Navigating the Maze of Versions: A LeetCode Guide to "First Bad Version"

Uncover the secrets to acing the "First Bad Version" problem on LeetCode with our step-by-step solutions in Python, TypeScript, and Java.

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5 min read

Navigating the Maze of Versions: A LeetCode Guide to "First Bad Version"

Hello, fellow software engineers, whether you're sharpening your skills for the next engineering interview or diving into the depths of coding challenges for the first time. Today, I'm here to walk you through a common but intriguing problem that pops up in interviews: the "First Bad Version" challenge from LeetCode (LeetCode 278).

Let's decode this problem together, exploring not just a solution but understanding the why and how, making you ready for when this or similar challenges come your way.

Introduction to the Problem

Imagine you're a product manager, and your latest product version fails the quality check. Since versions build on each other, all versions after a bad version are also bad. Given n versions [1, 2, ..., n] and an API bool isBadVersion(version), your task is to find the first bad one, minimizing API calls.

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.

Example 2:

Input: n = 1, bad = 1
Output: 1

Constraints:

  • 1 <= bad <= n <= 2<sup>31</sup> - 1

Approach to Solving the Problem

Initially, I took an approach that, albeit correct in logic, was inefficient in its execution. I made the mistake of not fully considering the constraints, leading to unnecessary API calls. I used a binary search but added an extra step to check if the immediate previous version was not bad, which doubled the number of API calls in some cases.

# Correct but inefficient solution
def firstBadVersion(self, n: int) -> int:
        low = 0 # constraints say the low is one
        high = n
        while low <= high: # one extra loop than is necessary
            mid = (low + high) // 2
            if isBadVersion(mid):
                if not isBadVersion(mid - 1): # duplicated call to the API
                    return mid
                else:
                    high = mid - 1 
            else:
                low = mid + 1
        return -1 # the constraints say there will always be a solution

Big O Notation Analysis:
This incorrect approach had a time complexity of O(log n) due to binary search but with unnecessary additional API calls, impacting performance.

Refining Our Approach

The key to solving this problem efficiently lies in minimizing API calls. This can be achieved through a refined binary search strategy:

  1. Initialize two pointers, left = 1 and right = n.

  2. While left < right, find the midpoint and use isBadVersion(mid).

    • If true, the bad version is at mid or before it. Set right = mid.

    • If false, the first bad version is after mid. Set left = mid + 1.

  3. Conclude when left == right, which will be your first bad version.

Solution in Python

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

# Correct and efficient solution
def firstBadVersion(n):
    left, right = 1, n
    while left < right:
        mid = left + (right - left) // 2
        # Minimize API calls by efficient binary search
        if isBadVersion(mid):
            right = mid  # Focus on the left half
        else:
            left = mid + 1  # Focus on the right half
    return left  # The convergence point is the first bad version

Solution in TypeScript

var solution = function(isBadVersion: any) {
    return function(n: number): number {
        let left = 1;
        let right = n;
        while (left < right) {
            const mid = left + Math.floor((right - left) / 2);
            // Efficient binary search to reduce API calls
            if (isBadVersion(mid)) {
                right = mid;  // Narrow down to the left
            } else {
                left = mid + 1;  // Narrow down to the right
            }
        }
        return left;  // Found the first bad version
    };
};

Solution in Java

public int firstBadVersion(int n) {
    int left = 1;
    int right = n;
    while (left < right) {
        int mid = left + (right - left) / 2;
        // Apply binary search to minimize API usage
        if (isBadVersion(mid)) {
            right = mid;  // The search continues on the left side
        } else {
            left = mid + 1;  // The search shifts to the right side
        }
    }
    return left;  // Identifies the first bad version
}

Conclusion

In solving the "First Bad Version" problem, the essence lies not just in finding the solution but in optimizing the process. By refining our binary search approach, we ensure minimal API calls, showcasing the kind of efficiency and problem-solving prowess sought in software engineering interviews.

Whether you're experienced or new to coding interviews, understanding the rationale behind each step and the importance of optimization can significantly impact your performance. I hope this walkthrough not only helps you solve this specific problem but also enhances your overall approach to tackling algorithmic challenges.

Happy coding, and may you find all your bad versions swiftly and efficiently!


I hope this post serves as a helpful guide in your interview preparation journey, offering both insights and practical solutions to a common interview challenge. If you have any questions or need further clarifications, feel free to ask. Good luck with your interviews!

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