How do you solve Check if There is a Valid Path in a Grid on LeetCode?

Check if There is a Valid Path in a Grid (LeetCode #1391) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Understanding the street types and their connections is crucial.

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

The optimal solution for Check if There is a Valid Path in a Grid runs in O(m * n) time and uses O(m * n) space. The time complexity remains O(m * n) as we may still visit every cell. The space complexity is O(m * n) due to the visited array and the queue used for BFS.

What is the brute force solution for Check if There is a Valid Path in a Grid?

The brute force approach for Check if There is a Valid Path in a Grid is Brute Force, with O(m * n) time complexity and O(m * n) space complexity. We can use Depth-First Search (DFS) to explore all possible paths from the starting cell (0, 0) to the destination cell (m-1, n-1). This approach checks every possible route but may take a long time for larger grids. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Check if There is a Valid Path in a Grid?

Check if There is a Valid Path in a Grid uses the following concepts: Array, Depth-First Search, Breadth-First Search, Union-Find, Matrix. The recommended patterns to study are: Graph Traversal, Breadth-First Search, Depth-First Search.

What are the interview tips for Check if There is a Valid Path in a Grid?

Explain your thought process while coding. Consider edge cases, such as the smallest grid size. Discuss the trade-offs between DFS and BFS.

What are common mistakes when solving Check if There is a Valid Path in a Grid?

Not considering all possible directions when moving between cells. Failing to mark cells as visited, leading to infinite loops.

#1391

Check if There is a Valid Path in a Grid

Medium

ArrayDepth-First SearchBreadth-First SearchUnion-FindMatrixGraph TraversalBreadth-First SearchDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n)	O(m * n)
Space	O(m * n)	O(m * n)

💡

Intuition

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

We can optimize the brute-force approach by using a queue for Breadth-First Search (BFS) to explore the grid iteratively. This allows us to explore all possible paths more efficiently and avoid deep recursion.

⚙️

Algorithm

5 steps

1Step 1: Initialize a queue and add the starting cell (0, 0).
2Step 2: While the queue is not empty, pop the front cell.
3Step 3: Check if the current cell is the destination cell (m-1, n-1).
4Step 4: For the current cell, determine possible moves based on the street type.
5Step 5: If a move is valid and the next cell has not been visited, add it to the queue.

solution.py20 lines

1from collections import deque
2
3def hasValidPath(grid):
4    m, n = len(grid), len(grid[0])
5    visited = set()
6    queue = deque([(0, 0)])
7    directions = {1: [(0, 1), (0, -1)], 2: [(1, 0), (-1, 0)], 3: [(0, -1), (1, 0)], 4: [(0, 1), (1, 0)], 5: [(0, -1), (-1, 0)], 6: [(0, 1), (-1, 0)]}
8
9    while queue:
10        x, y = queue.popleft()
11        if (x, y) in visited:
12            continue
13        if (x, y) == (m-1, n-1):
14            return True
15        visited.add((x, y))
16        for dx, dy in directions[grid[x][y]]:
17            nx, ny = x + dx, y + dy
18            if 0 <= nx < m and 0 <= ny < n and (nx, ny) not in visited:
19                queue.append((nx, ny))
20    return False

ℹ

Complexity note: The time complexity remains O(m * n) as we may still visit every cell. The space complexity is O(m * n) due to the visited array and the queue used for BFS.

1Understanding the street types and their connections is crucial.
2Using BFS can help avoid deep recursion and stack overflow issues.

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