What is the time complexity of Check if There is a Valid Path in a Grid?

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

What data structures are used to solve Check if There is a Valid Path in a Grid?

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

What are the key insights for solving Check if There is a Valid Path in a Grid?

Insight 1: Representing the street connections as possible moves from a cell allows for traversal using graph traversal algorithms. Insight 2: A Breadth-First Search (BFS) or Depth-First Search (DFS) algorithm can be used to explore the grid.

Which companies ask LeetCode 1391. Check if There is a Valid Path in a Grid in interviews?

This problem is frequently asked at Robinhood.

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Check if There is a Valid Path in a Grid - LeetCode 1391 Solution

Q: What is the space complexity of Check if There is a Valid Path in a Grid?

The space complexity is O(m*n).

Check if There is a Valid Path in a Grid - Complete Solution Guide

Check if There is a Valid Path in a Grid is LeetCode problem 1391, 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 an m x n grid . Each cell of grid represents a street. The street of grid[i][j] can be: 1 which means a street connecting the left cell and the right cell. 2 which means a street connecting the upper cell and the lower cell. 3 which means a street connecting the left cell and the lower cell. 4 which means a street connecting the right cell and the lower cell. 5 which means a street connecting the left cell and the upper cell. 6 which means a street connecting the right cell and the

Detailed Explanation

The problem asks us to determine if there exists a valid path from the top-left cell (0, 0) to the bottom-right cell (m-1, n-1) in a given m x n grid. Each cell in the grid represents a street with a specific direction (1-6). We can only move along the streets; we cannot change the direction of any street. A valid path is one that starts at (0, 0) and ends at (m-1, n-1) while following the allowed street directions.

Solution Approach

The provided solution uses a Breadth-First Search (BFS) approach to explore the grid. It maintains a queue of cells to visit and a set to keep track of visited cells. Starting from (0, 0), the algorithm explores valid neighbor cells based on the street type of the current cell. A cell is considered a valid neighbor if it's within the grid boundaries, hasn't been visited, and if its street type is compatible with the move from the current cell.

Step-by-Step Algorithm

Step 1: Initialize a queue with the starting cell (0, 0) and a set to track visited cells, adding (0, 0) to the visited set.
Step 2: Define a mapping (dictionary/hashmap) of street types to possible moves (directions as (row, col) offsets).
Step 3: While the queue is not empty, dequeue a cell (r, c).
Step 4: If the dequeued cell is the destination (m-1, n-1), return true (a valid path exists).
Step 5: Get the street type of the current cell.
Step 6: Iterate through the possible moves for the current street type.
Step 7: For each move (dr, dc), calculate the neighbor's coordinates (nr, nc).
Step 8: Check if the neighbor is within the grid boundaries (0 <= nr < m and 0 <= nc < n).
Step 9: Check if the neighbor has not been visited.
Step 10: Check if the neighbor's street type is compatible with the move from the current cell. This is done by checking if the opposite move (-dr, -dc) is a valid move for the neighbor's street type.
Step 11: If all the above conditions are met, mark the neighbor as visited and enqueue it.
Step 12: If the queue becomes empty and the destination has not been reached, return false (no valid path exists).

Key Insights

Insight 1: Representing the street connections as possible moves from a cell allows for traversal using graph traversal algorithms.
Insight 2: A Breadth-First Search (BFS) or Depth-First Search (DFS) algorithm can be used to explore the grid.
Insight 3: A critical check is to ensure that the neighbor street is compatible with the move we just made. For example, if we move right from cell A to cell B, cell B must allow a connection coming from the left.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

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

Companies

Asked at: Robinhood.

Check if There is a Valid Path in a Grid | LeetCode 1391

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

class Solution {
    public boolean hasValidPath(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        Map<Integer, Set<Pair>> moves = Map.of(
            1, Set.of(new Pair(0, 1), new Pair(0, -1)),   // right, left
            2, Set.of(new Pair(1, 0), new Pair(-1, 0)),   // down, up
            3, Set.of(new Pair(1, 0), new Pair(0, -1)),   // down, left
            4, Set.of(new Pair(1, 0), new Pair(0, 1)),    // down, right
            5, Set.of(new Pair(-1, 0), new Pair(0, -1)),  // up, left
            6, Set.of(new Pair(-1, 0), new Pair(0, 1))    // up, right
        );

        Deque<Pair> q = new ArrayDeque<>();
        q.offer(new Pair(0, 0));
        Set<Pair> visited = new HashSet<>();
        visited.add(new Pair(0, 0));

        while (!q.isEmpty()) {
            Pair curr = q.poll();
            int r = curr.r;
            int c = curr.c;

            if (r == m - 1 && c == n - 1) {
                return true;
            }

            int current_street_type = grid[r][c];

            for (Pair move : moves.get(current_street_type)) {
                int dr = move.r;
                int dc = move.c;
                int nr = r + dr;
                int nc = c + dc;

                if (0 <= nr && nr < m && 0 <= nc && nc < n) {
                    Pair neighbor = new Pair(nr, nc);
                    if (!visited.contains(neighbor)) {
                        int neighbor_street_type = grid[nr][nc];

