            return;
        }

        for (int i = 0; i < nums.size(); i++) {
            if (current.contains(nums.get(i))) {
                continue;
            }
            current.add(nums.get(i));
            List<int[]> nextNums = new ArrayList<>(nums);
            permuteHelper(nums, current, result);
            current.remove(current.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 2850: Minimum Moves to Spread Stones Over Grid - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Breadth-First Search Matrix

Companies

GeicoGuidewire

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2850

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Iterate through the grid and identify cells with zero stones (zeros) and cells with more than one stone (extras). For each cell with more than one stone, calculate the number of excess stones (grid[r][c] - 1) and add the cell's coordinates to the `extras` list as many times as there are excess stones.
2
Step 2: If there are no `extras` (all cells have exactly one stone), return 0 as no moves are needed.
3
Step 3: Generate all possible permutations of the `extras` list. Each permutation represents a potential assignment of excess stones to empty cells.
4
Step 4: For each permutation, calculate the total Manhattan distance. This is done by iterating through the `zeros` and the current permutation of `extras`, calculating the Manhattan distance between each corresponding pair of cells, and summing up these distances.
5
Step 5: Keep track of the minimum total Manhattan distance encountered across all permutations.
6
Step 6: After evaluating all permutations, return the minimum total Manhattan distance.
7
Step 7: In Java version, implement a permutation helper function to generate all permutations manually.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n!)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the permutation generation. In the worst case, where the number of zeros (and extras) is n, the number of permutations is n!. Calculating the Manhattan distance for each permutation takes O(n) time. Therefore, the overall time complexity is O(n! * n), which simplifies to O(n!) because factorial grows faster. The space complexity arises from storing the coordinates of the zero and extra stone locations. The Java version's permutation generation uses recursion which might have stack usage of O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Hungarian Algorithm (Minimum Weight Bipartite Matching)

Model the problem as a bipartite graph where one set of nodes represents the empty cells and the other set represents the excess stones. The weight of an edge between an empty cell and an excess stone is the Manhattan distance between them. The goal is to find a minimum weight matching, which corresponds to the minimum number of moves.

Time: O(n^3), where n is the number of empty cells (or excess stones).Space: O(n^2) to store the adjacency matrix.

Trade-offs: The Hungarian Algorithm would be more efficient than the brute-force approach if the number of empty cells and excess stones were larger. However, since the grid size is fixed at 3x3, the brute-force approach is often acceptable, especially considering the code simplicity. For larger grids or a higher constraint on time complexity, Hungarian algorithm is preferred.

Dynamic Programming

While less intuitive, DP could be explored with memoization to avoid recalculating distances. However, defining a state for DP in this problem can be complex, and the benefits may not outweigh the complexity for a 3x3 grid.

Time: Difficult to achieve better than O(n! * n) practically for this constrained grid size.Space: Potentially O(2^n) or worse depending on the state definition.

Trade-offs: Unlikely to be a practical or better solution than the existing brute-force method, especially given the constraints.

✨

Key Insights

1Insight 1: The total number of stones is fixed at 9, which is equal to the number of cells in the 3x3 grid. Therefore, the problem is about redistributing existing stones, not adding or removing any.
2Insight 2: Cells with more than one stone (excess stones) need to move stones to cells with zero stones. We need to find the optimal pairing between these 'excess' and 'zero' cells.
3Insight 3: The Manhattan distance (sum of absolute differences in row and column indices) represents the minimum number of moves required to move a stone between two cells. We need to minimize the sum of Manhattan distances for all stone transfers.
4Insight 4: Since the grid size is small (3x3), a brute-force approach involving permutations is feasible despite its high time complexity. This is because the number of excess stones, and thus the number of permutations, will be relatively small.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: The grid already has one stone in each cell. The code correctly handles this by checking if the `extras` list is empty. If it is, it returns 0, indicating no moves are needed.
Edge case 2: All stones are concentrated in one cell. The code will correctly identify all other cells as having zero stones and calculate the minimum moves required to distribute the stones.
Edge case 3: Input grid is invalid (sum of grid elements is not 9). The problem description states that the sum is guaranteed to be 9, so the code doesn't explicitly handle this case. However, the algorithm would still function; it may just provide wrong output without producing an error.
Edge case 4: The number of zeros and extras are not equal. Because the total number of stones is 9 which is the same as number of cells, the number of zeros and extras must be equal. The code does not have explicit check but its logic will fail if their numbers are not equal.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the Manhattan distance between cells. It's crucial to use the sum of absolute differences in row and column indices.

✗Common mistake 2: Not handling the case where the grid already has one stone in each cell (returning 0 in this case).

✗Common mistake 3: Failing to consider all possible permutations of excess stones, which would lead to a suboptimal solution.

✗Common mistake 4: Off-by-one errors when calculating the number of excess stones (remember to subtract 1 from the cell's value).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation and task assignment where resources (stones) need to be moved between locations (cells) to balance the workload. Examples include assigning workers to tasks or distributing inventory across warehouses.

🏦Real-world scenario 2: Robotics and path planning where a robot needs to move objects from one location to another while minimizing the distance traveled. This can be applied in warehouse automation, manufacturing, and autonomous navigation.

🎮Real-world scenario 3: Logistics and transportation: Optimizing delivery routes by minimizing the distance traveled by vehicles to distribute goods.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Moves to Spread Stones Over Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?