---
# System prepended metadata

title: LC 509. Fibonacci Number
tags: [easy, leedcode, DP, python, c++]

---

# 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
```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++
```cpp=
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
x
