The O(n) Club: Longest Common Prefix for Devs Who Love Autocomplete More Than Typing

The O(n) Club: Longest Common Prefix for Devs Who Love Autocomplete More Than Typing

⚡ TL;DR

The mission: Find the longest prefix all strings in your Java array have in common. The brute force solution: keep trimming until everybody agrees. For tiny data or masochists only:

// Naive and proud
public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    String prefix = strs[0];
    for (int i = 1; i < strs.length; i++) {
        while (strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length() - 1);
            if (prefix.isEmpty()) return "";
        }
    }
    return prefix;
}

🧠 How Devs Usually Mess This Up

This problem is a magnet for rookie mistakes, like:

• Thinking a “common prefix” can actually start anywhere (spoiler: it can’t)
• Trusting all strings are the same length (the index out-of-bounds boogeyman thanks you)
• Forgetting the “empty string inside means the answer is empty” rule, or skipping the empty input scenario

If your code skips these, hope your interviewer brought popcorn.

🔍 What’s Actually Going On

Imagine being an autocomplete robot at Stack Overflow HQ: six million user queries an hour, all with slightly different gibberish. You don’t scan every character of every variant—you find the common prefix with the minimum of fuss.

Basic: march down the strings until a mismatch. Smarter: sort the list, then compare just the first and last words. If "alphabet" and "alpaca" don’t agree, nothing else will past that point. Want to be an algorithm hipster? Try binary search on the prefix length, or divide-and-conquer (if you secretly enjoy recursion headaches).

🛠️ PseudoCode

• Step 1: Short-circuit the drama: Empty array? Return “”.
  

  if (strs == null || strs.length == 0) return "";

• Step 2: Sort strings. Outliers drift to ends.
  

  Arrays.sort(strs);

• Step 3: Compare the start of the first and last item. That’s all you need.
  

  String first = strs[0];
  String last = strs[strs.length - 1];
  int i = 0;
  while (i < first.length() && i < last.length() && first.charAt(i) == last.charAt(i)) {
      i++;
  }

• Step 4: Hand back everything before the first spat.
  

  return first.substring(0, i);

💻 The Code

// Sort and compare method
public String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";
    Arrays.sort(strs);
    String first = strs[0];
    String last = strs[strs.length - 1];
    int minLength = Math.min(first.length(), last.length());
    int i = 0;
    while (i < minLength && first.charAt(i) == last.charAt(i)) {
        i++;
    }
    return first.substring(0, i);
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Empty Strings: If any string is empty, your prefix is instantly empty. Period.
• Solo Input: One string? That’s your prefix. Also, check for null/empty arrays. (Trust issues matter.)
• Don’t Trip Over Indices: substring and charAt calls will mutiny if you go out of bounds.
• Time Complexity: Sorting is O(n log n), prefix check is O(m), where n = array length, m = word length. Fancy but fast.
• Space Complexity: Mostly in-place. Java’s sort might sneakily use a little more behind your back, but it won’t max out your RAM.

🧠 Interviewer Brain-Tattoo

“A real common prefix shows up at the very start—just like the engineer who brings coffee to the morning stand-up.”