What is the time complexity of Minimum Moves to Reach Target with Rotations?

The optimal solution has a time complexity of O(n^2).

What is the space complexity of Minimum Moves to Reach Target with Rotations?

The space complexity is O(n^2).

What data structures are used to solve Minimum Moves to Reach Target with Rotations?

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

What are the key insights for solving Minimum Moves to Reach Target with Rotations?

Insight 1: The problem can be modeled as a shortest path problem on a graph where each node represents the snake's state (row, column, orientation), and edges represent valid moves between states. Insight 2: Breadth-First Search (BFS) is an efficient algorithm for finding the shortest path in an unweighted graph, making it suitable for this problem.

Which companies ask LeetCode 1210. Minimum Moves to Reach Target with Rotations in interviews?

This problem is frequently asked at Kakao.

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Minimum Moves to Reach Target with Rotations - LeetCode 1210 Solution

Minimum Moves to Reach Target with Rotations - Complete Solution Guide

Minimum Moves to Reach Target with Rotations is LeetCode problem 1210, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In an n*n grid, there is a snake that spans 2 cells and starts moving from the top left corner at (0, 0) and (0, 1) . The grid has empty cells represented by zeros and blocked cells represented by ones. The snake wants to reach the lower right corner at (n-1, n-2) and (n-1, n-1) . In one move the snake can: Move one cell to the right if there are no blocked cells there. This move keeps the horizontal/vertical position of the snake as it is. Move down one cell if there are no blocked cells there.

Detailed Explanation

The problem presents a snake in an n x n grid, starting at the top-left corner (0, 0) and (0, 1), and asks for the minimum number of moves required for the snake to reach the bottom-right corner (n-1, n-2) and (n-1, n-1). The snake can move right, down, or rotate (clockwise or counter-clockwise) subject to certain constraints. The grid contains empty cells (0) and blocked cells (1), which restrict the snake's movements. The snake occupies two adjacent cells, either horizontally or vertically. The goal is to find the shortest path for the snake to reach the target, returning -1 if no path exists.

Solution Approach

The provided code uses Breadth-First Search (BFS) to find the minimum number of moves. Each state of the snake (row, column, orientation) is enqueued, and valid moves (right, down, rotate) are explored. A `visited` set prevents revisiting states. The algorithm continues until the target state is reached or the queue is empty (no path found).

Step-by-Step Algorithm

Step 1: Initialize the starting state (0, 0, 0) and target state (n-1, n-2, 0). Create a queue for BFS and a set to track visited states.
Step 2: Enqueue the starting state with a move count of 0 and add it to the visited set.
Step 3: While the queue is not empty, dequeue a state and its corresponding move count.
Step 4: If the dequeued state is the target state, return the move count.
Step 5: Based on the snake's orientation (horizontal or vertical), explore possible moves: right, down, or rotate. Check for valid moves by checking grid boundaries and blocked cells.
Step 6: For each valid move, create the new state, check if it has been visited. If not, add the new state to the visited set and enqueue it with an incremented move count.
Step 7: If the queue becomes empty before reaching the target state, return -1, indicating no path exists.

Key Insights

Insight 1: The problem can be modeled as a shortest path problem on a graph where each node represents the snake's state (row, column, orientation), and edges represent valid moves between states.
Insight 2: Breadth-First Search (BFS) is an efficient algorithm for finding the shortest path in an unweighted graph, making it suitable for this problem.
Insight 3: Representing the snake's state as a tuple (row, column, orientation) and using a set to track visited states is crucial to avoid cycles and ensure efficient exploration.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: Kakao.

Minimum Moves to Reach Target with Rotations | LeetCode 1210

Hard

Array Breadth-First Search Matrix

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.HashSet;
import java.util.Queue;
import java.util.Set;

class Solution {
    public int minimumMoves(int[][] grid) {
        int n = grid.length;
        
        int[] start = {0, 0, 0}; // (row, col, orientation), 0: horizontal, 1: vertical
        int[] target = {n - 1, n - 2, 0};
        
        Queue<Pair> queue = new ArrayDeque<>();
        queue.offer(new Pair(start, 0));
        Set<String> visited = new HashSet<>();
        visited.add(start[0] + "," + start[1] + "," + start[2]);
        
        while (!queue.isEmpty()) {
            Pair current = queue.poll();
            int[] state = current.state;
            int moves = current.moves;
            
            if (state[0] == target[0] && state[1] == target[1] && state[2] == target[2]) {
                return moves;
            }
            
            int r = state[0];
            int c = state[1];
            int orientation = state[2];
            
            if (orientation == 0) { // Horizontal
                // Move right
                if (c + 2 < n && grid[r][c + 2] == 0) {
                    int[] nextState = {r, c + 1, 0};
                    String nextStateStr = nextState[0] + "," + nextState[1] + "," + nextState[2];
                    if (!visited.contains(nextStateStr)) {
                        visited.add(nextStateStr);
                        queue.offer(new Pair(nextState, moves + 1));
                    }
                }
                
                // Move down or Rotate clockwise
                if (r + 1 < n && grid[r + 1][c] == 0 && grid[r + 1][c + 1] == 0) {
                    // Move down
                    int[] downState = {r + 1, c, 0};
                    String downStateStr = downState[0] + "," + downState[1] + "," + downState[2];
                    if (!visited.contains(downStateStr)) {
                        visited.add(downStateStr);
                        queue.offer(new Pair(downState, moves + 1));
                    }
                    
