935. Knight Dialer

Description

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 diagaram:

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 numbers of length n we can dial.

You are allowed to place the knight on any numeric cell initially and then you should perform n - 1 jumps to dial a number of length n. All jumps should be valid knight jumps.

As the answer may be very large, return the answer modulo 10⁹ + 7.

Example 1:

Input: n = 1
Output: 10
Explanation: 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
Explanation: 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
Explanation: Please take care of the mod.

Constraints:

1 <= n <= 5000

Solutions

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:

Current Digit	Previous Digits
0	4, 6
1	6, 8
2	7, 9
3	4, 8
4	0, 3, 9
5
6	0, 1, 7
7	2, 6
8	1, 3
9	2, 4

We can use a recurrence approach to calculate the number of different phone numbers of length $n$. Let $f[i]$ represent the number of different phone numbers of length $i$. Initially, $f[1] = 1$. For phone numbers of length $i$, we can calculate them based on phone numbers of length $i - 1$. Therefore, we can derive the recurrence relations:

\[\begin{aligned} g[0] & = f[4] + f[6] \\ g[1] & = f[6] + f[8] \\ g[2] & = f[7] + f[9] \\ g[3] & = f[4] + f[8] \\ g[4] & = f[0] + f[3] + f[9] \\ g[6] & = f[0] + f[1] + f[7] \\ g[7] & = f[2] + f[6] \\ g[8] & = f[1] + f[3] \\ g[9] & = f[2] + f[4] \end{aligned}\]

Then, we update $f$ to $g$ and continue calculating the phone numbers of the next length until we calculate the number of phone numbers of length $n$.

Finally, we sum all the elements in $f$ and take the result modulo $10^9 + 7$ to get the answer.

The time complexity is $O(n)$, where $n$ is the length of the phone number. The space complexity is $O(

\Sigma

)$, where $\Sigma$ is the set of digits, and in this problem $

\Sigma

= 10$.

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.:

\[\begin{bmatrix} F_0 & F_1 & F_2 \cdots F_9 \end{bmatrix} \times \textit{base} = \begin{bmatrix} F_0' & F_1' & F_2' \cdots F_9' \end{bmatrix}\]

Since $F_i’ = \sum_{j} F_j$, where $j$ is the previous digit of $i$, the first column of the matrix $\textit{base}$ is:

\[\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\]

Similarly, we can derive the entire matrix $\textit{base}$ as follows:

\[\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\]

We define the initial matrix $res = \begin{bmatrix} 1 & 1 & 1 \cdots 1 \end{bmatrix}$, and multiply it by the matrix $\textit{base}$ raised to the power of $n - 1$ to obtain $T(n)$. Finally, we sum all elements in $T(n)$ and take the result modulo $10^9 + 7$ to get the answer. The matrix $\textit{base}^{n - 1}$ can be computed using matrix exponentiation, which has a time complexity of $O(\log n)$.

The time complexity is $O(\log n)$, and the space complexity is $O(

\Sigma

^2)$, where $\Sigma$ is the set of digits, and in this problem $

\Sigma

= 10$.

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int knightDialer(int n) {
        if (n == 1) {
            return 10;
        }
        long[] f = new long[10];
        Arrays.fill(f, 1);
        while (--n > 0) {
            long[] t = new long[10];
            t[0] = f[4] + f[6];
            t[1] = f[6] + f[8];
            t[2] = f[7] + f[9];
            t[3] = f[4] + f[8];
            t[4] = f[0] + f[3] + f[9];
            t[6] = f[0] + f[1] + f[7];
            t[7] = f[2] + f[6];
            t[8] = f[1] + f[3];
            t[9] = f[2] + f[4];
            for (int i = 0; i < 10; ++i) {
                f[i] = t[i] % MOD;
            }
        }
        long ans = 0;
        for (long v : f) {
            ans = (ans + v) % MOD;
        }
        return (int) ans;
    }
}


