Problem Statement:
Explanation:
- You are climbing a staircase with n steps.
- You can climb either 1 or 2 steps at a time.
- You need to determine how many distinct ways you can climb to the top of the staircase.
Sample Data:
- Input: n = 3
- Output: 3 (1+1+1, 1+2, 2+1)
Questions to be asked to the interviewer:
- Can n be zero or negative? What should the return value be in that case?
- Are there any constraints on the value of n?
Edge cases for this problem:
- When n is 0, there are no steps, so the output should be 0.
- When n is 1, there is only one way to climb the staircase.
Available Approaches to Solve this Problem:
- Recursive Approach: A naive recursive solution calculating the ways for each step.
- Memoization: Top-down recursive approach with storing previously calculated values.
- Dynamic Programming: Bottom-up iterative approach with storing solutions to subproblems.
My Approach to Solve this Problem:
I will use the Dynamic Programming approach to solve this problem efficiently by building the solution from the bottom up and avoiding redundant calculations.
def climbStairs(n):
? ? if n == 1:
? ? ? ? return 1
? ? if n == 2:
? ? ? ? return 2
? ? dp = [0] * (n + 1)
? ? dp[1] = 1
? ? dp[2] = 2
? ? for i in range(3, n + 1):
? ? ? ? dp[i] = dp[i - 1] + dp[i - 2]
? ? return dp[n]
Explain the Solution to a Non-Programmer:
- Imagine you're at the bottom of a staircase with n steps.
- You can climb the staircase by taking one step at a time or two steps at a time.
- To find the total number of ways to climb the stairs, think about how you could climb a smaller staircase, and then build up from there.
- For example, with 1 step, there's only one way to climb. With 2 steps, there are two ways (1+1 or 2).
- For 3 steps, you can build on the previous solutions by adding one more step to the one-step solution or adding two more steps to the two-step solution.
- You continue this pattern, always building on the previous two solutions, until you reach the total number of steps.
Code in Detail :
- Line 2-4: If there's only one step, return 1. If there are two steps, return 2. These are the base cases.
- Line 5: Create a list dp with n+1 elements initialized to 0. This will store the solutions to the subproblems.
- Line 6-7: Set the base cases in the dp list. There's one way to climb 1 step and two ways to climb 2 steps.
- Line 8-10: Iterate from 3 to n, and for each step, calculate the total ways to climb by adding the ways to climb the previous two steps.
- Line 11: Return the total ways to climb n steps from the dp list.
This approach leverages previous solutions to build the final result, making it both intuitive and efficient.
Pattern :
This problem follows the pattern of Dynamic Programming, where we build the solution to a problem based on the solutions to its subproblems.
Big O Notation:
- Time Complexity: O(n)
- O(n), as we are iterating through the stairs from 3 to ?
- n.
- Space Complexity: O(n)
- O(n), as we are storing the number of ways to climb each stair in a list of length ?+1
- n+1.
Points to Remember to Solve this Problem in the Future:
- Recognize that this is a problem where earlier results can be reused to build up the solution.
- Implement the base cases first, then use them to calculate the subsequent solutions.
Code Line by Line with Some Sample Data:
- Input: n = 3
- Line 2-4: As n is not 1 or 2, we skip these conditions.
- Line 5: Initialize dp = [0, 0, 0, 0]
- Line 6-7: Set base cases dp[1] = 1 and dp[2] = 2
- Line 8-10: Iterate from 3 to 3:dp[3] = dp[2] + dp[1] = 2 + 1 = 3
- Line 11: Return dp[3], which is 3.
Key Syntax Used in this Solution:
- Iterating through a Range of Numbers: for i in range(3, n + 1):
- Initializing a List with a Specific Size: dp = [0] * (n + 1)
- Accessing and Modifying Elements in a List: dp[i] = dp[i - 1] + dp[i - 2]
These syntax points, combined with understanding the pattern of building upon subproblems, provide a clear pathway to solving this problem in the future.
