What is the time complexity of Knight Dialer?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Knight Dialer?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Knight Dialer?

Insight 1: The problem can be solved using dynamic programming, where the state represents the number of ways to reach a particular digit after a certain number of jumps. Insight 2: We can represent the possible moves from each digit using a predefined adjacency list or matrix. This eliminates the need to calculate valid moves for each step.

Which companies ask LeetCode 935. Knight Dialer in interviews?

This problem is frequently asked at Bridgewater Associates, Snap.

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Knight Dialer - LeetCode 935 Solution

Knight Dialer - Complete Solution Guide

Knight Dialer is LeetCode problem 935, 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

The chess knight has a unique movement , it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an L ). The possible movements of chess knight are shown in this diagram: A chess knight can move as indicated in the chess diagram below: We have a chess knight and a phone pad as shown below, the knight can only stand on a numeric cell (i.e. blue cell). Given an integer n , return how many distinct phone n

Detailed Explanation

The "Knight Dialer" problem asks you to determine the number of distinct phone numbers of length 'n' that can be dialed using a chess knight's movement pattern on a standard phone dial pad (0-9). The knight starts on any digit, and each subsequent digit must be reachable by a valid knight move. The answer needs to be returned modulo 10^9 + 7 to prevent integer overflow.

Solution Approach

The provided solution utilizes dynamic programming. The core idea is to build a table (represented by `dp` arrays in different languages) where `dp[i]` stores the number of distinct phone numbers of a given length that end on digit `i`. The algorithm iterates from length 1 up to 'n', updating the `dp` table in each iteration. For each digit, it sums the number of ways to reach that digit from all the digits that a knight could have jumped from in the previous step. The sum of all elements in the `dp` table at the end of the iterations gives the final result, modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Define the valid knight moves from each digit. This is usually done using an array or list, `moves`, where `moves[i]` contains a list of digits reachable from digit `i` in one knight's move.
Step 2: Initialize a `dp` array of size 10, where `dp[i]` represents the number of ways to reach digit `i`. Initially, all `dp[i]` are set to 1 because there is one way to start at any digit for n = 1.
Step 3: Iterate from 1 to n-1 (since the initial state covers the case n=1). In each iteration, create a `next_dp` array to store the updated counts for the next jump.
Step 4: For each digit `digit` from 0 to 9, iterate through all the possible `next_move` destinations from that digit, obtained from the `moves` array. Update `next_dp[next_move]` by adding `dp[digit]` (modulo 10^9 + 7) to account for all the ways to reach the current digit.
Step 5: After iterating through all the digits, update `dp` with the values from `next_dp`. This prepares the `dp` array for the next jump.
Step 6: After the loop finishes, calculate the sum of all the values in `dp` (modulo 10^9 + 7). This sum represents the total number of distinct phone numbers of length 'n' that can be dialed.
Step 7: Return the final sum (modulo 10^9 + 7).

Key Insights

Insight 1: The problem can be solved using dynamic programming, where the state represents the number of ways to reach a particular digit after a certain number of jumps.
Insight 2: We can represent the possible moves from each digit using a predefined adjacency list or matrix. This eliminates the need to calculate valid moves for each step.
Insight 3: Since we only need the number of ways for the *previous* jump to calculate the number of ways for the *current* jump, we can optimize space complexity by using only two arrays (or in some language implementations, one array). This helps keep the memory footprint low, especially for larger values of 'n'.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Bridgewater Associates, Snap.

Knight Dialer | LeetCode 935

Medium

Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int knightDialer(int n) {
        int MOD = (int)1e9 + 7;

        int[][] moves = {
            {4, 6},
            {6, 8},
            {7, 9},
            {4, 8},
            {0, 3, 9},
            {},
            {0, 1, 7},
            {2, 6},
            {1, 3},
            {2, 4},
        };

        // dp[i] represents the number of ways to end a sequence on digit i
        int[] dp = new int[10];
        for (int i = 0; i < 10; i++) {
            dp[i] = 1;
        }

