LeetCode 278: First Bad Version Solution in Python – A Step-by-Step Guide

What Is LeetCode 278: First Bad Version?

At its core, LeetCode 278: First Bad Version is about finding a breaking point in a sequence. You’re given an integer n, representing versions 1 through n, and an API function isBadVersion(version) that tells you whether a specific version is bad (True) or good (False). The catch? Once a version goes bad, all versions after it are bad too. Your job is to pinpoint the first bad version—the smallest number where isBadVersion returns True—while minimizing how many times you call that API. Think of it like finding the first spoiled apple in a row where every apple after it is rotten too. This problem is a great intro to optimization techniques, sharing DNA with classics like LeetCode 704: Binary Search, but with a twist: we’re hunting for a transition point rather than a specific value.

Problem Statement

• Input: An integer n (total number of versions) and the isBadVersion(version) API.
• Output: An integer—the first bad version.
• Rules: Versions are numbered 1 to n; if a version is bad, all subsequent versions are bad; aim to call isBadVersion as few times as possible.

Constraints

• n ranges from 1 to 2³¹ - 1 (a massive number, so efficiency matters!).

Examples

Let’s look at a few test cases to get the hang of it:

• Input: n = 5, first bad version = 4
• isBadVersion(3) = False, isBadVersion(4) = True, isBadVersion(5) = True
• Output: 4
• Input: n = 1, first bad version = 1
• isBadVersion(1) = True
• Output: 1
• Input: n = 3, first bad version = 2
• isBadVersion(1) = False, isBadVersion(2) = True, isBadVersion(3) = True
• Output: 2

These examples show a pattern: we’re looking for the moment the sequence flips from good (False) to bad (True). Let’s figure out how to find it efficiently.

Understanding the Problem: Pinpointing the Bug

To crack LeetCode 278 in Python, we need to locate the smallest version where isBadVersion returns True, knowing that everything after it follows suit. For instance, if n=5 and the first bad version is 4, the sequence looks like this: [F, F, F, T, T]. Our target is that first T. The simplest idea might be to check each version from 1 to n, but with n potentially reaching billions (2³¹ - 1), that could mean billions of API calls—way too slow! Instead, we’ll explore two solutions: 1. Best Solution (Binary Search): Runs in O(log n) time, making only a handful of calls. 2. Alternative Solution (Linear Search): Takes O(n) time, checking every version step-by-step.

The binary search is the star here, and we’ll break it down extra carefully for anyone new to coding. Let’s start there!

Best Solution: Binary Search

Why This Is the Best Solution

The binary search approach is hands-down the winner for LeetCode 278 because it’s blazing fast—O(log n) time—and super light on resources, using just O(1) space. It cuts the number of isBadVersion calls down to a logarithmic scale, meaning even for n = 1,000,000, it takes only about 20 checks instead of a million. For beginners, it might sound tricky, but it’s like a clever guessing game—and once you get it, you’ll feel like a coding wizard!

How It Works

Imagine you’re searching for a hidden treasure in a long row of boxes, numbered 1 to n, and you know that once you find a “bad” box, all the ones after it are bad too. Instead of opening every box, you’d start in the middle, check it, and then decide: “Is the treasure before or after this point?” That’s binary search in a nutshell—it halves your search area with each step until you zero in on the first bad version. Here’s the process, explained simply:

• Step 1: Define the Search Range:
• Set two pointers: left = 1 (the start) and right = n (the end).
• These mark the range where the first bad version hides.

• Step 2: Run the Guessing Game:
• While left < right (meaning there’s still a range to search):
• Calculate the middle point: mid = left + (right - left) // 2. (This avoids overflow for big numbers.)
• Call isBadVersion(mid):
• If True, the first bad version is at mid or earlier, so move right = mid.
• If False, the first bad version is after mid, so move left = mid + 1.
• Keep going until left and right meet at the answer.

• **Step 3: Why This Finds the *First* Bad Version**:
• The sequence is like a light switch: all False (good) before the first True (bad), then all True after.
• By setting right = mid when mid is bad, we keep mid in play—it might be the first bad one.
• When mid is good, left = mid + 1 skips past it, narrowing in on the flip point.

• Step 4: The Answer:
• When left == right, that’s your first bad version!

It’s like flipping through a book to find where the story turns sour—split, check, repeat!

Step-by-Step Example

Let’s walk through it with n = 5, first bad version = 4. The sequence is [F, F, F, T, T].

• Initial Setup: left = 1, right = 5.
• Iteration 1:
• Mid = 1 + (5 - 1) // 2 = 3.
• isBadVersion(3) = False (good).
• Move left = 3 + 1 = 4, right = 5.
• Iteration 2:
• Mid = 4 + (5 - 4) // 2 = 4.
• isBadVersion(4) = True (bad).
• Move right = 4, left = 4.
• Done: left = right = 4, and isBadVersion(4) = True, isBadVersion(3) = False. Answer: 4.

