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08.01. Three Steps Problem

Description

A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs. The result may be large, so return it modulo 1000000007.

Example1:

 Input: n = 3

 Output: 4

Example2:

 Input: n = 5

 Output: 13

Note:

  1. 1 <= n <= 1000000

Solutions

Solution 1: Recursion

Thinking

A step of \(1\), \(2\), or \(3\) is allowed. Bare recursion over the three choices recomputes the same suffixes many times.

The count at \(i\) depends only on \(i-1,i-2,i-3\), so a linear recurrence with a modulus is enough.

Three rolling variables \(a,b,c\) hold consecutive terms; after \(n-1\) updates \(a\) is \(f(n)\). Extra space is constant.

We define \(f[i]\) as the number of ways to reach the \(i\)-th step, initially \(f[1]=1\), \(f[2]=2\), \(f[3]=4\). The answer is \(f[n]\).

The recursion formula is \(f[i] = f[i-1] + f[i-2] + f[i-3]\).

Since \(f[i]\) is only related to \(f[i-1]\), \(f[i-2]\), \(f[i-3]\), we can use three variables \(a\), \(b\), \(c\) to store the values of \(f[i-1]\), \(f[i-2]\), \(f[i-3]\), reducing the space complexity to \(O(1)\).

The time complexity is \(O(n)\), where \(n\) is the given integer. The space complexity is \(O(1)\).

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class Solution:
    def waysToStep(self, n: int) -> int:
        a, b, c = 1, 2, 4
        mod = 10**9 + 7
        for _ in range(n - 1):
            a, b, c = b, c, (a + b + c) % mod
        return a

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class Solution {
    public int waysToStep(int n) {
        final int mod = (int) 1e9 + 7;
        int a = 1, b = 2, c = 4;
        for (int i = 1; i < n; ++i) {
            int t = a;
            a = b;
            b = c;
            c = (((a + b) % mod) + t) % mod;
        }
        return a;
    }
}

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class Solution {
public:
    int waysToStep(int n) {
        const int mod = 1e9 + 7;
        int a = 1, b = 2, c = 4;
        for (int i = 1; i < n; ++i) {
            int t = a;
            a = b;
            b = c;
            c = (((a + b) % mod) + t) % mod;
        }
        return a;
    }
};

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func waysToStep(n int) int {
    const mod int = 1e9 + 7
    a, b, c := 1, 2, 4
    for i := 1; i < n; i++ {
        a, b, c = b, c, (a+b+c)%mod
    }
    return a
}

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impl Solution {
    pub fn ways_to_step(n: i32) -> i32 {
        let (mut a, mut b, mut c) = (1, 2, 4);
        let m = 1000000007;
        for _ in 1..n {
            let t = a;
            a = b;
            b = c;
            c = (((a + b) % m) + t) % m;
        }
        a
    }
}

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/**
 * @param {number} n
 * @return {number}
 */
var waysToStep = function (n) {
    let [a, b, c] = [1, 2, 4];
    const mod = 1e9 + 7;
    for (let i = 1; i < n; ++i) {
        [a, b, c] = [b, c, (a + b + c) % mod];
    }
    return a;
};

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int waysToStep(int n) {
    const int mod = 1e9 + 7;
    int a = 1, b = 2, c = 4;
    for (int i = 1; i < n; ++i) {
        int t = a;
        a = b;
        b = c;
        c = (((a + b) % mod) + t) % mod;
    }
    return a;
}

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class Solution {
    func waysToStep(_ n: Int) -> Int {
        let mod = Int(1e9) + 7
        var a = 1, b = 2, c = 4
        if n == 1 { return a }
        if n == 2 { return b }
        if n == 3 { return c }

        for _ in 1..<n {
            let t = a
            a = b
            b = c
            c = ((a + b) % mod + t) % mod
        }
        return a
    }
}

Solution 2: Matrix Quick Power to Accelerate Recursion

Thinking

The linear scan is \(\Theta(n)\) and spends most of its time on modular adds when \(n\) is large.

The order-\(3\) recurrence is a \(1\times 3\) state times a \(3\times 3\) matrix, which matrix exponentiation evaluates in \(O(\log n)\) multiplications.

We set \(F(n)\) to represent a \(1 \times 3\) matrix \(\begin{bmatrix} F_{n - 1} & F_{n - 2} & F_{n - 3} \end{bmatrix}\), where \(F_{n - 1}\), \(F_{n - 2}\) and \(F_{n - 3}\) respectively represent the number of ways to reach the \(n - 1\)-th, \(n - 2\)-th and \(n - 3\)-th steps.

