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 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.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version. 

Explanation

Use binary search to find first bad version with O(logn) time complexity.

Java Solution

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int start = 1;
        int end = n;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (isBadVersion(mid)) {
                end = mid;
            } else {
                start = mid;    
            }            
        }
        
        if (isBadVersion(start)) {
            return start;
        }
        return end;
    }
}

Python Solution

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        start = 1
        end = n
        
        while start + 1 < end:
            mid = start + (end - start) // 2
            
            if isBadVersion(mid) == True:
                end = mid
            else:
                start = mid
        
        if isBadVersion(start):
            return start
        else:
            return end

• Time complexity: O(logn). The search space is halved each time, so the time complexity is O(logn).
• Space complexity: O(1)