First Bad Version

Problem Statement

In the context of software development, managing different versions of a product is critical, especially when identifying defective (bad) versions that may affect subsequent releases. As a product manager in such a scenario, your objective is to find the earliest bad version from a sequential list of versions labeled from 1 to n. Once a version is identified as bad, all subsequent versions are also deemed bad due to cumulative updates and dependencies.

You are provided with a helper function isBadVersion(version) that returns a boolean indicating whether a given version is bad. The challenge lies in minimizing the number of API calls made to isBadVersion() to efficiently determine the first bad version in the sequence.

Examples

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 <= 231 - 1

Approach and Intuition

The problem is essentially about finding the first occurrence of a bad version when versions are arranged sequentially. Given the nature of the problem where all versions subsequent to a bad version are also bad, we realize we can employ a binary search technique to optimize our search:

1. Start with two pointers: left initialized at 1, and right initialized at n (the total number of versions).
2. Calculate the midpoint mid using the formula mid = left + (right - left) / 2.
3. Use the isBadVersion(mid) function to check if the midpoint version is bad:
   • If isBadVersion(mid) is false, it means that the first bad version must be after mid, thus set left to mid + 1.
   • If isBadVersion(mid) is true, it could be the first bad version, but there might be earlier bad versions. Thus, set right to mid.
4. Continue the process until the left pointer matches the right pointer, which will then point to the first bad version.

This binary search approach reduces the potential number of isBadVersion() calls from n (in a worst-case linear scan) to log(n), improving efficiency significantly.

By analyzing examples:

• For n = 5 and the first bad version at 4, the method gradually narrows down the range from 1-5 to 4-5, pinpointing version 4 using just three checks.
• In a minimal case where n = 1 and it itself is bad, the approach quickly identifies the version as bad in a single check.

The constraint 1 <= bad <= n <= 2^31 - 1 assures us that our approach is efficient even for large values of n, fitting within reasonable computational limits.

Solutions

public int findBadVersion(int total) {
    int start = 1;
    int end = total;
    while (start < end) {
        int middle = start + (end - start) / 2;
        if (isBadVersion(middle)) {
            end = middle;
        } else {
            start = middle + 1;
        }
    }
    return start;
}