Problem Description

You are given an n x n integer matrix board representing a Snakes and Ladders game board. The cells are numbered from 1 to n² in a special pattern called "Boustrophedon style":

• Start from the bottom-left corner (board[n-1][0]) with number 1
• Fill numbers left-to-right for the bottom row
• For the next row up, fill numbers right-to-left
• Continue alternating direction for each row moving upward

You start at square 1 and want to reach square n². On each turn:

• Roll a 6-sided die to move forward between 1 to 6 squares
• From current square curr, you can move to any square numbered between curr + 1 and min(curr + 6, n²)
• If you land on a square with a snake or ladder (where board[r][c] != -1), you must immediately move to the destination square indicated by board[r][c]
• If you land on a square without a snake or ladder (board[r][c] == -1), you stay on that square
• Important: You only take a snake or ladder once per dice roll - if the destination has another snake or ladder, you don't follow it

The goal is to find the minimum number of dice rolls needed to reach square n². If it's impossible to reach the final square, return -1.

For example, in a 2x2 board [[-1,4],[-1,3]]:

• Squares are numbered: [[4,3],[1,2]] (following Boustrophedon pattern)
• If you roll to move from square 1 to square 2, and square 2 has a ladder to square 3, you go to square 3 but don't follow the subsequent ladder from square 3 to square 4

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The Snakes and Ladders board can be viewed as a graph where each square is a node, and edges exist from each square to the squares reachable by rolling a die (1-6 steps forward). Snakes and ladders create additional directed edges.

Is it a tree?

• No: The graph is not a tree because there can be multiple paths to reach the same square (through different dice rolls or via snakes/ladders), creating cycles in the graph structure.

Is the problem related to directed acyclic graphs?

• No: The graph contains cycles (you can potentially reach the same square through different paths), so it's not a DAG.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of dice rolls (edges traversed) to reach from square 1 to square n², which is essentially finding the shortest path in an unweighted graph.

Is the graph Weighted?

• No: Each move (dice roll) has the same cost of 1, regardless of how many squares you advance or whether you take a snake/ladder. This makes it an unweighted graph problem.

BFS

• Yes: For unweighted shortest path problems, BFS is the optimal choice.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for solving the Snakes and Ladders problem, which aligns perfectly with finding the minimum number of moves in an unweighted graph.

Intuition

The key insight is recognizing that this problem is fundamentally about finding the shortest path in a graph. Each square on the board is a node, and each possible move (rolling the die) creates an edge to another node.

Why BFS works perfectly here:

1. Level-by-level exploration: BFS explores all squares reachable in 1 roll, then all squares reachable in 2 rolls, and so on. This guarantees we find the minimum number of rolls.
2. Unweighted edges: Each dice roll counts as exactly 1 move, regardless of whether we move 1 square or 6 squares, or take a snake/ladder. This makes it an unweighted graph problem where BFS finds the optimal solution.

The tricky parts to handle:

• Board numbering: The Boustrophedon (zigzag) pattern means we need to convert between square numbers (1 to n²) and board coordinates [row][column]. For odd-numbered rows from the bottom, we read right-to-left.
• Snakes and ladders: These are just "teleportation" edges. When we land on a square with value ≠ -1, we immediately jump to that destination square instead.
• One-time teleportation: We only follow one snake/ladder per dice roll. If the destination has another snake/ladder, we don't chain them.

The BFS approach naturally handles all these complexities:

• Start from square 1
• For each square we can currently reach, try all possible dice rolls (1-6)
• If we land on a snake/ladder, go to its destination
• Mark visited squares to avoid redundant exploration
• The first time we reach square n², we've found the shortest path

This is more efficient than DFS because DFS might explore longer paths first, while BFS guarantees we find the shortest path as soon as we reach the destination.

Learn more about Breadth-First Search patterns.

Solution Approach

The implementation follows a standard BFS pattern with some special handling for the board's unique numbering system.