Now try n = 3, first bad version = 2 ([F, T, T]):

• Initial: left = 1, right = 3.
• Iteration 1:
• Mid = 1 + (3 - 1) // 2 = 2.
• isBadVersion(2) = True.
• right = 2, left = 1.
• Iteration 2:
• Mid = 1 + (2 - 1) // 2 = 1.
• isBadVersion(1) = False.
• left = 2, right = 2.
• Done: Answer: 2.

See? It’s like a dance—left, right, meet in the middle!

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down so anyone can follow:

Copy code# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n: int) -> int:
        # Start with the full range: 1 to n
        left = 1
        right = n

        # Keep guessing while there’s a range to search
        while left < right:
            # Pick the middle version (safe math to avoid overflow)
            mid = left + (right - left) // 2

            # Check if mid is bad
            if isBadVersion(mid):
                # If bad, look at mid or earlier
                right = mid
            else:
                # If good, look after mid
                left = mid + 1

        # When left meets right, that’s the first bad version
        return left

• Lines 8-9: left is 1 (first version), right is n (last version)—our starting range.
• Line 11: Loop runs while left < right—stops when they point to the same spot.
• Line 13: mid splits the range. We use left + (right - left) // 2 instead of (left + right) // 2 to avoid integer overflow with huge n.
• Line 15-19: If isBadVersion(mid) is True, the first bad version is mid or before, so right = mid. If False, it’s after mid, so left = mid + 1.
• Line 21: Once left == right, that’s our answer—return it!

• Time Complexity: O(log n)—each step halves the range.
• Space Complexity: O(1)—just three variables.

This is your go-to solution—fast, clean, and clever!

Alternative Solution: Linear Search

Why an Alternative Approach?

Sometimes simplicity is king. The linear search checks every version from 1 to n until it hits the first bad one. It’s O(n) time—slower than binary search—but it’s so straightforward that it’s a great way to understand the problem before diving into optimization. Think of it as a warm-up lap!

How It Works

Picture yourself walking down a line of versions, tapping each one: “Good? Good? Good? Bad!” The moment you hit “bad,” you stop—that’s your first bad version. Here’s the breakdown:

• Step 1: Start at version 1.
• Step 2: Loop through 1 to n, calling isBadVersion(i).
• Step 3: When you get True, return that version—it’s the first bad one.

It’s like tasting cookies until you find a burnt one—slow but sure.

Step-by-Step Example

For n = 5, first bad version = 4 ([F, F, F, T, T]):

• i=1: isBadVersion(1) = False.
• i=2: False.
• i=3: False.
• i=4: True—stop!
• Return: 4.

Code for Linear Search Approach

Copy codeclass Solution:
    def firstBadVersion(self, n: int) -> int:
        # Check each version one by one
        for i in range(1, n + 1):
            if isBadVersion(i):
                return i
        # If all checked and no bad found (edge case), return n
        return n

• Time Complexity: O(n)—could check every version.
• Space Complexity: O(1)—just the loop variable.

It gets the job done, but it’s not winning any speed awards!

Comparing the Two Solutions

• Binary Search (Best):
• Pros: O(log n) time, minimal API calls, scales well.
• Cons: Takes a moment to wrap your head around.
• Linear Search (Alternative):
• Pros: Dead simple, no brain twists.
• Cons: O(n) time, too many calls for big n.

Binary search is the clear champ for efficiency.

Additional Examples and Edge Cases

• Single Version: n = 1, bad at 1 → 1. Both work fine.
• Last Version Bad: n = 3, bad at 3 → 3. No issues.
• First Version Bad: n = 5, bad at 1 → 1. Both handle it.

The solutions are robust across the board!

Complexity Breakdown

• Binary Search:
• Time: O(log n)—logarithmic steps.
• Space: O(1)—constant variables.
• Linear Search:
• Time: O(n)—linear steps.
• Space: O(1)—still constant.

Binary’s speed shines here.

Key Takeaways

• Binary Search: Split, check, conquer—fast and fun!
• Linear Search: Slow and steady—great for learning.
• Optimization: Fewer API calls = better solution.
• Python Power: Master loops and logic with [Python Basics](/python/basics).

Final Thoughts: Debug Like a Pro

LeetCode 278: First Bad Version is a fantastic problem to sharpen your debugging skills in Python. The binary search solution is your high-speed ticket to success, while the linear search offers a gentle intro. Loved this? Check out LeetCode 704: Binary Search or LeetCode 35: Search Insert Position for more search adventures. Ready to test your skills? Head over to Solve LeetCode 278 on LeetCode and catch that first bad version today!