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Implement an algorithm to find the longest common prefix among an array of strings.

Python/Codingmedium0.6 min readPremium

**Why Prefix Matters in Data Engineering:** Common in key design (e.g., S3 prefix for partitioning), schema evolution (shared column prefixes), and deduplication logic. **Scalability Trade-offs:** (1) Horizontal scanning: O(S) where S = total chars—optimal for small-to-medium....

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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
Disney+ Hotstar
Key Concepts Tested
partitionspark

Why This Question Matters

This medium-level Python/Coding question appears frequently in data engineering interviews at companies like Disney+ Hotstar. While less common, it tests deeper understanding that distinguishes strong candidates. Mastering the underlying concepts (partition, spark) 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
126 words

Why Prefix Matters in Data Engineering: Common in key design (e.g., S3 prefix for partitioning), schema evolution (shared column prefixes), and deduplication logic.

Scalability Trade-offs: (1) Horizontal scanning: O(S) where S = total chars—optimal for small-to-medium. (2) Vertical scanning: stops early on first mismatch—better when strings diverge quickly. (3) Sort-and-compare: O(n log n * m)—avoids full scan but sort dominates for large n. (4) Trie: O(S) build, O(m) query—when you have many prefix queries over same set.

Cost: At Hotstar-scale, prefix logic on millions of event keys—use Spark's reduceByKey with a custom prefix reducer; avoid collect() of all strings.

def longest_common_prefix(strs):
if not strs: return ''
prefix = strs[0]
for s in strs[1:]:
while not s.startswith(prefix):
prefix = prefix[:-1]
if not prefix: return ''
return prefix

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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 a curated database of 1,863+ real data engineering interview questions across 7 categories, verified by industry professionals.

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