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Solving Snake & Ladder Problem in Python: Coding Interview Guide

Updated: at 05:12 AM

The “Snake & Ladder Problem” is a common technical coding interview question asked by companies like Google, Facebook, Amazon, and Microsoft during their recruiting process. This problem tests a candidate’s skills in graph theory, breadth-first search algorithms, object-oriented programming, and solving problems proactively.

In this comprehensive guide, we will examine the Snake & Ladder problem statement, analyze the problem-solving techniques, implement an object-oriented solution in Python, and discuss tips to ace this interview question.

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Overview of the Snake & Ladder Problem

The Snake & Ladder problem emulates the classic board game where players roll a six-sided die to navigate from the start of the board to the end, moving ahead or back based on landing on spaces containing ladders and snakes.

In the coding interview version, the candidate must write a program to find the minimum number of dice rolls required to reach the final position on the board, given the board configuration of snakes, ladders, and normal spaces.

This question tests a coder’s ability to:

To solve this, we will use the following problem-solving techniques:

Step-by-Step Python Solution

Below we will walk through a step-by-step solution to the Snake & Ladder problem in Python.

1. Model the Problem Domain

We first need to model the key components of the problem using OOP principles:


The board will contain N rows and N columns of squares where N is the size of the board. Each square will have:


The dice will represent a six-sided die with possible values 1 to 6. It needs:


The player keeps track of:

The player requires:


Each snake contains:


Each ladder contains:

2. Implement the Board Class

We will implement the Board class to model the game board:

import random

class Board:

  def __init__(self, snakes, ladders, size):
    self.board = [[0 for _ in range(size)] for _ in range(size)]
    self.snakes = snakes
    self.ladders = ladders
    self.final_pos = size * size

    # Initialize board squares
    for row in range(size):
      for col in range(size):
        self.board[row][col] = row * size + col + 1

    # Set final square flag
    self.board[size-1][size-1] = -1

  def move(self, position, dice_value):

The __init__() method initializes the 2D matrix to represent the board, sets the final square flag, and maps position numbers to the matrix indices.

We encapsulate the snake and ladder configurations as constructor parameters so the board can be customized.

The move() method will apply the snakes, ladders and dice logic to calculate the next position. We will implement it next.

3. Add Snakes, Ladders and Dice Logic

Let’s update the Board class to apply the effects of snakes, ladders and dice when moving:

# Inside Board class

def move(self, position, dice_value):

  next_pos = position + dice_value

  # Wrap around for cyclic board
  if next_pos > self.final_pos:
    next_pos -= self.final_pos

  # Apply snake/ladder logic
  if next_pos in self.snakes:
    next_pos = self.snakes[next_pos]
  elif next_pos in self.ladders:
    next_pos = self.ladders[next_pos]

  return next_pos

We increment the current position by the dice value to get the next position. To simulate a cyclic board, we wrap around if the position exceeds the final square.

Using dict lookups, we map the next position to the end of a snake or ladder appropriately.

This encapsulates all the game logic required for a single dice roll turn.

4. Implement the Player Class

The Player class will model a player with a current position and number of dice rolls:

class Player:

  def __init__(self, board, position=0):
    self.board = board
    self.position = position
    self.no_plays = 0

  def take_turn(self, dice):
    dice_value = dice.roll()
    self.no_plays += 1
    self.position = self.board.move(self.position, dice_value)

The take_turn() method rolls the dice, increments dice rolls counter, and updates player’s position using the underlying Board logic.

Now we need a way to simulate the game by exploring all possible dice roll sequences using breadth-first search until we find the shortest path.

from collections import deque

def simulate(player):

  queue = deque()
  queue.append((player, 0)) # Tuple of (player, no_of_turns)

  visited = set() # Track states already visited

  while queue:

    current_player, turns = queue.popleft()

    # Check if current state is final position
    if current_player.position == current_player.board.final_pos:
      return turns

    curr_state = (current_player.position, turns)

    if curr_state in visited:


    # Generate all possible dice rolls
    for i in range(1,7):
      player_copy = copy(current_player)
      queue.append((player_copy, turns+1))

  return -1 # Unreachable final position

We use a BFS queue to explore all possible game states in sequence. We check if the current player state has reached the final position, returning the number of turns if true.

To prevent infinite loops, we store the visited (position, turns) states in a hash set.

For each state, we generate new player instances for all possible dice rolls and add them to the queue with incremented number of turns.

This allows us to simulate playing the game iteratively until we find the shortest path.

6. Initialize and Run the Simulation

Finally, we need to initialize the board, player, and dice instances and run the simulation:

# Set board size
board_size = 30

# Set snake and ladder configurations
snakes = {...}
ladders = {...}

# Create components
board = Board(snakes, ladders, board_size)
player = Player(board)
dice = Dice()


This will output the minimum number of dice rolls needed to complete the game.

Analyzing Solution Complexity

The above BFS approach provides an optimal solution as we are exploring all possible dice roll sequences in increasing order of turns.

Time Complexity: O(N!)

In the worst case where no shortcuts are available, the time complexity is O(N!) where N is the board size since we need to enumerate all permutations of dice rolls.

Having shortcuts reduces the complexity as paths are cut short.

Space Complexity: O(N)

We need O(N) space for the BFS queue, visited set, and recursive stack depth.

Tips for Acing the Coding Interview

Here are some tips to ace the Snake & Ladder interview question:

Practical Applications

The Snake & Ladder problem demonstrates fundamental concepts like graphs, search algorithms, OOP, recursion, and dynamic programming that are applicable in many domains:

Path Finding - Finding shortest path in route planning, network routing.

AI - Game simulations, state space exploration in AI agents.

Operations Research - Logistics optimization.

Bioinformatics - DNA sequence alignment.

Machine Learning - Prediction of sequence/time series data.


In this guide, we covered a step-by-step object-oriented approach to solving the Snake & Ladder coding interview problem in Python using breadth-first search.

Key concepts included graph modeling, OOP principles, BFS traversal, hash tables, recursion, and modular design. Mastering these foundations will help crack not just this problem but a variety of other algorithm and data structure problems.

With diligent practice of core CS fundamentals, strong analytical skills, and clear communication, you can successfully solve challenging technical interview questions like the Snake & Ladder problem.