You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Note: Given n will be a positive integer.
Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Input: 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
Although the above answer is in lexicographical order, your answer could be in any order you want.
The problem can be solved in a recursive way:
Below is the white board:
class Solution: def climbstairs(self, n): if n == 1: return 1 if n == 2: return 2 return self.climbstairs(n-1) + self.climbstairs(n-2) def test(): cs = Solution().climbstairs(5) print(cs) test()
- Time complexity is O(2**n), which will TLE (Time Limit Exceeded)
One way to improve this is via memoization.
class Solution2: def climbstairs_m(self, n): memo = [-1] * (n+1) if n == 1: memo[n] = 1 if n == 2: memo[n] = 2 if memo[n] == -1: memo[n] = self.climbstairs_m(n-1) + self.climbstairs_m(n-2) return memo[n] def test(): csm = Solution2().climbstairs_m(5) print(csm) test()
The above two solutions will TLE. Another way to optimize is via dynamic programming. Now think about the base case, if there are 0 stairs, there is 1 way: don’t move; if there are 1 stails there’s 1 way. so memo = memo =1
class Solution3: def climbStairs_dp(self, n): memo = [-1] * (n+1) memo = 1 memo = 1 for i in range(2,n+1): memo[i] = memo[i-2] + memo[i-1] return memo[n] def test(): csdp=Solution3().climbStairs_dp(5) print(csdp) test()
Time and space complexities are both O(n) for dynamic programming version.