                    // Rotate clockwise
                    int[] rotateState = {r, c, 1};
                    String rotateStateStr = rotateState[0] + "," + rotateState[1] + "," + rotateState[2];
                    if (!visited.contains(rotateStateStr)) {
                        visited.add(rotateStateStr);
                        queue.offer(new Pair(rotateState, moves + 1));
                    }
                }
            } else { // Vertical
                // Move down
                if (r + 2 < n && grid[r + 2][c] == 0) {
                    int[] nextState = {r + 1, c, 1};
                    String nextStateStr = nextState[0] + "," + nextState[1] + "," + nextState[2];
                    if (!visited.contains(nextStateStr)) {
                        visited.add(nextStateStr);
                        queue.offer(new Pair(nextState, moves + 1));
                    }
                }
                
                // Move right or Rotate counter-clockwise
                if (c + 1 < n && grid[r][c + 1] == 0 && grid[r + 1][c + 1] == 0) {
                    // Move right
                    int[] rightState = {r, c + 1, 1};
                    String rightStateStr = rightState[0] + "," + rightState[1] + "," + rightState[2];
                    if (!visited.contains(rightStateStr)) {
                        visited.add(rightStateStr);
                        queue.offer(new Pair(rightState, moves + 1));
                    }
                    
                    // Rotate counter-clockwise
                    int[] rotateState = {r, c, 0};
                    String rotateStateStr = rotateState[0] + "," + rotateState[1] + "," + rotateState[2];
                    if (!visited.contains(rotateStateStr)) {
                        visited.add(rotateStateStr);
                        queue.offer(new Pair(rotateState, moves + 1));
                    }
                }
            }
        }
        
        return -1;
    }
    
    private static class Pair {
        int[] state;
        int moves;
        
        public Pair(int[] state, int moves) {
            this.state = state;
            this.moves = moves;
        }
    }
}

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Problem Info

Description

Problem 1210: Minimum Moves to Reach Target with Rotations - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Matrix

Companies

Kakao

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1210

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: Initialize the starting state (0, 0, 0) and target state (n-1, n-2, 0). Create a queue for BFS and a set to track visited states.
2
Step 2: Enqueue the starting state with a move count of 0 and add it to the visited set.
3
Step 3: While the queue is not empty, dequeue a state and its corresponding move count.
4
Step 4: If the dequeued state is the target state, return the move count.
5
Step 5: Based on the snake's orientation (horizontal or vertical), explore possible moves: right, down, or rotate. Check for valid moves by checking grid boundaries and blocked cells.
6
Step 6: For each valid move, create the new state, check if it has been visited. If not, add the new state to the visited set and enqueue it with an incremented move count.
7
Step 7: If the queue becomes empty before reaching the target state, return -1, indicating no path exists.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity stems from the BFS algorithm potentially exploring every cell in the grid, multiplied by the two possible orientations of the snake. The space complexity is determined by the size of the `visited` set and the queue, both of which can grow linearly with the number of cells.

📖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

A* Search

A* search is an informed search algorithm that can potentially improve the search efficiency compared to BFS. It uses a heuristic function to estimate the distance to the target and prioritizes nodes based on this estimate.

Time: Potentially better than O(n^2) depending on the heuristic. Worst case O(n^2) if the heuristic isn't effective.Space: O(n^2)

Trade-offs: A* search can be faster if a good heuristic is available. However, a poorly chosen heuristic can lead to worse performance than BFS. Requires designing an admissible heuristic function.

✨

Key Insights

1Insight 1: The problem can be modeled as a shortest path problem on a graph where each node represents the snake's state (row, column, orientation), and edges represent valid moves between states.
2Insight 2: Breadth-First Search (BFS) is an efficient algorithm for finding the shortest path in an unweighted graph, making it suitable for this problem.
3Insight 3: Representing the snake's state as a tuple (row, column, orientation) and using a set to track visited states is crucial to avoid cycles and ensure efficient exploration.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The grid has no valid path to the target. The BFS algorithm correctly returns -1 when the queue becomes empty without reaching the target.
Edge case 2: The grid contains blocked cells that prevent certain moves. The code explicitly checks for these blocked cells before making a move, ensuring the snake only moves through empty cells.
Edge case 3: The snake starts in a position where reaching the target is impossible due to the placement of obstacles. The algorithm will exhaust all possibilities and correctly return -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking for grid boundaries before making a move. This can lead to array index out-of-bounds errors.

✗Common mistake 2: Not checking for blocked cells before making a move. This will lead to incorrect path calculations.

✗Common mistake 3: Forgetting to mark states as visited, resulting in infinite loops and incorrect results.

✗Common mistake 4: Incorrectly implementing the rotation logic, leading to invalid snake states.

✅

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: Pathfinding for robots or autonomous vehicles in a constrained environment. The snake's movement restrictions can model the physical limitations of the robot.

🏦Real-world scenario 2: Game development, where an AI-controlled character needs to navigate a complex maze or environment with specific movement constraints. Similar problems arise in puzzle games where the player must move a specific object from start to end.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 9, 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 approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 Reach Target with Rotations"?

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

🤖 Need help with this problem?