        // We need to perform n - 1 jumps for a number of length n
        for (int i = 0; i < n - 1; i++) {
            int[] next_dp = new int[10];
            for (int digit = 0; digit < 10; digit++) {
                for (int next_move : moves[digit]) {
                    if (next_move >= 0 && next_move < 10) {
                    next_dp[next_move] = (int)(((long)next_dp[next_move] + dp[digit]) % MOD);
                    }
                }
            }
            dp = next_dp;
        }

        long sum = 0;
        for (int val : dp) {
            sum += val;
        }

        return (int)(sum % MOD);
    }
}

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Problem Info

Description

Problem 935: Knight Dialer - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Bridgewater AssociatesSnap

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:935

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

7 steps

1
Step 1: Define the valid knight moves from each digit. This is usually done using an array or list, `moves`, where `moves[i]` contains a list of digits reachable from digit `i` in one knight's move.
2
Step 2: Initialize a `dp` array of size 10, where `dp[i]` represents the number of ways to reach digit `i`. Initially, all `dp[i]` are set to 1 because there is one way to start at any digit for n = 1.
3
Step 3: Iterate from 1 to n-1 (since the initial state covers the case n=1). In each iteration, create a `next_dp` array to store the updated counts for the next jump.
4
Step 4: For each digit `digit` from 0 to 9, iterate through all the possible `next_move` destinations from that digit, obtained from the `moves` array. Update `next_dp[next_move]` by adding `dp[digit]` (modulo 10^9 + 7) to account for all the ways to reach the current digit.
5
Step 5: After iterating through all the digits, update `dp` with the values from `next_dp`. This prepares the `dp` array for the next jump.
6
Step 6: After the loop finishes, calculate the sum of all the values in `dp` (modulo 10^9 + 7). This sum represents the total number of distinct phone numbers of length 'n' that can be dialed.
7
Step 7: Return the final sum (modulo 10^9 + 7).

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the loop that iterates from 1 to n-1, representing the jumps. The space complexity is constant because regardless of the size of `n`, the number of possible starting positions (digits 0-9) and the adjacency information between them remain fixed.

📖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

Matrix Exponentiation

The problem can also be solved using matrix exponentiation. We can represent the possible moves as an adjacency matrix. Then, raising this matrix to the power of (n-1) and summing the elements in the first row gives the answer. This approach is beneficial for very large values of n.

Time: O(log n)Space: O(1)

Trade-offs: Matrix exponentiation provides logarithmic time complexity, making it faster for very large values of 'n'. However, the constant factors involved in matrix multiplication might make it slower than the dynamic programming approach for smaller values of 'n'. It's also more complex to implement.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming, where the state represents the number of ways to reach a particular digit after a certain number of jumps.
2Insight 2: We can represent the possible moves from each digit using a predefined adjacency list or matrix. This eliminates the need to calculate valid moves for each step.
3Insight 3: Since we only need the number of ways for the *previous* jump to calculate the number of ways for the *current* jump, we can optimize space complexity by using only two arrays (or in some language implementations, one array). This helps keep the memory footprint low, especially for larger values of 'n'.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly handles this case by initializing the `dp` array with 1 for each digit, and the sum of the array (which is 10) is returned.
Edge case 2: Large values of n. The solution uses modulo arithmetic (10^9 + 7) to prevent integer overflow when calculating the number of ways to reach a digit. Without the modulo operation, the intermediate sums could exceed the maximum integer value, leading to incorrect results.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not applying the modulo operation. Without taking the modulo, the number of ways to dial a phone number can quickly exceed the maximum integer value, leading to incorrect results.

✗Common mistake 2: Incorrectly defining the valid knight moves. A mistake in defining the moves array will lead to exploring invalid paths and therefore an incorrect count.

✗Common mistake 3: Using a large data type only in the final return. Intermediate calculations can still overflow, so the modulo operation needs to be applied within the loop to prevent overflow.

✅

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: Route planning in a constrained environment. The problem is analogous to finding the number of paths a robot can take in a grid-like environment with specific movement constraints.

🏦Real-world scenario 2: Modeling network hops. The knight's movements can be seen as hops within a network. The algorithm can be used to determine the number of possible paths to reach a particular node in the network after a certain number of hops, given connection constraints.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 6, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

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 "Knight Dialer"?

For this medium-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?