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70. Climbing Stairs

Difficulty: Easy

Related Topics: Math, Dynamic Programming, Memoization

You are climbing a staircase. It takes n steps to reach the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Example 1:

Input: n = 2 Output: 2 Explanation: There are two ways to climb to the top.

  1. 1 step + 1 step
  2. 2 steps

Example 2:

Input: n = 3 Output: 3 Explanation: There are three ways to climb to the top.

  1. 1 step + 1 step + 1 step
  2. 1 step + 2 steps
  3. 2 steps + 1 step

First Attempt:

Code:

class Solution:
    def climbStairs(self, k: int) -> int:
        def climb(k):
            if k==0:
                return 1
            elif k==1:
                return 2
            else:
                return climb(k-1) + climb((k-2))
        return climb(k-1)

Feedback:

I found that this question is same as fibonacci numbers question, so I tried to use recursion to solve this problem, but it resulted time limit exit, because as number gets bigger, it recursively calls huge numbers of functions. I should use dynamic programming to solve this problem.

Second Attempt:

Code:

class Solution:
    def climbStairs(self, k: int) -> int:
        stairs = [0 for i in range(k)]
        curr = 2
        if k==1:
            return 1
        elif k==2:
            return 2
        else:
            stairs[0] = 1
            stairs[1] = 2
            while curr < k:
                stairs[curr] = stairs[curr-1] + stairs[curr-2]
                curr+=1
            return stairs[k-1]

Feedback:

I used dynamic programming to solve this time. I assigned a list with 0s for range of k. Then stored the values correctly using same logic to solve fibonacci numbers problem.

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