Climbing Stairs Leetcode Solution

In this post, you will know how to solve the Climbing Stairs Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

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

Climbing Stairs Leetcode Solutions 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 Solutions 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 Solutions 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  BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

Next: Valid Number Leetcode Solution