#935

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Knight Dialer

Solve the Knight Dialer problem using state transition dynamic programming to efficiently calculate valid number paths for any given length.

dynamic programming

Problem Statement

Given a 3x3 phone pad where each number is placed on specific cells, a knight starts on any of these 10 numeric cells. The knight moves in an L-shape, either 2 squares vertically and 1 horizontally or 2 squares horizontally and 1 vertically. For each length n, compute the number of distinct valid numbers that the knight can dial following these movements, ensuring all transitions respect the constraints of the pad.

The goal is to find an efficient way to compute the number of valid paths of length n. The knight can make these transitions multiple times, but each move must be valid within the pad's bounds. This is a typical dynamic programming problem where we focus on state transitions and avoid recomputation of intermediate results.

Examples

Example 1

Input: n = 1

Output: 10

We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.

Example 2

Input: n = 2

Output: 20

All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]

Example 3

Input: n = 3131

Output: 136006598

Please take care of the mod.

Constraints

1 <= n <= 5000

Solution Approach

State Transition Dynamic Programming

The knight's movement from any digit can be expressed as a state transition. We define a DP array where each element represents the number of ways to reach each digit. For each move, we update the DP state based on valid knight movements from the previous step.

Modular Arithmetic

Given the constraints, the result may grow large, so we use modular arithmetic to keep the results within the required bounds, ensuring we compute the result modulo 10^9 + 7.

Time and Space Complexity Optimization

By using a rolling DP array instead of maintaining the entire DP matrix for every step, we reduce space complexity to O(1) while still ensuring an optimal time complexity of O(n), making the solution efficient for large inputs.

Complexity Analysis

Metric	Value
Time	O(\log{}n)
Space	O(1)

Time complexity is O(n) due to the linear progression of the knight’s movements. Space complexity is reduced to O(1) by using only two arrays (current and previous states), optimizing memory usage while maintaining the same time efficiency.

What Interviewers Usually Probe

The candidate should demonstrate an understanding of state transition dynamic programming for solving path problems.
They should address space optimization through rolling arrays and modular arithmetic in large number problems.
Look for clarity in their explanation of the knight’s movement rules and how they lead to state transitions in the dynamic programming solution.

Common Pitfalls or Variants

Common pitfalls

Not handling the modular arithmetic correctly, leading to overflow or incorrect results.
Failing to optimize space complexity, potentially using an O(n) space solution instead of the required O(1).
Misunderstanding the knight's movement rules, resulting in incorrect valid paths.

Follow-up variants

Consider different movement rules or additional constraints (e.g., limiting the knight to specific rows or columns).
Modify the problem to ask for all the possible valid numbers rather than just the count.
Increase the pad size or modify the grid structure to test for scalability of the solution.

FAQ

What is the Knight Dialer problem?

The Knight Dialer problem asks for the number of valid phone numbers a knight can dial on a standard phone pad, following specific movement rules of a knight in chess.

How do you solve the Knight Dialer problem using dynamic programming?

By treating each digit as a state and transitioning between states based on knight moves, you can use dynamic programming to compute the number of valid paths for any given length n.

Why do we use modular arithmetic in the Knight Dialer problem?

Modular arithmetic is used to prevent overflow and to keep results within the bounds required by the problem, typically modulo 10^9 + 7.

What is the time and space complexity of the Knight Dialer problem?

The time complexity is O(n) and the space complexity is O(1), as the solution uses a rolling DP array to compute the result efficiently.

What are common pitfalls in solving the Knight Dialer problem?

Common pitfalls include not using modular arithmetic correctly, not optimizing space complexity, and misunderstanding the knight's movement rules.

terminal

Solution

Solution 1: Recurrence

According to the problem description, we need to calculate the number of different phone numbers of length $n$. Each digit can only follow certain fixed digits, which we can list as follows:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

class Solution:
    def knightDialer(self, n: int) -> int:
        f = [1] * 10
        for _ in range(n - 1):
            g = [0] * 10
            g[0] = f[4] + f[6]
            g[1] = f[6] + f[8]
            g[2] = f[7] + f[9]
            g[3] = f[4] + f[8]
            g[4] = f[0] + f[3] + f[9]
            g[6] = f[0] + f[1] + f[7]
            g[7] = f[2] + f[6]
            g[8] = f[1] + f[3]
            g[9] = f[2] + f[4]
            f = g
        return sum(f) % (10**9 + 7)

Solution 2: Matrix Exponentiation to Accelerate Recurrence

Let's denote $T(n)$ as a $1 \times 10$ matrix $\begin{bmatrix} F_0 & F_1 & F_2 \cdots F_9 \end{bmatrix}$, where $F_i$ represents the number of phone numbers ending with digit $i$. We want to derive $T(n)$ from $T(n - 1)$. In other words, we need a matrix $\textit{base}$ such that $T(n - 1) \times \textit{base} = T(n)$, i.e.:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

class Solution:
    def knightDialer(self, n: int) -> int:
        f = [1] * 10
        for _ in range(n - 1):
            g = [0] * 10
            g[0] = f[4] + f[6]
            g[1] = f[6] + f[8]
            g[2] = f[7] + f[9]
            g[3] = f[4] + f[8]
            g[4] = f[0] + f[3] + f[9]
            g[6] = f[0] + f[1] + f[7]
            g[7] = f[2] + f[6]
            g[8] = f[1] + f[3]
            g[9] = f[2] + f[4]
            f = g
        return sum(f) % (10**9 + 7)

Continue Practicing

#931 minimum falling path sum #940 distinct subsequences ii #943 find the shortest superstring