**Why Preserve Special Chars:** Log parsing (reverse for suffix search), obfuscation (mask emails but keep structure), and data cleaning where delimiters must stay. **Two-Pointer:** Swap only when both i and j point to letters. Advance when hitting non-letter. O(n), single...
This medium-level Python/Coding question appears frequently in data engineering interviews at companies like EY. While less common, it tests deeper understanding that distinguishes strong candidates. Mastering the underlying concepts (join) will help you answer variations of this question confidently.
Break this problem into components. Identify the core trade-offs involved, then walk the interviewer through your reasoning step by step. Demonstrate awareness of edge cases and production considerations - this is what separates good answers from great ones.
Why Preserve Special Chars: Log parsing (reverse for suffix search), obfuscation (mask emails but keep structure), and data cleaning where delimiters must stay.
Two-Pointer: Swap only when both i and j point to letters. Advance when hitting non-letter. O(n), single pass. Convert to list for in-place; join for result.
Edge Cases: Empty string, all special chars, Unicode (use str.isalpha()—handles most; unicodedata for full Unicode).
def reverse_preserve_special(s):
lst, i, j = list(s), 0, len(s)-1
while i < j:
if not lst[i].isalpha(): i += 1
elif not lst[j].isalpha(): j -= 1
else:
lst[i], lst[j] = lst[j], lst[i]
i += 1; j -= 1
return ''.join(lst)
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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 1 company. DataEngPrep.tech maintains a curated database of 1,863+ real data engineering interview questions across 7 categories, verified by industry professionals.