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509. Fibonacci Number

Description

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

 

Example 1:

Input: n = 2
Output: 1
Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

Input: n = 3
Output: 2
Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

Input: n = 4
Output: 3
Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

 

Constraints:

  • 0 <= n <= 30

Solutions

Solution 1: Recurrence

We define two variables $a$ and $b$, initially $a = 0$ and $b = 1$.

Next, we perform $n$ iterations. In each iteration, we update the values of $a$ and $b$ to $b$ and $a + b$, respectively.

Finally, we return $a$.

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 fib(self, n: int) -> int:
        a, b = 0, 1
        for _ in range(n):
            a, b = b, a + b
        return a

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class Solution {
    public int fib(int n) {
        int a = 0, b = 1;
        while (n-- > 0) {
            int c = a + b;
            a = b;
            b = c;
        }
        return a;
    }
}

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class Solution {
public:
    int fib(int n) {
        int a = 0, b = 1;
        while (n--) {
            int c = a + b;
            a = b;
            b = c;
        }
        return a;
    }
};

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func fib(n int) int {
    a, b := 0, 1
    for i := 0; i < n; i++ {
        a, b = b, a+b
    }
    return a
}

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function fib(n: number): number {
    let [a, b] = [0, 1];
    while (n--) {
        [a, b] = [b, a + b];
    }
    return a;
}

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impl Solution {
    pub fn fib(n: i32) -> i32 {
        let mut a = 0;
        let mut b = 1;
        for _ in 0..n {
            let t = b;
            b = a + b;
            a = t;
        }
        a
    }
}

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/**
 * @param {number} n
 * @return {number}
 */
var fib = function (n) {
    let [a, b] = [0, 1];
    while (n--) {
        [a, b] = [b, a + b];
    }
    return a;
};

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class Solution {
    /**
     * @param Integer $n
     * @return Integer
     */
    function fib($n) {
        $a = 0;
        $b = 1;
        for ($i = 0; $i < $n; $i++) {
            $temp = $a;
            $a = $b;
            $b = $temp + $b;
        }
        return $a;
    }
}

Solution 2: Matrix Exponentiation

We define $\textit{Fib}(n)$ as a $1 \times 2$ matrix $\begin{bmatrix} F_n & F_{n - 1} \end{bmatrix}$, where $F_n$ and $F_{n - 1}$ are the $n$-th and $(n - 1)$-th Fibonacci numbers, respectively.

We want to derive $\textit{Fib}(n)$ from $\textit{Fib}(n - 1) = \begin{bmatrix} F_{n - 1} & F_{n - 2} \end{bmatrix}$. In other words, we need a matrix $\textit{base}$ such that $\textit{Fib}(n - 1) \times \textit{base} = \textit{Fib}(n)$, i.e.:

$$ \begin{bmatrix} F_{n - 1} & F_{n - 2} \end{bmatrix} \times \textit{base} = \begin{bmatrix} F_n & F_{n - 1} \end{bmatrix} $$

Since $F_n = F_{n - 1} + F_{n - 2}$, the first column of the matrix $\textit{base}$ is:

$$ \begin{bmatrix} 1 \ 1 \end{bmatrix} $$

The second column is:

$$ \begin{bmatrix} 1 \ 0 \end{bmatrix} $$

Thus, we have:

$$ \begin{bmatrix} F_{n - 1} & F_{n - 2} \end{bmatrix} \times \begin{bmatrix}1 & 1 \ 1 & 0\end{bmatrix} = \begin{bmatrix} F_n & F_{n - 1} \end{bmatrix} $$

We define the initial matrix $res = \begin{bmatrix} 1 & 0 \end{bmatrix}$, then $F_n$ is equal to the second element of the first row of the result matrix obtained by multiplying $res$ with $\textit{base}^{n}$. We can solve this using matrix exponentiation.

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

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