What is the time complexity of Snakes and Ladders?

The optimal solution has a time complexity of O(N^2).

What is the space complexity of Snakes and Ladders?

The space complexity is O(N^2).

What data structures are used to solve Snakes and Ladders?

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

What are the key insights for solving Snakes and Ladders?

Insight 1: The core problem is a shortest path problem, ideally suited for Breadth-First Search (BFS). Insight 2: Mapping the 1D square number to 2D board coordinates is crucial and requires handling the boustrophedon pattern correctly.

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Snakes and Ladders - LeetCode 909 Solution

Q: Which companies ask LeetCode 909. Snakes and Ladders in interviews?

This problem is frequently asked at Anduril, Cisco, National Instruments, PhonePe, Pinterest and 2 more companies.

Snakes and Ladders - Complete Solution Guide

Snakes and Ladders is LeetCode problem 909, 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 n x n integer matrix board where the cells are labeled from 1 to n 2 in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0] ) and alternating direction each row. You start on square 1 of the board. In each move, starting from square curr , do the following: Choose a destination square next with a label in the range [curr + 1, min(curr + 6, n 2 )] . This choice simulates the result of a standard 6-sided die roll : i.e., there are always at most

Detailed Explanation

The Snakes and Ladders problem presents a game board represented by an n x n matrix. The board is numbered from 1 to n^2 in a boustrophedon (snake-like) pattern. The goal is to find the minimum number of dice rolls to reach the square n^2, starting from square 1. Each roll allows you to move from your current square `curr` to a square `next` within the range `[curr + 1, min(curr + 6, n^2)]`. If a square has a snake or a ladder (indicated by a value other than -1), you must move to the destination of that snake or ladder. The objective is to find the shortest path to the final square.

Solution Approach

The solution utilizes Breadth-First Search (BFS) to find the shortest path from square 1 to square n^2. A queue is used to store the squares to be visited, along with the number of moves taken to reach them. The algorithm explores possible moves from each square (simulating a dice roll), checking for snakes or ladders at each step, and ensuring visited squares are not revisited.

Step-by-Step Algorithm

Step 1: Initialize a queue with the starting square (1) and the number of moves (0). Also, initialize a set to keep track of visited squares to avoid cycles.
Step 2: While the queue is not empty, dequeue the current square and its corresponding move count.
Step 3: If the current square is the target square (n^2), return the move count.
Step 4: Iterate through possible dice rolls (1 to 6) from the current square, calculating the next potential square.
Step 5: If the next square is within the board's bounds, determine its row and column coordinates using the `get_coords` function (which accounts for the boustrophedon pattern).
Step 6: Check if the next square has a snake or ladder. If it does, update the destination to the snake/ladder's target square; otherwise, the destination is the originally calculated next square.
Step 7: If the destination square has not been visited, add it to the visited set and enqueue it with an incremented move count.
Step 8: If the queue becomes empty and the target square is not reached, it means there's no path, so return -1.

Key Insights

Insight 1: The core problem is a shortest path problem, ideally suited for Breadth-First Search (BFS).
Insight 2: Mapping the 1D square number to 2D board coordinates is crucial and requires handling the boustrophedon pattern correctly.
Insight 3: Snakes and ladders act as 'shortcuts' or 'jumps' in the graph, but only one jump is allowed per turn. Avoiding cycles is important to prevent infinite loops.

Complexity Analysis

Time Complexity: O(N^2)

Space Complexity: O(N^2)

Topics

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

Companies

Asked at: Anduril, Cisco, National Instruments, PhonePe, Pinterest, Zomato, thoughtspot.

Snakes and Ladders | LeetCode 909

Medium

Array Breadth-First Search Matrix

Time: O(N^2)Space: O(N^2)

Solution Code

Available in:

import java.util.LinkedList;
import java.util.Queue;
import java.util.HashSet;
import java.util.Set;

class Solution {
    public int snakesAndLadders(int[][] board) {
        int n = board.length;
        int target = n * n;

        // Helper function to get coordinates from square number
        class Pair {
            int row, col;
            Pair(int row, int col) {
                this.row = row;
                this.col = col;
            }
        }

        // Helper function to get coordinates from square number
        function Interface get_coords {
           Pair apply(int square);
        }

        get_coords getCoords = (square) -> {
            int s_zero_based = square - 1;
            int row_from_bottom = s_zero_based / n;
            int row = (n - 1) - row_from_bottom;
            int col_offset = s_zero_based % n;
            int col;
            if (row_from_bottom % 2 == 0) {
                col = col_offset;
            } else {
                col = (n - 1) - col_offset;
            }
            return new Pair(row, col);
        };