Data Structures Used:

• q (deque): A queue to store squares to be explored in BFS order
• vis (set): Tracks visited squares to avoid revisiting them
• ans: Counter for the number of dice rolls (levels in BFS)

Step-by-step Implementation:

1. Initialization: Start with square 1 in the queue and mark it as visited. Set ans = 0 for counting rolls, and calculate m = n * n as the target square.

2. BFS Level Processing: Process all squares at the current level (same number of rolls):

   for _ in range(len(q)):
       x = q.popleft()

   This ensures we process all squares reachable with the current number of rolls before incrementing.

3. Check for Target: If we've reached square m (which is n²), return the current number of rolls.

4. Explore Next Moves: From square x, try all possible dice rolls (1 to 6):

   for y in range(x + 1, min(x + 6, m) + 1):

   This generates all reachable squares from the current position.

5. Convert Square Number to Board Coordinates: The tricky part - converting square number y to board position [i][j]:

   i, j = divmod(y - 1, n)  # Get row and column in logical order
   if i & 1:                # If odd row (0-indexed)
       j = n - j - 1        # Reverse column for zigzag pattern
   i = n - i - 1           # Flip row (board is bottom-up)

   • divmod(y - 1, n) gives us the logical row and column (0-indexed)
   • Odd rows (when i & 1 is true) go right-to-left, so we reverse the column
   • Rows are numbered bottom-to-top, so we flip the row index

6. Handle Snakes/Ladders: Check if the square has a snake or ladder:

   z = y if board[i][j] == -1 else board[i][j]

   If board[i][j] != -1, we must move to that destination instead of staying at y.

7. Add to Queue if Unvisited: Only explore squares we haven't visited:

   if z not in vis:
       vis.add(z)
       q.append(z)

   This prevents infinite loops and redundant exploration.

8. Increment Roll Count: After processing all squares at the current level, increment ans to move to the next level.

9. Return -1 if Unreachable: If the queue becomes empty without reaching the target, return -1.

The beauty of this BFS approach is that it guarantees the minimum number of rolls because it explores all possibilities level by level, and the first time we reach the target is guaranteed to be via the shortest path.

Example Walkthrough

Let's walk through a small example with a 3x3 board to illustrate the solution approach:

Board: [[-1,-1,-1],
        [-1, 9, 8],
        [-1,-1, 2]]

Square numbering (Boustrophedon style):
[7, 8, 9]  ← Row 2 (top)
[6, 5, 4]  → Row 1 (middle) 
[1, 2, 3]  ← Row 0 (bottom)

Board with snakes/ladders marked:
- Square 5 has a ladder to square 9
- Square 6 has a ladder to square 8
- Square 3 has a snake to square 2

BFS Execution:

Initial State:

• Queue: [1]
• Visited: {1}
• Rolls: 0

Roll 1 (Process square 1):

• From square 1, we can reach squares 2-7 (roll 1-6)
• Check each destination:
  • Square 2: No snake/ladder → stay at 2 → add to queue
  • Square 3: Has snake to 2 → already visited 2 → skip
  • Square 4: No snake/ladder → stay at 4 → add to queue
  • Square 5: Has ladder to 9 → go to 9 → add to queue ✓
  • Square 6: Has ladder to 8 → go to 8 → add to queue
  • Square 7: No snake/ladder → stay at 7 → add to queue
• Queue: [2, 4, 9, 8, 7]
• Visited: {1, 2, 4, 9, 8, 7}
• Rolls: 1

Check: Square 9 is in our queue! We reached the target in just 1 roll.

Answer: 1 roll

The key insight: Even though square 5 is at position 5, its ladder takes us directly to square 9 (the target) in just one dice roll. This demonstrates how BFS efficiently finds the shortest path by exploring all possibilities at each level.