We hope to derive \(F(n)\) based on \(F(n-1) = \begin{bmatrix} F_{n - 2} & F_{n - 3} & F_{n - 4} \end{bmatrix}\). That is to say, we need a matrix \(base\), so that \(F(n - 1) \times base = F(n)\), i.e.:

\[ \begin{bmatrix} F_{n - 2} & F_{n - 3} & F_{n - 4} \end{bmatrix} \times base = \begin{bmatrix} F_{n - 1} & F_{n - 2} & F_{n - 3} \end{bmatrix} \]

Since \(F_n = F_{n - 1} + F_{n - 2} + F_{n - 3}\), the matrix \(base\) is:

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

We define the initial matrix \(res = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix}\), then \(F_n\) equals the sum of all elements in the result matrix of \(res\) multiplied by \(base^{n - 4}\). It can be solved using matrix quick power.

The time complexity is \(O(\log n)\), and the space complexity is \(O(1)\).

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import numpy as np


class Solution:
    def waysToStep(self, n: int) -> int:
        if n < 4:
            return 2 ** (n - 1)
        mod = 10**9 + 7
        factor = np.asmatrix([(1, 1, 0), (1, 0, 1), (1, 0, 0)], np.dtype("O"))
        res = np.asmatrix([(4, 2, 1)], np.dtype("O"))
        n -= 4
        while n:
            if n & 1:
                res = res * factor % mod
            factor = factor * factor % mod
            n >>= 1
        return res.sum() % mod

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

    public int waysToStep(int n) {
        if (n < 4) {
            return (int) Math.pow(2, n - 1);
        }
        long[][] a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
        long[][] res = pow(a, n - 4);
        long ans = 0;
        for (long x : res[0]) {
            ans = (ans + x) % mod;
        }
        return (int) ans;
    }

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

    private long[][] pow(long[][] a, int n) {
        long[][] res = {{4, 2, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                res = mul(res, a);
            }
            a = mul(a, a);
            n >>= 1;
        }
        return res;
    }
}

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class Solution {
public:
    int waysToStep(int n) {
        if (n < 4) {
            return pow(2, n - 1);
        }
        vector<vector<ll>> a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
        vector<vector<ll>> res = qpow(a, n - 4);
        ll ans = 0;
        for (ll x : res[0]) {
            ans = (ans + x) % mod;
        }
        return ans;
    }

private:
    using ll = long long;
    const int mod = 1e9 + 7;
    vector<vector<ll>> mul(vector<vector<ll>>& a, vector<vector<ll>>& b) {
        int m = a.size(), n = b[0].size();
        vector<vector<ll>> c(m, vector<ll>(n));
        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] + a[i][k] * b[k][j] % mod) % mod;
                }
            }
        }
        return c;
    }

    vector<vector<ll>> qpow(vector<vector<ll>>& a, int n) {
        vector<vector<ll>> res = {{4, 2, 1}};
        while (n) {
            if (n & 1) {
                res = mul(res, a);
            }
            a = mul(a, a);
            n >>= 1;
        }
        return res;
    }
};

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const mod = 1e9 + 7

func waysToStep(n int) (ans int) {
    if n < 4 {
        return int(math.Pow(2, float64(n-1)))
    }
    a := [][]int{{1, 1, 0}, {1, 0, 1}, {1, 0, 0}}
    res := pow(a, n-4)
    for _, x := range res[0] {
        ans = (ans + x) % mod
    }
    return
}

func mul(a, b [][]int) [][]int {
    m, n := len(a), 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) % mod
            }
        }
    }
    return c
}

func pow(a [][]int, n int) [][]int {
    res := [][]int{{4, 2, 1}}
    for n > 0 {
        if n&1 == 1 {
            res = mul(res, a)
        }
        a = mul(a, a)
        n >>= 1
    }
    return res
}

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/**
 * @param {number} n
 * @return {number}
 */

const mod = 1e9 + 7;

var waysToStep = function (n) {
    if (n < 4) {
        return Math.pow(2, n - 1);
    }
    const a = [
        [1, 1, 0],
        [1, 0, 1],
        [1, 0, 0],
    ];
    let ans = 0;
    const res = pow(a, n - 4);
    for (const x of res[0]) {
        ans = (ans + x) % mod;
    }
    return ans;
};

function mul(a, b) {
    const [m, n] = [a.length, b[0].length];
    const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 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, n) {
    let res = [[4, 2, 1]];
    while (n) {
        if (n & 1) {
            res = mul(res, a);
        }
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}

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