                        Pair back_move = new Pair(-dr, -dc);
                        if (moves.get(neighbor_street_type).contains(back_move)) {
                            visited.add(neighbor);
                            q.offer(neighbor);
                        }
                    }
                }
            }
        }

        return false;
    }
    
    static class Pair {
        int r;
        int c;

        public Pair(int r, int c) {
            this.r = r;
            this.c = c;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return r == pair.r && c == pair.c;
        }

        @Override
        public int hashCode() {
            int result = r;
            result = 31 * result + c;
            return result;
        }
    }
}

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Problem Info

Description

Problem 1391: Check if There is a Valid Path in a Grid - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Union Find Matrix

Companies

Robinhood

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1391

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

12 steps

1
Step 1: Initialize a queue with the starting cell (0, 0) and a set to track visited cells, adding (0, 0) to the visited set.
2
Step 2: Define a mapping (dictionary/hashmap) of street types to possible moves (directions as (row, col) offsets).
3
Step 3: While the queue is not empty, dequeue a cell (r, c).
4
Step 4: If the dequeued cell is the destination (m-1, n-1), return true (a valid path exists).
5
Step 5: Get the street type of the current cell.
6
Step 6: Iterate through the possible moves for the current street type.
7
Step 7: For each move (dr, dc), calculate the neighbor's coordinates (nr, nc).
8
Step 8: Check if the neighbor is within the grid boundaries (0 <= nr < m and 0 <= nc < n).
9
Step 9: Check if the neighbor has not been visited.
10
Step 10: Check if the neighbor's street type is compatible with the move from the current cell. This is done by checking if the opposite move (-dr, -dc) is a valid move for the neighbor's street type.
11
Step 11: If all the above conditions are met, mark the neighbor as visited and enqueue it.
12
Step 12: If the queue becomes empty and the destination has not been reached, return false (no valid 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(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

Time Complexity: In the worst case, the BFS algorithm visits each cell in the grid once. Therefore, the time complexity is O(m*n), where m is the number of rows and n is the number of columns. Space Complexity: The queue and the visited set can, in the worst case, contain all the cells in the grid. Therefore, the space complexity is O(m*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

Depth-First Search (DFS)

DFS could be used instead of BFS. The logic for checking valid moves and compatibility of streets remains the same. However, DFS uses recursion or a stack.

Time: O(m*n)Space: O(m*n) in the worst case (due to recursion stack)

Trade-offs: BFS guarantees finding the shortest path (in terms of the number of cells visited), while DFS may not. DFS can potentially lead to stack overflow errors if the grid is very large due to the recursive calls.

Union Find

Union Find could potentially be used. Create a Union Find data structure and connect adjacent cells if they are compatible. At the end, check if the starting cell and the ending cell belong to the same connected component.

Time: O(m*n * α(m*n)), where α is the inverse Ackermann function, which grows very slowly, so it's practically O(m*n)Space: O(m*n)

Trade-offs: Union Find can be more complex to implement for this problem compared to BFS and DFS. The performance difference is likely minimal in this case, so the increased complexity may not be worthwhile.

✨

Key Insights

1Insight 1: Representing the street connections as possible moves from a cell allows for traversal using graph traversal algorithms.
2Insight 2: A Breadth-First Search (BFS) or Depth-First Search (DFS) algorithm can be used to explore the grid.
3Insight 3: A critical check is to ensure that the neighbor street is compatible with the move we just made. For example, if we move right from cell A to cell B, cell B must allow a connection coming from the left.

📚

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: A grid with only one cell. The solution correctly identifies this as a valid path if m=1 and n=1, as the starting and ending points are the same.
Edge case 2: A grid where no valid path exists because the street directions don't allow movement. The BFS will explore all reachable cells and correctly return false.
Edge case 3: A grid where the starting cell has no valid connections. The BFS will terminate quickly because no neighbors can be enqueued, returning false.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for grid boundaries when calculating the coordinates of neighbor cells. This can lead to out-of-bounds errors.

✗Common mistake 2: Not checking the compatibility of the neighbor's street type with the current move. It is crucial to ensure the move is allowed in both directions.

✗Common mistake 3: Not properly marking visited cells, leading to infinite loops in the BFS/DFS traversal.

✅

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 in a city grid with one-way streets. Determining if a route exists from one point to another, respecting the direction of the streets.

🏦Real-world scenario 2: Routing network packets through a network where each node has specific connection capabilities. Determining if a path exists between two nodes, given the network topology.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

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 10, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Check if There is a Valid Path in a 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 depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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