Fibonacci with dynamic programming

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

• 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)
}

• 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]=0
    dp[1]=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;
}