This approach uses recursion to explore the different ways to reach the top. To optimize, we use memoization to store results of subproblems to avoid redundant calculations. This ensures that each subproblem is solved only once, dramatically improving efficiency.
Time Complexity: O(n), as each state is only computed once.
Space Complexity: O(n), due to the memoization array.
1class Solution:
2 def __init__(self):
3 self.memo = {}
4
5 def climbStairs(self, n: int) -> int:
6 if n <= 2:
7 return n
8 if n in self.memo:
9 return self.memo[n]
10 self.memo[n] = self.climbStairs(n-1) + self.climbStairs(n-2)
11 return self.memo[n]
12
13solution = Solution()
14print(solution.climbStairs(3))This Python code uses a class with a dictionary for memoization. The climbStairs method calculates the number of ways to climb the staircase, storing previously computed results in the self.memo dictionary to avoid recomputation.
This approach builds from the base up using a table to store results at each step. Starting with known base cases, each subsequent result is built by combining previous results. This eliminates the recursive overhead, making it a very efficient algorithm.
Time Complexity: O(n), since each value is calculated sequentially in a loop.
Space Complexity: O(n), for storing the results in the dp array.
1public class Solution {
2 public int climbStairs(int n) {
3 if (n <= 2) return n;
4 int[] dp = new int[n + 1];
5 dp[1] = 1;
6 dp[2] = 2;
7 for (int i = 3; i <= n; i++) {
8 dp[i] = dp[i - 1] + dp[i - 2];
9 }
10 return dp[n];
11 }
12 public static void main(String[] args) {
13 Solution sol = new Solution();
14 System.out.println(sol.climbStairs(3));
15 }
16}This Java solution uses an array to perform a dynamic programming approach iteratively. Building upon basic cases, this method efficiently calculates the answer using an dp array to store results, avoiding recursive complexity.