# LC 509. Fibonacci Number

### [Problem link](https://leetcode.com/problems/fibonacci-number/)

###### tags: `leedcode` `python` `c++` `easy` `DP`

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

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

Given <code>n</code>, calculate <code>F(n)</code>.

**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:**

- <code>0 <= n <= 30</code>

## Solution 1 - DP
#### Python

class Solution:
    def fib(self, n: int) -> int:
        if n <= 1:
            return n
        dp = [0, 1]

        for i in range(2, n + 1):
            dp[0], dp[1] = dp[1], dp[0] + dp[1]

        return dp[1]

#### C++

class Solution {
public:
    int fib(int n) {
        vector<int> dp = {0, 1};

        if (n <= 1) {
            return dp[n];
        }

        for (int i = 2; i <= n; i++) {
            int tmp = dp[0] + dp[1];
            dp[0] = dp[1];
            dp[1] = tmp;
        }
        return dp[1];
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
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