70. Climbing Stairs

https://leetcode.com/problems/climbing-stairs/

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.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

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

---

Intuition
How can we reach stair n

1. One step from n - 1 or
2. Two steps from n - 2

# ways to reach n = # ways to reach (n -1) + # ways to reach (n - 2)

F(n) = F(n - 1) + F(n -2)

Recursion with memoization.

We do not need all past entries, just two last ones are sufficient - constant space

--

Time - O(n)

Space - O(1)

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Related problems
min-cost-climbing-stairs

	class Solution {
	int climbStairs(int n) {
	if (n < 3) {
	return n;
	}
	int first = 1, second = 2, third = 0;
	for (int i = 3; i <=n; i++) {
	third = first + second;
	first = second;
	second = third;
	}
	return third;
	}
	}

view raw 70.ClimbingStairs.java hosted with ❤ by GitHub

	class Solution:
	def climbStairs(self, n: int) -> int:
	if n < 3:
	return n
	first, second, third = 1, 2, 3
	for i in range(3, n + 1):
	third = first + second
	first = second
	second = third
	return third

view raw 70.ClimbingStairsIterative.py hosted with ❤ by GitHub

	class Solution:
	def __init__(self):
	self.cache = {}
	def climbStairs(self, n: int) -> int:
	if n < 3:
	return n
	if n not in self.cache:
	self.cache[n] = self.climbStairs(n - 1) + self.climbStairs(n - 2)
	return self.cache[n]

view raw 70.ClimbingStairsMemo.py hosted with ❤ by GitHub