**Approach**: Two passes—first to count, second to find first with count 1. **Code**: `def first_non_repeating(s): counts = {}; [counts.update({c: counts.get(c, 0) + 1}) or None for c in s]; return next((c for c in s if counts[c] == 1), None)`. Simpler: `for c in s: counts[c] =...
Red Flag: O(n²) solution with nested loops. Pro-Move: 'I use a single-pass with OrderedDict (char→count) and a set of seen-repeats; return first char not in repeats. Handles unicode and large inputs.'
This easy-level Python/Coding question appears frequently in data engineering interviews at companies like Altimetrik, Infosys. 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 passes—first to count, second to find first with count 1. Code: def first_non_repeating(s): counts = {}; [counts.update({c: counts.get(c, 0) + 1}) or None for c in s]; return next((c for c in s if counts[c] == 1), None). Simpler: for c in s: counts[c] = counts.get(c,0)+1; return next((c for c in s if counts[c]==1), None). Complexity: O(n) time, O(k) space (k = distinct chars). Scalability: Single pass is possible with OrderedDict to preserve order; two-pass is clearer and sufficient for typical strings.
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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.