Pascals Triangle
Given an integer `numRows`, return the first `numRows` of Pascal's triangle.
Why Interviewers Ask This
Apple interviewers use Pascal's Triangle to assess a candidate's ability to recognize recursive patterns and translate mathematical definitions into efficient iterative code. They specifically evaluate attention to edge cases, such as handling numRows equal to zero or one, and the capacity to optimize space complexity by reusing existing rows rather than allocating new memory structures unnecessarily.
How to Answer This Question
1. Clarify requirements immediately: Ask about constraints on numRows and confirm that the output should be a list of lists containing integers. 2. Analyze the pattern: Verbally explain how each number is the sum of the two numbers directly above it, noting the boundary conditions where the first and last elements are always 1. 3. Choose an iterative strategy: Propose building the triangle row by row, initializing the first row as [1], then generating subsequent rows based on the previous one. 4. Optimize for space: Mention that you can construct the current row in place or using a temporary array to avoid O(n^2) extra space if possible, though O(n^2) total space is required for the output itself. 5. Implement with edge case checks: Write clean code that handles numRows < 1 gracefully before entering the main loop. 6. Walk through a trace: Manually execute the logic with numRows = 5 to demonstrate correctness and show your thought process clearly to the interviewer.
Key Points to Cover
- Explicitly identifying the recurrence relation (current = prev[i-1] + prev[i])
- Demonstrating awareness of boundary conditions for the first and last elements
- Choosing an iterative solution over recursion to prevent stack overflow risks
- Handling edge cases like numRows equals 0 or 1 proactively
- Explaining the time complexity as O(n^2) due to the triangular structure
Sample Answer
To solve this efficiently, I would start by acknowledging that Pascal's Triangle is fundamentally built on the principle that every element is the sum of the two elements directly above it. First, I need to handle the base cases. If numRows is less than or equal to zero, I return an empty list. If it is exactly one, I return [[1]]. For larger values, I initialize the result list with the first row, which is simply [1]. Then, I iterate from the second row up to numRows. In each iteration, I create a new row starting with 1. Next, I look at the previous row I just added to the result. I calculate the intermediate values by summing adjacent pairs from that previous row. Finally, I append 1 to the end of the current row. This approach ensures that we only access valid indices and naturally handle the boundaries. For example, if numRows is 5, the third row becomes [1, 2, 1] because 1+1=2. This algorithm runs in O(numRows^2) time since we visit every cell once, and uses O(1) extra space beyond the output storage, which aligns well with Apple's focus on performance and memory efficiency. I would also ensure the code is readable, perhaps using a helper function or clear variable names like 'prevRow' to make the logic self-documenting during the review.
Common Mistakes to Avoid
- Failing to check for non-positive input values, leading to runtime errors
- Attempting to allocate a fixed 2D array of size n*n instead of dynamic lists
- Using recursion without memoization, causing exponential time complexity
- Forgetting to add the trailing 1 to each row, resulting in incorrect triangle shapes
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