# 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.
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
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.

Then 4 is the first bad version.

@tag-array


# 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 + (right-left) / 2, which is good Avoided overflow problem.

# Code

Java

• 

/* The isBadVersion API is defined in the parent class VersionControl.

public class Solution extends VersionControl {

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;

end = mid;
} else {
start = mid + 1;
}
}

return start;
}
}
}

class VersionControl{
// stub
// stub
return false;
}
}

• // OJ: https://leetcode.com/problems/first-bad-version/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
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

class Solution(object):
"""
:type n: int
:rtype: int
"""
lo = 1
hi = n
while lo < hi:
mid = lo + (hi - lo) / 2