Longest Common Prefix: Five Algorithms Compared

🔢 Problem Statement

LeetCode 14 – Longest Common Prefix
Given an array of strings, return the longest common prefix among them. If no common prefix exists, return an empty string ("").

• Example 1
  Input: ["flower", "flow", "flight"] → Output: "fl"
• Example 2
  Input: ["dog", "racecar", "car"] → Output: ""

Constraints

• 1 ≤ strs.length ≤ 200
• 0 ≤ strs[i].length ≤ 200
• strs[i] contains only lowercase English letters if non-empty.

👨‍💻 Solution Strategies

We’ll explore and compare five distinct approaches, from brute force to divide & conquer:

Where:

• n = number of strings
• m = length of the shortest string
• S = total number of characters across all strings

📝 Detailed Breakdown

1. Horizontal Scan

Compare each character position across all strings until a mismatch occurs.

for i in range(min_len):
    ch = strs[0][i]
    if all(s[i] == ch for s in strs):
        prefix += ch
    else:
        break

👍 When to use: Very small input size or whiteboard interviews—simple and intuitive.

2. Zip-based Scan

Transpose the list of strings with zip(*strs) and use set to detect divergence.

for letters in zip(*strs):
    if len(set(letters)) == 1:
        prefix += letters[0]
    else:
        break

👍 When to use: Clean Pythonic code; moderate input sizes, readability prioritized.

3. Binary Search on Prefix Length

Binary‐search the possible prefix length range [1..min_len], validating with startswith().

low, high = 1, min_len
while low <= high:
    mid = (low + high) // 2
    if is_common_prefix(mid):
        best = strs[0][:mid]
        low = mid + 1
    else:
        high = mid - 1

👍 When to use: Strings are long (large m), and individual comparisons are expensive.

4. Sort & Compare (First vs Last)

Sort strs lexicographically; only the first and last items need comparison.

strs.sort()
first, last = strs[0], strs[-1]
for i, (a, b) in enumerate(zip(first, last)):
    if a != b:
        return first[:i]
return first

👍 When to use: Default choice—smallest code footprint and best benchmark performance.

5. Divide & Conquer

Recursively split the array, compute LCP for each half, then merge.

def lcp_range(l, r):
    if l == r:
        return strs[l]
    mid = (l + r) // 2
    left = lcp_range(l, mid)
    right = lcp_range(mid + 1, r)
    # merge two prefixes
    return common_prefix(left, right)

👍 When to use: Academic deep dives or parallel execution frameworks.

📊 Benchmark Results

Average runtime per call (seconds):

🧑‍🏫 When to Use Each Approach

• Sort & Compare: Go-to for most real-world cases—minimal code, fastest in practice.
• Binary Search: When strings are very long and each startswith check is expensive.
• Zip-based Scan: Super readable and concise; perfect for moderate input sizes.
• Horizontal Scan: Ultra-clear and memory-light; good for tiny lists or interview whiteboard.
• Divide & Conquer: Great for teaching or parallel frameworks—even though recursion adds overhead.

💡 Key Takeaways

• Constant‐time factors matter: two O(S) scans can differ drastically in practice.
• Benchmark on realistic data shapes before optimizing.
• Keep code simple unless a clear bottleneck exists.

⏱️ Time Invested

• Analysis & design: 15 min
• Implementation & benchmarking: 30 min
• Writing & review: 10 min

Total: ~ 55 min

🔗 Resources

• ▶ Run on LeetCode
• 🔗 My Submission
• 💻 Source Code on GitHub

✍️ Enjoyed this deep dive? Let’s connect on LinkedIn or explore more on GitHub.