Climbing Stairs
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
Constraints:
1 <= n <= 45
My Solution
This is a dynamic programming problem. We can use memoization. We need a cache to store the output of the overlapping subproblems and at each iteration climbStairs(n) = climbStairs(n-1) + climbStairs(n-2), referring to the fact that we can climb either 1 or 2 stairs at a time. If we already computed for n, we just return the cached result.
Time complexity is O(n) because there are at most n subproblems. Space complexity is also O(n) to cache the subproblems.
class Solution {
public int climbStairs(int n) {
int[] cache = new int[n+1];
return dp(n, cache);
}
public int dp(int n, int[] cache) {
if (n == 1) {
return 1;
}
if (n == 2) {
return 2;
}
if (cache[n] != 0) {
return cache[n];
}
int out = dp(n-2, cache) + dp(n-1, cache);
cache[n] = out;
return out;
}
}