Coin Change
Given an array of coins and an amount, return the fewest number of coins that you need to make up that amount. Use Dynamic Programming.
Why Interviewers Ask This
Tesla evaluates candidates on this problem to assess optimization skills and logical rigor under constraints. They specifically look for the ability to transition from a naive recursive solution to an efficient Dynamic Programming approach, demonstrating how you handle overlapping subproblems. This mirrors real-world challenges in battery management systems where minimizing resource usage is critical for performance and safety.
How to Answer This Question
1. Clarify requirements: Confirm if coins are unlimited and what happens if the amount cannot be formed. 2. Analyze complexity: Briefly explain why a greedy approach fails here using a counter-example like coins [1, 3, 4] and target 6. 3. Define state: Propose dp[i] as the minimum coins needed for amount i. 4. Establish recurrence: Explain that dp[i] = min(dp[i - coin]) + 1 for all valid coins. 5. Implement bottom-up: Describe iterating from 1 to the target amount, filling the array iteratively. 6. Optimize space: Mention that while O(amount) space is standard, discuss if further optimization is possible. 7. Trace example: Walk through a small input manually to verify logic before coding.
Key Points to Cover
- Explicitly explaining why a greedy algorithm fails demonstrates deep conceptual understanding
- Defining the recurrence relation clearly shows mastery of Dynamic Programming fundamentals
- Handling edge cases like impossible amounts proves robustness and attention to detail
- Analyzing time and space complexity aligns with Tesla's focus on system efficiency
- Walking through a manual trace validates logic before writing code
Sample Answer
To solve the Coin Change problem efficiently, I would first rule out a greedy strategy because it doesn't guarantee the optimal solution. For instance, with coins [1, 3, 4] and a target of 6, a greedy approach picks 4 then two 1s for three coins, whereas the optimal is two 3s. Since we need the absolute minimum count, Dynamic Programming is the correct path. I would define a DP array where dp[i] represents the fewest coins needed to make amount i. We initialize dp[0] to 0 and all other entries to infinity. Then, for every amount from 1 up to the target, we iterate through each available coin. If the coin value is less than or equal to the current amount, we update dp[current] by taking the minimum of its current value and dp[current - coin] plus one. This ensures we always store the best combination found so far. After filling the table, if dp[target] remains infinity, we return -1 indicating the amount is impossible; otherwise, we return the value at dp[target]. This approach runs in O(amount * number_of_coins) time and uses O(amount) space, which fits Tesla's requirement for scalable, high-performance algorithms in embedded systems.
Common Mistakes to Avoid
- Assuming a greedy approach works without verifying optimality for all inputs
- Failing to initialize the DP array correctly, leading to incorrect base cases
- Not handling the scenario where the target amount cannot be formed by any coin combination
- Confusing the recurrence relation by adding the coin value instead of incrementing the count
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