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In this post, you will find the solution for the **Climbing Stairs** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Climbing Stairs– LeetCode Problem

Climbing Stairs– LeetCode Problem

**Problem:**

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

**Example 2:**

Input:n = 3Output:3Explanation: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:**

Climbing Stairs– LeetCode Solutions

Climbing Stairs in C++:

class Solution { public: int climbStairs(int n) { int prev1 = 1; // dp[i - 1] int prev2 = 1; // dp[i - 2] for (int i = 2; i <= n; ++i) { const int dp = prev1 + prev2; prev2 = prev1; prev1 = dp; } return prev1; } };

Climbing Stairs in Java:

class Solution { public int climbStairs(int n) { int prev1 = 1; // dp[i - 1] int prev2 = 1; // dp[i - 2] for (int i = 2; i <= n; ++i) { final int dp = prev1 + prev2; prev2 = prev1; prev1 = dp; } return prev1; }

Climbing Stairs in Python:

class Solution: def climbStairs(self, n: int) -> int: # dp[i] := # of distinct ways to climb to i-th stair dp = [1, 1] + [0] * (n - 1) for i in range(2, n + 1): dp[i] = dp[i - 1] + dp[i - 2] return dp[n] © 2022 GitHub, Inc.