The O(n) Club: First Bad Version: Detective Work for Burnt Toast Builds

The O(n) Club: First Bad Version — Detective Work for Burnt Toast Builds

⚡ TL;DR

Bugs appeared, and now you’re stuck playing detective. Find the first bad software version out of n—without asking the API 2 billion times. Binary search saves your sanity and your weekend. Here’s the quick fix in Java:

// Wish all diagnostics were this easy.
public int firstBadVersion(int n) {
    int left = 1, right = n;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (isBadVersion(mid)) {
            right = mid; // Might be THE bad build
        } else {
            left = mid + 1; // Safe, go forward
        }
    }
    return left;
}

🧠 How Devs Usually Mess This Up

Panic, coffee, and a brute-force linear scan: check every single version, one at a time. You’ll discover the bug by the time your beard grows in, but you’ll need a new cloud bill and a support group. Or you’ll go binary search but forget how off-by-one errors work. Returning mid + 1 or using the wrong loop boundaries gets you a new best friend: the infinite loop. Rookies, assemble.

🔍 What’s Actually Going On

Picture this: a row of toasters by their production age. At some point, one starts delivering only burnt toast, and every toaster after it is also ruined. You don’t want to sniff every one. You want to find the first smoldering culprit—fast. Enter binary search: every isBadVersion(mid) check halves the number of suspects. Bad at mid? The crime happened there or earlier. Good at mid? Skip ahead. When left == right, you’ve got your toast assassin.

🛠️ PseudoCode

• Start: left = 1, right = n
• While left < right:
• Calculate: mid = left + (right - left) / 2 (avoids integer explosion for big n)
• If isBadVersion(mid) is true: set right = mid (mid could be our burned toast).
• Else: set left = mid + 1 (move on, the bread’s still good here).
• After loop: left pinpoints the first failure (and sore spot on your dev team).

💻 The Code

// Because Java rules interviews (and hearts?)
class VersionControl {
    boolean isBadVersion(int version) {
        // simulation; implementation hidden by LeetCode
        return version >= 12345;
    }
}
 public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int left = 1, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Off-by-one agony: Forget whether to return mid or mid + 1? Just let the loop finish where left == right and return left. Peace.
• Boundary chaos: Use left < right not left <= right, unless you enjoy while(true) as a lifestyle.
• Indexing battle: LeetCode gives you 1-based versions. Java arrays? Zero-based. Don’t mix drinks.
• Overflow drama: Use the safe mid calculation. Otherwise, with big n, say hi to integer wraparound and bye to your score.
• First-or-last fails: Test with a bug in version 1 and a bug in version n. Catch your code hallucinating fake results.
• Time/space: O(log n) checks, O(1) space. If this burns CPU, check your anti-virus settings—not the code.

🧠 Interviewer Brain-Tattoo

“Using a for-loop here is like printing your resume on a napkin. Just binary search it and let the API budget breathe.”