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

// Related Topics: Dynamic Programming

// Difficulty: Easy

/*

Initial thoughts:

Considering the fact that the ways we can reach step i equals

the ways we can reach step i-1 + the ways we can reach step i-2

we approach the problem like calculating the Fibonacci number

by calculating every next step based on the last two, starting from

our two base cases where for n === 1 there is one way and for n === 2

there are two ways to reach step n.

Time complexity: O(n)

Space complexity: O(1)

*/

/**

* @param {number} n

* @return {number}

*/

const climbStairs = n => {

if (n === 1) return 1;

let first = 1,

second = 2;

for (let i = 3; i <= n; i++) {

const third = first + second;

first = second;

second = third;

}

return second;

};