To solve Pascal's Triangle II, directly compute only the requested row using dynamic programming rather than building the full triangle. Iteratively update a single array from the end toward the beginning to maintain correct state transitions. This approach minimizes memory while accurately reflecting the combinatorial sums for the rowIndex-th row.
Problem Statement
Given a non-negative integer rowIndex, return the rowIndex-th row (0-indexed) of Pascal's Triangle. Each element in the row is the sum of the two elements directly above it in the previous row.
For example, rowIndex = 3 should return [1,3,3,1], and rowIndex = 0 should return [1]. Implement an approach that efficiently calculates the row using array-based dynamic programming and proper state transitions.
Examples
Example 1
Input: rowIndex = 3
Output: [1,3,3,1]
Example details omitted.
Example 2
Input: rowIndex = 0
Output: [1]
Example details omitted.
Example 3
Input: rowIndex = 1
Output: [1,1]
Example details omitted.
Constraints
- 0 <= rowIndex <= 33
Solution Approach
Iterative Array Update
Initialize an array of size rowIndex + 1 with 1s and iterate from the second row to rowIndex. Update each element from end to start to avoid overwriting needed values.
In-Place Dynamic Programming
Use a single array to store current row values, modifying elements in place. For each row, update elements from right to left based on the sum of the current element and the one before it.
Space Optimization
Avoid constructing the entire triangle by maintaining only one array for the current row. This reduces space complexity to O(rowIndex) while still using the standard state transition logic.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(rowIndex^2) due to nested iteration to update each element, and space complexity is O(rowIndex) because only a single row array is maintained.
What Interviewers Usually Probe
- Check if you can compute only the requested row without building the full triangle.
- Expect discussion on in-place updates and avoiding overwriting previous row values.
- Consider edge cases like rowIndex = 0 or rowIndex = 1 to confirm correct base cases.
Common Pitfalls or Variants
Common pitfalls
- Overwriting elements from left to right, which corrupts intermediate sums.
- Attempting to build the full Pascal's Triangle when only one row is needed.
- Ignoring base cases, leading to index out-of-bounds errors or incorrect first row values.
Follow-up variants
- Return the triangle up to rowIndex instead of only one row.
- Compute a specific element in rowIndex-th row using combinatorial formula.
- Modify the algorithm to handle large rowIndex values efficiently with BigInteger types.
How GhostInterview Helps
- Highlights the correct dynamic programming state transitions for efficient single-row computation.
- Provides step-by-step array update examples for interview-ready reasoning.
- Identifies subtle overwriting pitfalls that often cause wrong outputs in Pascal's Triangle II.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the best approach for Pascal's Triangle II using dynamic programming?
Use a single array updated from right to left, applying state transitions to compute only the requested row.
How do I handle rowIndex = 0 or rowIndex = 1?
Return [1] for rowIndex = 0 and [1,1] for rowIndex = 1, handling base cases before iteration.
Can I reduce space usage when generating Pascal's Triangle II?
Yes, by maintaining a single array and updating it in place, space complexity is O(rowIndex).
Why does updating from left to right fail?
It overwrites values needed for computing subsequent elements, breaking the state transition logic.
Is there a combinatorial formula alternative?
Yes, each element can be computed using C(rowIndex, k), but iterative DP is often simpler for interviews.
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