// Solution 2
class Solution {
    private final int mod = (int) 1e9 + 7;
    private final int[][] base = {{0, 0, 0, 0, 1, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
        {0, 0, 0, 0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
        {1, 0, 0, 1, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        {1, 1, 0, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 1, 0, 0, 0, 1, 0, 0, 0},
        {0, 1, 0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 1, 0, 0, 0, 0, 0}};

    public int knightDialer(int n) {
        int[][] res = pow(base, n - 1);
        int ans = 0;
        for (int x : res[0]) {
            ans = (ans + x) % mod;
        }
        return ans;
    }

    private int[][] mul(int[][] a, int[][] b) {
        int m = a.length, n = b[0].length;
        int[][] c = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < b.length; ++k) {
                    c[i][j] = (int) ((c[i][j] + 1L * a[i][k] * b[k][j] % mod) % mod);
                }
            }
        }
        return c;
    }

    private int[][] pow(int[][] a, int n) {
        int[][] res = new int[1][a.length];
        Arrays.fill(res[0], 1);
        while (n > 0) {
            if ((n & 1) == 1) {
                res = mul(res, a);
            }
            a = mul(a, a);
            n >>= 1;
        }
        return res;
    }
}

using ll = long long;

class Solution {
public:
    int knightDialer(int n) {
        if (n == 1) return 10;
        int mod = 1e9 + 7;
        vector<ll> f(10, 1ll);
        while (--n) {
            vector<ll> t(10);
            t[0] = f[4] + f[6];
            t[1] = f[6] + f[8];
            t[2] = f[7] + f[9];
            t[3] = f[4] + f[8];
            t[4] = f[0] + f[3] + f[9];
            t[6] = f[0] + f[1] + f[7];
            t[7] = f[2] + f[6];
            t[8] = f[1] + f[3];
            t[9] = f[2] + f[4];
            for (int i = 0; i < 10; ++i) f[i] = t[i] % mod;
        }
        ll ans = accumulate(f.begin(), f.end(), 0ll);
        return (int) (ans % mod);
    }
};


// Solution 2
class Solution {
public:
    int knightDialer(int n) {
        const int mod = 1e9 + 7;
        vector<vector<int>> base = {
            {0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
            {0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
            {0, 0, 0, 0, 0, 0, 0, 1, 0, 1},
            {0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
            {1, 0, 0, 1, 0, 0, 0, 0, 0, 1},
            {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
            {1, 1, 0, 0, 0, 0, 0, 1, 0, 0},
            {0, 0, 1, 0, 0, 0, 1, 0, 0, 0},
            {0, 1, 0, 1, 0, 0, 0, 0, 0, 0},
            {0, 0, 1, 0, 1, 0, 0, 0, 0, 0}};
        vector<vector<int>> res = pow(base, n - 1, mod);
        return accumulate(res[0].begin(), res[0].end(), 0LL) % mod;
    }

private:
    vector<vector<int>> mul(const vector<vector<int>>& a, const vector<vector<int>>& b, int mod) {
        int m = a.size(), n = b[0].size();
        vector<vector<int>> c(m, vector<int>(n, 0));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < b.size(); ++k) {
                    c[i][j] = (c[i][j] + (1LL * a[i][k] * b[k][j]) % mod) % mod;
                }
            }
        }
        return c;
    }

    vector<vector<int>> pow(vector<vector<int>>& a, int n, int mod) {
        int size = a.size();
        vector<vector<int>> res(1, vector<int>(size, 1));
        while (n > 0) {
            if (n % 2 == 1) {
                res = mul(res, a, mod);
            }
            a = mul(a, a, mod);
            n /= 2;
        }
        return res;
    }
};

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


# Solution 2
import numpy as np

base = [
    (0, 0, 0, 0, 1, 0, 1, 0, 0, 0),
    (0, 0, 0, 0, 0, 0, 1, 0, 1, 0),
    (0, 0, 0, 0, 0, 0, 0, 1, 0, 1),
    (0, 0, 0, 0, 1, 0, 0, 0, 1, 0),
    (1, 0, 0, 1, 0, 0, 0, 0, 0, 1),
    (0, 0, 0, 0, 0, 0, 0, 0, 0, 0),
    (1, 1, 0, 0, 0, 0, 0, 1, 0, 0),
    (0, 0, 1, 0, 0, 0, 1, 0, 0, 0),
    (0, 1, 0, 1, 0, 0, 0, 0, 0, 0),
    (0, 0, 1, 0, 1, 0, 0, 0, 0, 0),
]


