LC278. First Bad Version¶

Problem Description¶

LeetCode Problem 278: 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.

Clarification¶

[1, 2, ..., n] versions
find out the first bad version
requirement: minimize the number of calls to the API

Assumption¶

Solution¶

Approach - Binary Search¶

Each version in [1, 2, ..., n] has its status in good (\(0\)) or bad (\(1\)):

\[\begin{bmatrix}1 & 2 & \cdots & i & i + 1 & \cdots n \\ 0 & 0 & \cdots & 0 & 1 & \cdots & 1 \end{bmatrix}\]

The status array consists of just two statuses and is sorted. So we can use binary search to find the first bad version effectively.

Python

class Solution:
    def firstBadVersion(self, n: int) -> int:
        left, right = 0, n

        while left < right:
            mid = (left + right) // 2

            if isBadVersion(mid):
                right = mid
            else:
                left = mid + 1

        return left

Complexity Analysis¶

Time complexity: \(O(\log n)\)
Since using binary search, the time complexity is \(O(\log n)\).
Space complexity: \(O(1)\)
Only use two index variables.