112 lines
2.2 KiB
Markdown
112 lines
2.2 KiB
Markdown
# Leetcode Climbing-Chairs
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#### 2022-07-20 14:37
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> ##### Algorithms:
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> #algorithm #dynamic_programming
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#dynamic_programming
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```
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##### Links:
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- [Link to problem](https://leetcode.com/problems/climbing-stairs/submissions/)
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___
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### Problem
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You are climbing a staircase. It takes `n` steps to reach the top.
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Each time you can either climb `1` or `2` steps. In how many distinct ways can you climb to the top?
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#### Examples
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**Example 1:**
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**Input:** n = 2
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**Output:** 2
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**Explanation:** There are two ways to climb to the top.
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1. 1 step + 1 step
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2. 2 steps
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**Example 2:**
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**Input:** n = 3
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**Output:** 3
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**Explanation:** There are three ways to climb to the top.
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1. 1 step + 1 step + 1 step
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2. 1 step + 2 steps
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3. 2 steps + 1 step
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#### Constraints
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- `1 <= n <= 45`
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### Thoughts
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> [!summary]
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> This is a #dynamic_programming problem. our solution relies on other solutions
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I had come up with an recursive solution, with a little bit of optimizations.
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#### First iteration:
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Base case:
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- n is 0, 1, 2:
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return n
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Pseudo code:
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answer = climb(n - 1) + climb(n - 2);
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This is a brute force-y solution, because there are too much combinations, and we are re-calculating solutions.
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#### Second iteration:
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Base case:
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- n = 1, 2, **3**
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Return n
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Note that n == 3 is added, because we will calculate n - 3, which otherwise will make n = 0
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and when n == 0, it should return 1.
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> [!tips] Optimization
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> We can optimize it because some solutions can be
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> calculated and used for both 1 step and 2 steps.
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>
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> how many ways to climb: one step or two steps
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> 1 + 1 = 2
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> 0 + 2 = 2
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> 1 + 2 = 3
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> -> 2 x climb two steps, and 1 x climb 3 steps
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### Solution
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Optimized version:
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```cpp
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class Solution {
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public:
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int climbStairs(int n) {
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// Dymanic programming, using solution from the past
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// Recursive, base case: n == 0, 1, 2, 3
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if (n <= 3) {
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return n;
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}
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// how many ways to climb: one step or two steps
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// 1 + 1 = 2
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// 0 + 2 = 2
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// 1 + 2 = 3
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// -> 2 * climb two steps, and 1 * climb 3 steps
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return 2 * climbStairs(n - 2) + climbStairs(n - 3);
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}
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};
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``` |