class Solution:
    def knightDialer(self, n: int) -> int:
        factor = np.asmatrix(base, np.dtype("O"))
        res = np.asmatrix([[1] * 10], np.dtype("O"))
        n -= 1
        mod = 10**9 + 7
        while n:
            if n & 1:
                res = res * factor % mod
            factor = factor * factor % mod
            n >>= 1
        return res.sum() % mod

func knightDialer(n int) int {
	if n == 1 {
		return 10
	}
	f := make([]int, 10)
	for i := range f {
		f[i] = 1
	}
	mod := int(1e9) + 7
	for i := 1; i < n; i++ {
		t := make([]int, 10)
		t[0] = f[4] + f[6]
		t[1] = f[6] + f[8]
		t[2] = f[7] + f[9]
		t[3] = f[4] + f[8]
		t[4] = f[0] + f[3] + f[9]
		t[6] = f[0] + f[1] + f[7]
		t[7] = f[2] + f[6]
		t[8] = f[1] + f[3]
		t[9] = f[2] + f[4]
		for j, v := range t {
			f[j] = v % mod
		}
	}
	ans := 0
	for _, v := range f {
		ans = (ans + v) % mod
	}
	return ans
}


// Solution 2
const mod = 1e9 + 7

func knightDialer(n int) int {
	base := [][]int{
		{0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
		{0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
		{0, 0, 0, 0, 0, 0, 0, 1, 0, 1},
		{0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
		{1, 0, 0, 1, 0, 0, 0, 0, 0, 1},
		{0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
		{1, 1, 0, 0, 0, 0, 0, 1, 0, 0},
		{0, 0, 1, 0, 0, 0, 1, 0, 0, 0},
		{0, 1, 0, 1, 0, 0, 0, 0, 0, 0},
		{0, 0, 1, 0, 1, 0, 0, 0, 0, 0},
	}

	res := pow(base, n-1)
	ans := 0
	for _, x := range res[0] {
		ans = (ans + x) % mod
	}
	return ans
}

func mul(a, b [][]int) [][]int {
	m := len(a)
	n := len(b[0])
	c := make([][]int, m)
	for i := range c {
		c[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			for k := 0; k < len(b); k++ {
				c[i][j] = (c[i][j] + a[i][k]*b[k][j]) % mod
			}
		}
	}
	return c
}

func pow(a [][]int, n int) [][]int {
	size := len(a)
	res := make([][]int, 1)
	res[0] = make([]int, size)
	for i := 0; i < size; i++ {
		res[0][i] = 1
	}

	for n > 0 {
		if n%2 == 1 {
			res = mul(res, a)
		}
		a = mul(a, a)
		n /= 2
	}

	return res
}

function knightDialer(n: number): number {
    const MOD: number = 1e9 + 7;

    if (n === 1) {
        return 10;
    }

    const f: number[] = new Array(10).fill(1);

    while (--n > 0) {
        const t: number[] = new Array(10).fill(0);

        t[0] = f[4] + f[6];
        t[1] = f[6] + f[8];
        t[2] = f[7] + f[9];
        t[3] = f[4] + f[8];
        t[4] = f[0] + f[3] + f[9];
        t[6] = f[0] + f[1] + f[7];
        t[7] = f[2] + f[6];
        t[8] = f[1] + f[3];
        t[9] = f[2] + f[4];

        for (let i = 0; i < 10; ++i) {
            f[i] = t[i] % MOD;
        }
    }

    let ans: number = 0;
    for (const v of f) {
        ans = (ans + v) % MOD;
    }

    return ans;
}


// Solution 2
const mod = 1e9 + 7;

function knightDialer(n: number): number {
    const base: number[][] = [
        [0, 0, 0, 0, 1, 0, 1, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 1, 0, 1, 0],
        [0, 0, 0, 0, 0, 0, 0, 1, 0, 1],
        [0, 0, 0, 0, 1, 0, 0, 0, 1, 0],
        [1, 0, 0, 1, 0, 0, 0, 0, 0, 1],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [1, 1, 0, 0, 0, 0, 0, 1, 0, 0],
        [0, 0, 1, 0, 0, 0, 1, 0, 0, 0],
        [0, 1, 0, 1, 0, 0, 0, 0, 0, 0],
        [0, 0, 1, 0, 1, 0, 0, 0, 0, 0],
    ];

