70. Climbing Stairs LeetCode Solution

In this guide, you will get 70. Climbing Stairs LeetCode Solution with the best time and space complexity. The solution to Climbing Stairs problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Climbing Stairs solution in C++
Climbing Stairs solution in Java
Climbing Stairs solution in Python
Additional Resources

70. Climbing Stairs LeetCode Solution image

Problem Statement of Climbing Stairs

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

70. Climbing Stairs LeetCode Solution in C++

class Solution {
 public:
  int climbStairs(int n) {
    // dp[i] := the number of ways to climb to the 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];
  }
};
/* code provided by PROGIEZ */

70. Climbing Stairs LeetCode Solution in Java

class Solution {
  public int climbStairs(int n) {
    // dp[i] := the number of ways to climb to the 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];
  }
}
// code provided by PROGIEZ

70. Climbing Stairs LeetCode Solution in Python

class Solution:
  def climbStairs(self, n: int) -> int:
    # dp[i] := the number of ways to climb to the 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]
 # code by PROGIEZ