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?

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Question Formatted question description: 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 Algorithm You can only climb 1 or 2 steps at a time, so the method to climb to the nth layer is either from the n-1 layer one step up, or from the n-2 layer 2 steps up, so the recursive formula is very easy We get: dp[n] = dp[n-1] + dp[n-2]. Code • • • • • • • • • public class Climbing_Stairs • class Solution : def climbStairs ( self , n : int ) -> int : a , b = 0 , 1 for _ in range ( n ): a , b = b , a + b return b ############ class Solution ( object ): def climbStairs ( self , n ): """ :type n: int :rtype: int """ if n <= 1 : return 1 pre , ppre = 1 , 1 for i in range ( 2 , n + 1 ): tmp = pre pre = ppre + pre ppre = tmp return pre •

LeetCode 70. Climbing Stairs

Description You are climbing a staircase. It takes nsteps to reach the top. Each time you can either climb 1or 2steps. 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 int: f = [] f.append(1) f.append(2) for i in range(2, n): f.append(f[i - 1] + f[i - 2]) return f[n - 1] • Time complexity:O(n). Single loop up ton. • Space complexity:O(n).dparray of sizenis used. Posted in Tagged Post navigation

Solved You are climbing a staircase. It takes n steps to

• • • • Question: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? For this problem, why does this solution work? class Solution { public: int climbStairs(int n) { int one =1; int two = 1; int temp; for(int i = 0; i < n-1;i++){ temp = one; one = one + 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? For this problem, why does this solution work? class Solution ; I thought that the return statement had to be inside the for loop otherwise it would return 1, but that isn't the case. Why does the return statement have to be outside of the for loop for this problem? Thanks!

Climbing Stairs

Problem Statement You are climbing a stair case. It takes nsteps 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 nwill 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 class Solution Complexity Analysis • Time complexity :O(n). Single loop upton .. • Space complexity :O(n).dp dparray of sizenis used. Explanation Dynamic Programming As we can see this problem can be broken into subproblems, and it contains the optimal substructure property i.e. its optimal solution can be constructed efficiently from optimal solutions of its subproblems, we can use dynamic programming to solve this problem. One can reachi thstep in one of the two ways: • Taking a single step from(i-1) thstep. • Taking a step of2from(i-2) thstep. So, the total number of ways to reachi this equal to sum of ways of reaching (i-1) thstep and ways of reaching(i-2) thstep. Letdp[i]denotes the number of ways to reach oni thstep: dp[i]= dp[i-1]+ dp[i-2] Feel free to post queries. May 2020 M T W T F S S 1 2 3 4 5 9 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Posts • July 30, 2020 • July 29, 2020 • July 27, 2020 • May 11, 2020 • May 10, 2020 • May 10, 2020 • May 10, 2020 • Ma...

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