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Ericsson·Software Engineer·Onsite - Coding / Algorithms·Junior

Junior
May 2026

Summary

Coding round at Ericsson for a software engineer role, pretty straightforward from what I can tell. Just the one problem and it was a classic.

Questions Asked (1)

Q1

Generate Pascal's Triangle up to n rows.

Algorithms & Data Structures
Author's notes

Straight from LeetCode, no modifications.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: confirm whether the output should be a list of lists or printed rows, and discuss edge cases like n=0. Then present a clear algorithm, such as building each row from the previous one using the relation triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j], and analyze time and space complexity.

Pro tip: Mention that you can optimize space to O(n) by generating each row in-place from right to left, but only if the interviewer cares about memory; otherwise, prioritize clarity. Also, relate the problem to real-world scenarios like binomial coefficients or dynamic programming to show depth.

1. Clarify requirements and edge cases

Ask whether n is guaranteed non-negative, what output format is expected (e.g., list of lists), and how to handle n=0 or n=1.

2. Choose an approach

Decide between iterative row-by-row construction (O(n^2) time, O(n^2) space) or space-optimized version (O(n) extra space). Explain the trade-offs.

3. Outline the algorithm

Describe how to build each row: first and last elements are 1; middle elements are sum of two elements above. Optionally, mention using combinatorial formula C(n,k) = n!/(k!(n-k)!) but note it's less efficient.

4. Analyze complexity

State time complexity O(n^2) and space complexity O(n^2) for the straightforward approach, or O(n) extra space for the optimized version.

5. Test with examples

Walk through a small example like n=5 to verify correctness, and mention potential pitfalls like integer overflow for large n (though not typical in interviews).

Key Points to Mention

  • The recurrence relation: each element is the sum of the two directly above it.
  • Edge cases: n=0 (empty output), n=1 (single row [1]).
  • Time complexity: O(n^2) because total elements = n(n+1)/2.
  • Space complexity: O(n^2) for storing all rows; can be reduced to O(n) extra space by generating rows in-place.
  • Alternative approach using binomial coefficients, but it's less efficient due to factorial computations.
  • Potential optimization: generate each row from right to left to use only O(n) space.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.