Binary Search Speedrun ⚡

For each of the Speedrun questions, you will be given a binary-search related problem and a corresponding multiple choice question. The multiple choice questions are related to the techniques and template(s) introduced in the binary search section.

It's recommended that you have gone through at least Find the First True in a Sorted Boolean Array, Monotonic Function and Binary Search Template and a number of example problems in the binary search section.

Quick question links:

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Question 1 of 8

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 <= 231 - 1

Question

Select the correct implementation:

def firstBadVersion(self, n: int) -> int:
    left, right = 1, n
    ans = -1

    while left <= right:
        mid = (left + right) // 2
        if isBadVersion(mid):
            ans = mid
            right = mid - 1
        else:
            left = mid + 1

    return ans

def firstBadVersion(self, n: int) -> int:
    left, right = 1, n
    ans = -1

    while left <= right:
        mid = (left + right) // 2
        if isBadVersion(mid):
            right = mid - 1
        else:
            ans = mid
            left = mid + 1

    return ans

def firstBadVersion(self, n: int) -> int:
    left, right = 1, n
    ans = -1

    while left < right:
        mid = (left + right) // 2
        if isBadVersion(mid):
            left = mid - 1
        else:
            ans = mid
            right = mid + 1

    return ans

def firstBadVersion(self, n: int) -> int:
    left, right = 1, n
    ans = -1

    while left < right:
        mid = (left + right) // 2
        ans = mid
        if isBadVersion(mid):
            left = mid - 1
        else:
            right = mid + 1

    return ans