Coordinate Conversion Example: Let's see how square 5 converts to board coordinates:

1. i, j = divmod(5-1, 3) → i=1, j=1 (logical position)
2. Row 1 is odd (0-indexed), so reverse column: j = 3-1-1 = 1 (stays same)
3. Flip row for bottom-up: i = 3-1-1 = 1
4. board[1][1] = 9 → ladder to square 9

This walkthrough shows how BFS naturally handles the complex board navigation and finds the optimal solution.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The algorithm uses BFS to traverse the board. In the worst case, we visit each cell on the board exactly once. Since the board has n × n = n² cells, and for each cell we explore up to 6 possible next positions (dice rolls 1-6), the total operations are bounded by O(n² × 6) = O(n²). The coordinate conversion operations divmod and index calculations are O(1) operations.

Space Complexity: O(n²)

The space complexity comes from two main sources:

1. The vis set stores visited positions, which in the worst case contains all n² cells on the board, requiring O(n²) space.
2. The queue q for BFS can contain at most O(n²) elements in the worst case when multiple cells are reachable at the same level.

Therefore, the total space complexity is O(n²), where n is the length of the side of the board.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Boustrophedon (Zigzag) Pattern Conversion

The most common pitfall is incorrectly converting between the square number (1 to n²) and the board coordinates [row][column]. Many developers make mistakes in:

• Forgetting that odd rows (from the bottom) go right-to-left
• Miscalculating which rows should be reversed
• Confusing 0-based vs 1-based indexing

Wrong Implementation:

# Common mistake: not handling the zigzag pattern correctly
row = (position - 1) // n
col = (position - 1) % n
# Missing the reversal for odd rows!

Correct Implementation:

row_from_bottom, col = divmod(position - 1, n)
if row_from_bottom & 1:  # Odd row from bottom
    col = n - col - 1     # Reverse column for right-to-left
row_from_top = n - row_from_bottom - 1  # Convert to top-down indexing

2. Following Multiple Snakes/Ladders in One Move

Another critical pitfall is recursively following snakes/ladders. When you land on a square with a snake/ladder, you should move to its destination, but if that destination also has a snake/ladder, you should NOT follow it.

Wrong Implementation:

final_position = next_position
while board[row][col] != -1:
    final_position = board[row][col]
    # Recalculate row, col for final_position
    # This is WRONG - only follow one snake/ladder per dice roll

Correct Implementation:

# Only check once - don't chain snakes/ladders
final_position = next_position if board[row][col] == -1 else board[row][col]

3. Visiting Logic Errors

Marking squares as visited at the wrong time can lead to incorrect results or missing optimal paths.

Wrong Implementation:

# Marking visited AFTER popping from queue
current = queue.popleft()
visited.add(current)  # Too late! Might add duplicates to queue

Correct Implementation:

# Mark as visited when ADDING to queue
if final_position not in visited:
    visited.add(final_position)  # Mark immediately
    queue.append(final_position)

4. Off-by-One Errors in Dice Roll Range

The dice roll should allow moves to squares [current + 1, min(current + 6, n²)], inclusive.

Wrong Implementation:

# Missing the +1 for inclusive range
for dice in range(1, 7):
    next_pos = current + dice
    if next_pos > n * n:  # Should use >= or continue instead of break
        break

Correct Implementation:

for dice in range(1, 7):
    next_pos = current + dice
    if next_pos > n * n:
        break  # Can break since subsequent rolls will also exceed

5. Testing Your Coordinate Conversion

A helpful debugging technique is to verify your coordinate conversion with a simple test:

def verify_conversion(n):
    """Print the board with square numbers to verify conversion logic"""
    board_numbers = [[0] * n for _ in range(n)]
    for num in range(1, n * n + 1):
        row_from_bottom, col = divmod(num - 1, n)
        if row_from_bottom & 1:
            col = n - col - 1
        row_from_top = n - row_from_bottom - 1
        board_numbers[row_from_top][col] = num
  
    # Should show the Boustrophedon pattern
    for row in board_numbers:
        print(row)

For a 3x3 board, this should output:

[7, 8, 9]
[6, 5, 4]
[1, 2, 3]