Explain this dynamic programming climbing n-stair code

Question

Problem is

\"You are climbing a stair case. Each time you can either make 1 step or 2 steps. The staircase has n steps. In how many distinct ways can you climb the s

Answer 1

Climbing Stairs Using DP

class Solution {
   public:
     int climbStairs(int n) {
    int dp[n+1];

    if (n <= 1)
        return 1;

    if (n ==2)
        return 2;

    dp[0] = 1;
    dp[1] = 1;
    dp[2] = 2;

    for (int i = 3; i <=n; i++){
        dp[i] = dp[i-1]+dp[i-2];
    }

    return dp[n];
   }};

Answer 2

Well, first you need to understand the recursive formula, and how we derived the iterative one from it.

The recursive formula is:

f(n) = f(n-1) + f(n-2)
f(0) = f(1) = 1

(f(n-1) for one step, f(n-2) for two steps, and the total numbers is the number of ways to use one of these options - thus the summation).

If you look carefully - this is also a well known series - the fibonacci numbers, and the solution is simply calculating each number buttom-up instead of re-calculating the recursion over and over again, resulting in much more efficient solution.