How do you solve Snakes and Ladders on LeetCode?

Snakes and Ladders (LeetCode #909) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using BFS ensures we find the shortest path efficiently.

What is the brute force solution for Snakes and Ladders?

The brute force approach for Snakes and Ladders is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating every possible move from the current square using a recursive method. We explore all paths until we either reach the end or exhaust all options, which can be inefficient for larger boards. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Snakes and Ladders?

Snakes and Ladders uses the following concepts: Array, Breadth-First Search, Matrix. The recommended patterns to study are: Breadth-First Search, Graph Traversal.

What are the interview tips for Snakes and Ladders?

Explain your thought process clearly while coding. Consider edge cases like small boards or no snakes/ladders. Practice BFS problems to get comfortable with the approach.

What are common mistakes when solving Snakes and Ladders?

Not handling the snake or ladder correctly. Forgetting to mark squares as visited, leading to infinite loops.

#909

Snakes and Ladders

Medium

ArrayBreadth-First SearchMatrixBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal approach uses Breadth-First Search (BFS) to explore the board level by level. This ensures that we find the shortest path to the end square efficiently, as BFS explores all possible moves from the current square before moving deeper.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue for BFS starting from square 1.
2Step 2: For each square, explore all possible moves (1 to 6) and check if there is a snake or ladder.
3Step 3: If we reach square n², return the number of rolls taken. If the queue is exhausted, return -1.

solution.py21 lines

1from collections import deque
2
3def snakesAndLadders(board):
4    n = len(board)
5    queue = deque([(1, 0)])  # (current square, rolls)
6    visited = set([1])
7    while queue:
8        curr, rolls = queue.popleft()
9        if curr == n * n:
10            return rolls
11        for move in range(1, 7):
12            next_square = curr + move
13            if next_square > n * n:
14                continue
15            r, c = divmod(next_square - 1, n)
16            if board[n - 1 - r][c] != -1:
17                next_square = board[n - 1 - r][c]
18            if next_square not in visited:
19                visited.add(next_square)
20                queue.append((next_square, rolls + 1))
21    return -1

ℹ

Complexity note: The time complexity is O(n²) because we may need to visit each square on the board, and the space complexity is O(n) due to the queue used for BFS.

1Using BFS ensures we find the shortest path efficiently.
2Understanding the board layout is crucial for correct indexing.

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