onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming - Complex approach
Space-Optimized Dynamic Programming - Complex approach
Matrix Exponentiation - Complex approach

Approach 1: Dynamic Programming

The key insight for solving this problem is to use dynamic programming to count the number of distinct phone numbers that can be dialed.

Let's define a 2D DP array dp[i][j] where:

i represents the number of jumps made so far (from 0 to n-1)
j represents the current digit (from 0 to 9)

The value dp[i][j] represents the number of distinct phone numbers of length i+1 that end with digit j.

The base cases are:

dp[0][j] = 1 for all j from 0 to 9 (there is one way to dial a number of length 1 for each digit)

For the recurrence relation, we need to consider all possible previous digits that can lead to the current digit through a valid knight move.

Let's define the possible knight moves for each digit:

From digit 0, the knight can move to digits 4 and 6
From digit 1, the knight can move to digits 6 and 8
From digit 2, the knight can move to digits 7 and 9
From digit 3, the knight can move to digits 4 and 8
From digit 4, the knight can move to digits 0, 3, and 9
From digit 5, the knight cannot move to any digit
From digit 6, the knight can move to digits 0, 1, and 7
From digit 7, the knight can move to digits 2 and 6
From digit 8, the knight can move to digits 1 and 3
From digit 9, the knight can move to digits 2 and 4

For each digit j and jump i, we sum up the number of ways to reach all possible previous digits after i-1 jumps:

dp[i][j] = sum(dp[i-1][prev]) for all prev that can jump to j

The final answer is the sum of dp[n-1][j] for all j from 0 to 9.

Algorithm:

Define the possible knight moves for each digit.
Initialize a 2D DP array dp of size n × 10, where dp[i][j] represents the number of distinct phone numbers of length i+1 that end with digit j.
Set the base cases: dp[0][j] = 1 for all j from 0 to 9.
For each jump i from 1 to n-1:
a. For each digit j from 0 to 9:
i. For each previous digit prev that can jump to j:
1. dp[i][j] += dp[i-1][prev]
ii. Apply modulo 10^9 + 7 to dp[i][j]
Calculate the final answer as the sum of dp[n-1][j] for all j from 0 to 9, modulo 10^9 + 7.

Time Complexity:

O(n)

We need to compute the DP values for each jump and each digit. Since there are n jumps and 10 digits, and for each digit, we consider a constant number of previous digits, the time complexity is O(n).

Space Complexity:

O(n)

We use a 2D DP array of size n × 10 to store the number of distinct phone numbers for each jump and each digit.

Approach 2: Space-Optimized Dynamic Programming

We can optimize the space complexity by observing that to compute the DP values for the current jump, we only need the DP values for the previous jump.

Instead of using a 2D DP array, we can use two 1D arrays: curr and prev, each of size 10, to store the DP values for the current jump and the previous jump, respectively.

After computing the DP values for the current jump, we update prev to curr and reset curr for the next jump.

The recurrence relation remains the same:

curr[j] = sum(prev[prev_digit]) for all prev_digit that can jump to j

The final answer is the sum of prev[j] for all j from 0 to 9 after the last jump.

Algorithm:

Define the possible knight moves for each digit.
Initialize two 1D arrays prev and curr, each of size 10, where prev[j] represents the number of distinct phone numbers of length i that end with digit j.
Set the base cases: prev[j] = 1 for all j from 0 to 9.
For each jump i from 1 to n-1:
a. Reset curr to all zeros.
b. For each digit j from 0 to 9:
i. For each previous digit prev_digit that can jump to j:
1. curr[j] += prev[prev_digit]
ii. Apply modulo 10^9 + 7 to curr[j]
c. Update prev = curr.
Calculate the final answer as the sum of prev[j] for all j from 0 to 9, modulo 10^9 + 7.

Time Complexity:

O(n)

The time complexity remains the same as the previous approach, as we still need to compute the DP values for each jump and each digit.

Space Complexity:

O(1)

We only use two 1D arrays of size 10, which is a constant space regardless of the input size.

Approach 3: Matrix Exponentiation

For large values of n, we can use matrix exponentiation to compute the result more efficiently.

We can represent the transitions between digits as a 10×10 adjacency matrix M, where M[i][j] = 1 if there is a valid knight move from digit i to digit j, and M[i][j] = 0 otherwise.

Let's define a column vector v of size 10, where v[i] = 1 for all i from 0 to 9, representing the base case.

Then, the number of distinct phone numbers of length n is given by the sum of all elements in the vector M^(n-1) * v.

We can compute M^(n-1) efficiently using binary exponentiation, which takes O(log n) time.

Algorithm:

Define the adjacency matrix M, where M[i][j] = 1 if there is a valid knight move from digit i to digit j, and M[i][j] = 0 otherwise.
Define a column vector v of size 10, where v[i] = 1 for all i from 0 to 9.
Compute M^(n-1) using binary exponentiation.
Compute the result vector res = M^(n-1) * v.
Calculate the final answer as the sum of all elements in res, modulo 10^9 + 7.

Time Complexity:

O(log n)

Using binary exponentiation, we can compute the matrix power in logarithmic time.

Space Complexity:

O(1)

We use a constant amount of space to store the matrices and vectors, regardless of the input size.

Pseudocode

solution.pseudo

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function knightDialer(n):
    MOD = 10^9 + 7
 
    // Define the possible knight moves for each digit
    moves = [
        [4, 6],       // From digit 0
        [6, 8],       // From digit 1
        [7, 9],       // From digit 2
        [4, 8],       // From digit 3
        [0, 3, 9],    // From digit 4
        [],           // From digit 5
        [0, 1, 7],    // From digit 6
        [2, 6],       // From digit 7
        [1, 3],       // From digit 8
        [2, 4]        // From digit 9
    ]
 
    // Initialize DP array
    dp = new 2D array of size n × 10, initialized with 0
 
    // Base cases
    for j from 0 to 9:
        dp[0][j] = 1
 
    // Fill DP array
    for i from 1 to n-1:
        for j from 0 to 9:
            for prev in moves[j]:
                dp[i][j] = (dp[i][j] + dp[i-1][prev]) % MOD
 
    // Calculate the final answer
    result = 0
    for j from 0 to 9:
        result = (result + dp[n-1][j]) % MOD
 
    return result

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Learning Objective

Understand different approaches to solve the knight dialer problem and analyze their efficiency.