Fibonacci with dynamic programming

# Fibonacci with dynamic programming

February 13, 2023 Name
Moch Lutfi
@kaptenupi

## Problem

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

Given `n`, calculate `F(n)`.

Example 1:

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

Example 2:

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

Example 3:

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

Constraints:

• `0 <= n <= 30`

## Solution

### Recursion

• Find a base recursive call, in this case is condition where input equals with 0 will return 0 and 1 will return 1
• Then from the description above, call the recursive with F(n-1) + F(n-2)
`func fibonacci(n int) int{ if n==0 || n==1{ return n } return fibonacci(n-1)+fibonacci(n-2)}`

### Dynamic programming

• First step similar with recursive, early return if n equals 0 or 1
• Use memoize to compose calculation
`func fib(n int) int { if n==0 || n==1{ return n } dp:=make([]int, n+1) dp=0 dp=1 for i:=2;i<n+1;i++{ dp[i] = dp[i-2]+dp[i-1] } return dp[n]}`

But let's remove unnecessary array

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