Min Cost Climbing Stairs
You are given an integer array `cost` where `cost[i]` is the cost to step on the $i$-th stair. You can either start at index 0 or index 1. Find the minimum cost to reach the top of the floor. Use DP.
Why Interviewers Ask This
Spotify asks this to evaluate your ability to identify overlapping subproblems and apply dynamic programming efficiently. They want to see if you can recognize that the optimal path to any stair depends only on the previous two states, allowing you to optimize space complexity from O(n) to O(1). This tests your core algorithmic thinking and code optimization skills under pressure.
How to Answer This Question
1. Clarify constraints: Confirm if the array is empty or if costs are non-negative. 2. Define the state: Explicitly state that dp[i] represents the minimum cost to reach step i. 3. Establish the recurrence relation: Explain that reaching step i requires coming from either i-1 or i-2, so dp[i] = min(dp[i-1], dp[i-2]) + cost[i]. 4. Handle base cases: Note that you can start at index 0 or 1, meaning the cost to reach those steps is just their respective costs. 5. Optimize space: Propose using two variables instead of an array since you only need the last two values. 6. Walk through a trace: Use a small example like [10, 15, 20] to demonstrate how the logic flows step-by-step before coding.
Key Points to Cover
- Explicitly defining the recurrence relation as min(prev, prev2) + current_cost
- Recognizing the base cases allow starting at either index 0 or 1
- Optimizing space complexity from O(n) to O(1) using two variables
- Correctly identifying that the final answer is the min of the last two calculated values
- Demonstrating a clear thought process through a manual trace of the example
Sample Answer
To solve the Min Cost Climbing Stairs problem, I would first clarify the input constraints, ensuring we handle edge cases like an empty array. My approach relies on Dynamic Programming because the optimal solution for any stair depends entirely on the solutions to the previous two stairs.
I define the state as the minimum cumulative cost to reach the current index. The recurrence relation is straightforward: to get to step i, you could have come from i-1 or i-2. Therefore, the cost to reach i is the minimum of the costs to reach i-1 and i-2, plus the cost of the current step itself. However, since the goal is to reach the top (which is beyond the last index), the final answer is the minimum of the costs to reach the last two steps.
For implementation, I'd start with base cases where the cost to reach index 0 is simply cost[0] and index 1 is cost[1]. Then, I would iterate from index 2 to the end, updating the current cost based on the previous two. Crucially, I would optimize space by maintaining only two variables, say 'prev' and 'curr', reducing space complexity to O(1). For example, with input [10, 15, 20], starting at index 0 costs 10, moving to index 1 costs 15. To reach index 2, we take min(10, 15) + 20 = 30. The final result is min(15, 30) = 15, representing the cheapest way to skip the expensive middle step. This approach ensures both time and space efficiency.
Common Mistakes to Avoid
- Failing to realize the final step doesn't require adding the cost of a non-existent top floor
- Using an O(n) space array when interviewers expect O(1) space optimization
- Incorrectly setting base cases by ignoring that one can start at index 1 without paying index 0's cost
- Confusing the cost to reach a step versus the cost to climb off a step
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