Climbing Stairs (Dynamic Programming)
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?
Why Interviewers Ask This
Uber interviewers use this problem to assess a candidate's ability to recognize overlapping subproblems and optimal substructure, the core pillars of dynamic programming. They specifically evaluate whether you can transition from an inefficient brute-force recursive solution to an optimized iterative approach with O(n) time and O(1) space complexity, demonstrating practical algorithmic efficiency skills required for high-scale ride-matching systems.
How to Answer This Question
1. Clarify constraints: Immediately ask about edge cases like n=0 or negative inputs to show attention to detail.
2. Define the recurrence relation: Explain that reaching step n is the sum of ways to reach (n-1) plus ways to reach (n-2), as you can only arrive from those two previous steps.
3. Start with recursion: Briefly outline the naive recursive approach to establish the baseline logic, acknowledging its exponential time complexity.
4. Optimize with DP: Propose using memoization or an iterative bottom-up approach. Explicitly state that since we only need the last two values, we can optimize space to O(1) by maintaining just two variables instead of an array.
5. Walk through an example: Trace the logic with a small number, such as n=4, showing how the values build up sequentially (1, 2, 3, 5).
6. Analyze complexity: Conclude by stating the final Time Complexity is O(n) and Space Complexity is O(1).
Key Points to Cover
- Identifying the Fibonacci recurrence relation immediately
- Acknowledging the inefficiency of naive recursion before optimizing
- Explicitly stating the O(n) time and O(1) space complexity trade-off
- Handling edge cases like n=1 or n=0 correctly
- Using clear variable names to represent the DP state transitions
Sample Answer
This problem is a classic introduction to Dynamic Programming because it demonstrates how breaking a complex problem into smaller, overlapping subproblems leads to significant efficiency gains. The core insight is that to reach step n, you must have come from either step n-1 or step n-2. Therefore, the total distinct ways to reach step n is simply the sum of the ways to reach those two preceding steps. Mathematically, this forms the Fibonacci sequence where dp[n] = dp[n-1] + dp[n-2].
If I were to implement this, I would first consider a naive recursive solution. While correct, it recalculates the same subproblems repeatedly, resulting in exponential time complexity, which is unacceptable for large inputs typical in Uber's backend services. Instead, I would use an iterative bottom-up approach. I would initialize two variables, say 'prev' and 'curr', representing the ways to reach the previous step and the current step respectively. By iterating from step 2 up to n, I update these variables in each loop: the new current value becomes the sum of the old prev and curr, and then I shift them forward. This ensures we only store the necessary history, reducing space complexity to O(1) while keeping time complexity at O(n). For instance, if n is 4, the sequence evolves as 1, 2, 3, 5, yielding 5 distinct ways. This approach balances readability with the strict performance requirements expected in production environments.
Common Mistakes to Avoid
- Implementing only the recursive solution without addressing the TLE (Time Limit Exceeded) risk
- Allocating an entire array for DP when only two variables are needed, wasting memory
- Failing to handle the base case where n is less than 2
- Not explaining the thought process behind why the problem fits the DP category
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