    const res = pow(base, n - 1);
    let ans = 0;
    for (const x of res[0]) {
        ans = (ans + x) % mod;
    }
    return ans;
}

function mul(a: number[][], b: number[][]): number[][] {
    const m = a.length;
    const n = b[0].length;
    const c: number[][] = Array.from({ length: m }, () => Array(n).fill(0));

    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            for (let k = 0; k < b.length; k++) {
                c[i][j] =
                    (c[i][j] + Number((BigInt(a[i][k]) * BigInt(b[k][j])) % BigInt(mod))) % mod;
            }
        }
    }
    return c;
}

function pow(a: number[][], n: number): number[][] {
    const size = a.length;
    let res: number[][] = Array.from({ length: 1 }, () => Array(size).fill(1));

    while (n > 0) {
        if (n % 2 === 1) {
            res = mul(res, a);
        }
        a = mul(a, a);
        n = Math.floor(n / 2);
    }

    return res;
}

public class Solution {
    public int KnightDialer(int n) {
        if (n == 1) return 10;
        int A = 4;
        int B = 2;
        int C = 2;
        int D = 1;
        int MOD = (int)1e9 + 7;
        for (int i = 0; i < n - 1; i++) {
            int tempA = A;
            int tempB = B;
            int tempC = C;
            int tempD = D;
            A = ((2 * tempB) % MOD + (2 * tempC) % MOD) % MOD;
            B = tempA;
            C = (tempA + (2 * tempD) % MOD) % MOD;
            D = tempC;
        }
        
        int ans = (A + B) % MOD;
        ans = (ans + C) % MOD;
        return (ans + D) % MOD;
    }
}


// Solution 2
public class Solution {
    private const int mod = 1000000007;
    private readonly int[][] baseMatrix = {
        new int[] {0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
        new int[] {0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
        new int[] {0, 0, 0, 0, 0, 0, 0, 1, 0, 1},
        new int[] {0, 0, 0, 0, 1, 0, 0, 0, 1, 0},
        new int[] {1, 0, 0, 1, 0, 0, 0, 0, 0, 1},
        new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
        new int[] {1, 1, 0, 0, 0, 0, 0, 1, 0, 0},
        new int[] {0, 0, 1, 0, 0, 0, 1, 0, 0, 0},
        new int[] {0, 1, 0, 1, 0, 0, 0, 0, 0, 0},
        new int[] {0, 0, 1, 0, 1, 0, 0, 0, 0, 0}
    };

    public int KnightDialer(int n) {
        int[][] res = Pow(baseMatrix, n - 1);
        int ans = 0;
        foreach (var x in res[0]) {
            ans = (ans + x) % mod;
        }
        return ans;
    }

    private int[][] Mul(int[][] a, int[][] b) {
        int m = a.Length, n = b[0].Length;
        int[][] c = new int[m][];
        for (int i = 0; i < m; i++) {
            c[i] = new int[n];
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < b.Length; k++) {
                    c[i][j] = (int)((c[i][j] + (long)a[i][k] * b[k][j]) % mod);
                }
            }
        }
        return c;
    }

    private int[][] Pow(int[][] a, int n) {
        int size = a.Length;
        int[][] res = new int[1][];
        res[0] = new int[size];
        for (int i = 0; i < size; i++) {
            res[0][i] = 1;
        }

        while (n > 0) {
            if (n % 2 == 1) {
                res = Mul(res, a);
            }
            a = Mul(a, a);
            n /= 2;
        }

        return res;
    }
}

935 - Knight Dialer

935. Knight Dialer

Description

Solutions

Solution 1: Recurrence

Solution 2: Matrix Exponentiation to Accelerate Recurrence

All Problems

All Solutions

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935. Knight Dialer

Description

Solutions

Solution 1: Recurrence

Solution 2: Matrix Exponentiation to Accelerate Recurrence

All Problems

All Solutions