**Approach**: Two-pass—first pass counts characters; second pass finds first with count 1. Preserves order of first occurrence. **Code**: def first_non_repeating(s): counts = {}; for c in s: counts[c] = counts.get(c, 0) + 1; return next((c for c in s if counts[c] == 1), None)....
Red Flag: Using a nested loop (O(n²)) or not handling edge cases (empty, all repeating). Pro-Move: 'I'd use Counter for readability; in production we'd add a docstring and handle the case where the input might be bytes vs. str for encoding safety.'
This easy-level Python/Coding question appears frequently in data engineering interviews at companies like Delivery Hero, Dunnhumby. While less common, it tests deeper understanding that distinguishes strong candidates. Mastering the underlying concepts (python) will help you answer variations of this question confidently.
Start by clearly defining the core concept being asked about. Interviewers want to see that you understand the fundamentals before diving into implementation details. Structure your answer with a definition, then explain the practical application with a concise example.
Approach: Two-pass—first pass counts characters; second pass finds first with count 1. Preserves order of first occurrence. Code: def first_non_repeating(s): counts = {}; for c in s: counts[c] = counts.get(c, 0) + 1; return next((c for c in s if counts[c] == 1), None). Alternative with Counter: def first_non_repeating(s): from collections import Counter; cnt = Counter(s); return next((c for c in s if cnt[c] == 1), None). Complexity: O(n) time, O(k) space (k = unique chars). Edge cases: Empty string → None; all repeating → None. Scalability: Single pass for count is optimal; two passes for order. Alternative: OrderedDict or collections.Counter; Counter is cleaner: from collections import Counter; cnt = Counter(s); return next((c for c in s if cnt[c] == 1), None). Best practice: Clarify if 'first' means left-to-right occurrence or insertion order; handle Unicode (char vs. code point).
This answer is partially locked
Unlock the full expert answer with code examples and trade-offs
Practice real interviews with AI feedback, track progress, and get interview-ready faster.
Pro starts at $24/mo - cancel anytime
Paste your answer and get instant AI feedback with a FAANG-level improved version.
Analyze My Answer — FreeAccording to DataEngPrep.tech, this is one of the most frequently asked Python/Coding interview questions, reported at 2 companies. DataEngPrep.tech maintains a curated database of 1,863+ real data engineering interview questions across 7 categories, verified by industry professionals.