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Home/Questions/Python/Coding/Reverse a string with special characters preserved.

Reverse a string with special characters preserved.

Python/Codingmedium0.5 min read

Reviewed by Aditya Kumar · Last reviewed 2026-03-24

**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...

🤖 Analyze Your Answer
Frequency
Low
Asked at 1 company
Category
179
questions in Python/Coding
Difficulty Split
127E|24M|28H
in this category
Total Bank
1,863
across 7 categories
Asked at these companies
EY
Key Concepts Tested
join

Why This Question Matters

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.

How to Approach This

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.

Expert Answer
103 words

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)

⚡
Pro Tip

Red Flag: Creating new string per swap (immutable). Pro-Move: 'We use list(s) for mutable buffer—O(1) swap vs O(n) string concat per char.'

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According to DataEngPrep.tech, this is one of the most frequently asked Python/Coding interview questions, reported at 1 company. DataEngPrep.tech maintains an editor-reviewed database of 1,863 data engineering interview questions across 7 categories.

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