First Bad Version - Leetcode Solution

Problem Statement:

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 will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Solution:

The problem statement is essentially asking us to perform a binary search on a sorted list of n elements. We know that all versions after a bad version are also bad. Therefore, if version mid is bad, we know that the first bad version must be in the range [1, mid]. Similarly, if version mid is not bad, we know that the first bad version must be in the range [mid + 1, n]. We can use this observation to perform a binary search.

Let's start by initializing the left and right pointers to the first and last versions, respectively. We'll use a while loop to keep dividing the range in half until we find the first bad version.

class Solution:
    def firstBadVersion(self, n):
        left = 1
        right = n
        while left < right:
            mid = left + (right - left) // 2
            if isBadVersion(mid):
                right = mid
            else:
                left = mid + 1
        return left

Initially, the range is [1, n]. We calculate the midpoint mid as shown above.

If mid is bad, we know that the first bad version must be in the range [1, mid], so we set right = mid. Otherwise, we know that the first bad version must be in the range [mid + 1, n], so we set left = mid + 1.

We repeat this process until the range is reduced to a single version. At this point, we have found the first bad version.

Note that we use the expression left + (right - left) // 2 to calculate the midpoint. This is done to prevent integer overflow. If we use the expression (left + right) // 2 and left + right is too large, we may get an incorrect result.

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