LeetCode 278: First Bad Version

Description

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.

Constraints

• 1 <= bad <= n <= 231 - 1

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

Solution 1: Brute-Force

Steps

1. The most intuitive and basic solution is to iterate from version 1 to version n sequentially and check one by one. If isBadVersion returns True, then immediately returns the current version.

Codes

def first_bad_version(n: int) -> int:
    for version in range(1, n + 1):
        if isBadVersion(version):
            return version

Complexities

• Time Complexity: O(N), where N is n.
• Space Complexity: O(1)

Solution 2: Binary Search

Thought

According to the requirements of the problem, we need to find a specific target within a range ([1, 2, ..., n]) where the elements in the array are sorted. This gives us enough hints to use binary search. During each search, we continually narrow down the search range, getting closer to the target we are looking for.

Steps

1. Use two pointers, left and right, to represent the left and right boundaries of the current search range.
2. When left is less than or equal to right:
   1. Calculate the midpoint mid of the current range as mid = (left + right) // 2.
   2. Check the version pointed to by mid. If it is a normal version (isBadVersion(mid) == False), it means that all versions before mid are also normal and can be skipped. Move the left pointer to the element immediately after mid.
   3. If mid points to a bad version (isBadVersion(mid) == True), it’s important to note that mid may not be the answer we’re looking for because we want to find the first bad version. In this case, we need to check if the version before mid (mid - 1) is also a bad version. If it is, it means that mid is not the answer, so we move the right pointer to the element before mid and repeat the above steps. On the other hand, if the version before mid is a normal version, it implies that mid is indeed the first bad version, so we can directly return mid.

Codes

def first_bad_version(self, n: int) -> int:
        left = 1
        right = n
        
        while left <= right:
            mid = (left + right) // 2
            if not isBadVersion(mid):
                left = mid + 1
            else:
                if mid > 1 and isBadVersion(mid - 1):
                    right = mid - 1
                else:
                    return mid

Complexities

• Time Complexity: O(logN), where N is n.
• Space Complexity: O(1)

References

• LeetCode link: First Bad Version – LeetCode