Climbing Stairs Leetcode Solution

In this post, we are going to solve the Climbing Stairs Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

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 = 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

Now, let’s see the leetcode solution of Climbing Stairs Leetcode Solution.

Climbing Stairs Leetcode Solution 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]

Climbing Stairs Leetcode Solution in CPP

class Solution {
 public:
  int climbStairs(int n) {
    // dp[i] := # of distinct ways to climb to i-th stair
    vector<int> dp(n + 1);
    dp[0] = 1;
    dp[1] = 1;

    for (int i = 2; i <= n; ++i)
      dp[i] = dp[i - 1] + dp[i - 2];

    return dp[n];
  }
};

Climbing Stairs Leetcode Solution in Java

class Solution {
  public int climbStairs(int n) {
    // dp[i] := # of distinct ways to climb to i-th stair
    int[] dp = new int[n + 1];
    dp[0] = 1;
    dp[1] = 1;

    for (int i = 2; i <= n; ++i)
      dp[i] = dp[i - 1] + dp[i - 2];

    return dp[n];
  }
}

Note: This problem Climbing Stairs is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Simplify Path Leetcode Solution