        Queue<int[]> q = new LinkedList<>();
        q.offer(new int[]{1, 0}); // (square, moves)
        Set<Integer> visited = new HashSet<>();
        visited.add(1);

        while (!q.isEmpty()) {
            int[] curr_data = q.poll();
            int curr = curr_data[0];
            int moves = curr_data[1];

            if (curr == target) {
                return moves;
            }

            for (int i = 1; i <= 6; i++) {
                int next_square_base = curr + i;

                if (next_square_base > target) {
                    break;
                }

                Pair coords = getCoords.apply(next_square_base);
                int r = coords.row;
                int c = coords.col;

                int destination = board[r][c] != -1 ? board[r][c] : next_square_base;

                if (!visited.contains(destination)) {
                    visited.add(destination);
                    q.offer(new int[]{destination, moves + 1});
                }
            }
        }

        return -1;
    }
}

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Problem Info

Description

Problem 909: Snakes and Ladders - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Matrix

Companies

AndurilCiscoNational InstrumentsPhonePePinterestZomatothoughtspot

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:909

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

8 steps

1
Step 1: Initialize a queue with the starting square (1) and the number of moves (0). Also, initialize a set to keep track of visited squares to avoid cycles.
2
Step 2: While the queue is not empty, dequeue the current square and its corresponding move count.
3
Step 3: If the current square is the target square (n^2), return the move count.
4
Step 4: Iterate through possible dice rolls (1 to 6) from the current square, calculating the next potential square.
5
Step 5: If the next square is within the board's bounds, determine its row and column coordinates using the `get_coords` function (which accounts for the boustrophedon pattern).
6
Step 6: Check if the next square has a snake or ladder. If it does, update the destination to the snake/ladder's target square; otherwise, the destination is the originally calculated next square.
7
Step 7: If the destination square has not been visited, add it to the visited set and enqueue it with an incremented move count.
8
Step 8: If the queue becomes empty and the target square is not reached, it means there's no path, so return -1.

💡

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(N^2)

💾

Space Complexity

O(N^2)

📝

Why?

The time complexity is derived from the fact that BFS explores each reachable node (square) at most once. Since the number of squares is N^2, and each square's neighbors are explored in O(1) time (at most 6 neighbors from dice roll), the overall time complexity is O(N^2). The space complexity arises from the `visited` set and the queue, both of which can grow up to the number of squares (N^2) in the worst-case scenario.

📖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

Dijkstra's Algorithm

Dijkstra's algorithm could be used as well to solve this problem. The graph is unweighted or uniformly weighted, making BFS more suitable because Dijkstra's algorithm uses a priority queue, which is more complex than the simple queue used in BFS.

Time: O(N^2 log N) if using a heap-based priority queueSpace: O(N^2)

Trade-offs: For this particular problem, Dijkstra's is generally less efficient than BFS due to the uniform 'cost' of each move. Dijkstra's Algorithm becomes more advantageous when different moves have different weights or costs. For instance, a move that utilizes a ladder could be given a weight of 0 if the primary objective is solely to reach the end as fast as possible, regardless of dice rolls.

✨

Key Insights

1Insight 1: The core problem is a shortest path problem, ideally suited for Breadth-First Search (BFS).
2Insight 2: Mapping the 1D square number to 2D board coordinates is crucial and requires handling the boustrophedon pattern correctly.
3Insight 3: Snakes and ladders act as 'shortcuts' or 'jumps' in the graph, but only one jump is allowed per turn. Avoiding cycles is important to prevent infinite loops.

📚

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: Board with no snakes or ladders. The algorithm correctly finds the shortest path by simply rolling the dice until the target is reached.
Edge case 2: Board where target is unreachable due to snakes and ladders creating a cycle. The algorithm correctly returns -1 because the BFS will terminate without finding the target.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly mapping the 1D square number to 2D board coordinates. Failing to account for the boustrophedon pattern leads to incorrect calculations and incorrect movement on the board.

✗Common mistake 2: Failing to use a `visited` set or similar mechanism to prevent cycles. Without this, the BFS can get stuck in infinite loops, especially if snakes and ladders create cyclical paths.

✅

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 video games where different terrain types have varying costs associated with them, similar to snakes/ladders offering 'free' jumps or obstacles hindering progress. Finding the most efficient path.

🏦Real-world scenario 2: Route planning in logistics or transportation networks where certain routes might represent shortcuts (like ladders) or detours (like snakes), and the goal is to find the quickest delivery route.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: June 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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 "Snakes and Ladders"?

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 breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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