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First Bad Version

Created: October 26, 2018 by [lek-tin]

Last updated: May 1, 2020

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.

Solution

Java

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

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        if (n < 1) {
            return 0;
        }
        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;
        }
        if (isBadVersion(end)) {
            return end;
        }
        return 0;
    }
}

Python

# 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, end = 1, n
        ## "start+1" ensures start and end never cross
        while start + 1 < end:
            mid = start + (end-start)//2
            if isBadVersion(mid):
                end = mid
            # If a version is ok, all the preceding versions should be ok too 😉
            else:
                start = mid + 1

        if isBadVersion(start):
            return start
        if isBadVersion(end):
            return end