Approach: Two-pass—count chars, then find first with count 1. Code: def first_non_repeating(s): counts = {}; [counts.update({c: counts.get(c, 0) + 1}) for c in s]; return next((c for c in s if counts[c] == 1), None). Or: from collections import Counter; counts = Counter(s);...
Red Flag: O(n²) solution (nested loops). Pro-Move: 'Counter is readable; for production we'd handle Unicode normalization (NFD) since users expect á and a to match'—shows production edge-case thinking.
This easy-level Python/Coding question appears frequently in data engineering interviews at companies like Delivery Hero, Dunnhumby, Fragma Data Systems. 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—count chars, then find first with count 1. Code: def first_non_repeating(s): counts = {}; [counts.update({c: counts.get(c, 0) + 1}) for c in s]; return next((c for c in s if counts[c] == 1), None). Or: from collections import Counter; counts = Counter(s); return next((c for c in s if counts[c] == 1), None). Complexity: O(n) time, O(k) space. Why: Single pass can't know if char is unique until full scan. Best practice: Clarify behavior for empty string (None); consider OrderedDict if order matters for first occurrence.
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According to DataEngPrep.tech, this is one of the most frequently asked Python/Coding interview questions, reported at 3 companies. DataEngPrep.tech maintains a curated database of 1,863+ real data engineering interview questions across 7 categories, verified by industry professionals.