Question
Formatted question description: https://leetcode.ca/all/278.html
278. First Bad Version
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.
@tagarray
Algorithm
Binary search.
It should be noted that there is a pitfall
in OJ, that is, if both left and right are particularly large, then left+right may overflow. Our solution is to become left + (rightleft) / 2
, which is good Avoided overflow problem.
Code
Java

public class First_Bad_Version { /* 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; // @note: cannot be <= // eg. n=5 then mid=3 then mid= 2 then start=1&end=1, infinite loop while (start < end) { int mid = start + (end  start) / 2; if (isBadVersion(mid)) { end = mid; } else { start = mid + 1; } } return start; } } } class VersionControl{ // stub public boolean isBadVersion(int n) { // stub return false; } }

// OJ: https://leetcode.com/problems/firstbadversion/ // Time: O(logN) // Space: O(1) class Solution { public: int firstBadVersion(int n) { int L = 1, R = n; while (L <= R) { int M = L + (R  L) / 2; if (isBadVersion(M)) R = M  1; else L = M + 1; } return L; } };

# The isBadVersion API is already defined for you. # @param version, an integer # @return a bool # def isBadVersion(version): class Solution(object): def firstBadVersion(self, n): """ :type n: int :rtype: int """ lo = 1 hi = n while lo < hi: mid = lo + (hi  lo) / 2 if isBadVersion(mid): hi = mid else: lo = mid + 1 return lo