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.

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

Select the correct implementation:

1def firstBadVersion(self, n: int) -> int:
2    left, right = 1, n
3    ans = -1
4
5    while left <= right:
6        mid = (left + right) // 2
7        if isBadVersion(mid):
8            ans = mid
9            right = mid - 1
10        else:
11            left = mid + 1
12
13    return ans

1def firstBadVersion(self, n: int) -> int:
2    left, right = 1, n
3    ans = -1
4
5    while left <= right:
6        mid = (left + right) // 2
7        if isBadVersion(mid):
8            right = mid - 1
9        else:
10            ans = mid
11            left = mid + 1
12
13    return ans

1def firstBadVersion(self, n: int) -> int:
2    left, right = 1, n
3    ans = -1
4
5    while left < right:
6        mid = (left + right) // 2
7        if isBadVersion(mid):
8            left = mid - 1
9        else:
10            ans = mid
11            right = mid + 1
12
13    return ans

1def firstBadVersion(self, n: int) -> int:
2    left, right = 1, n
3    ans = -1
4
5    while left < right:
6        mid = (left + right) // 2
7        ans = mid
8        if isBadVersion(mid):
9            left = mid - 1
10        else:
11            right = mid + 1
12
13    return ans


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