# Question

Formatted question description: https://leetcode.ca/all/278.html

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.

Example 1:

Input: n = 5, bad = 4
Output: 4
Explanation:
Then 4 is the first bad version.


Example 2:

Input: n = 1, bad = 1
Output: 1


Constraints:

• 1 <= bad <= n <= 231 - 1

# 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

• 

/* 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;
}
}

############

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

public class Solution extends VersionControl {
int left = 1, right = n;
while (left < right) {
int mid = (left + right) >>> 1;
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}

• // 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 an integer

class Solution:
"""
:type n: int
:rtype: int
"""
left, right = 1, n
while left < right:
mid = (left + right) >> 1
right = mid
else:
left = mid + 1
return left

############

# @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
hi = mid
else:
lo = mid + 1
return lo


• /**
* Forward declaration of isBadVersion API.
* @return 	 	      true if current version is bad
*			          false if current version is good
*/

left, right := 1, n
for left < right {
mid := (left + right) >> 1
right = mid
} else {
left = mid + 1
}
}
return left
}

• /**
*
* @param {integer} version number
* @return {boolean} whether the version is bad
*     ...
* };
*/

/**
* @return {function}
*/
var solution = function (isBadVersion) {
/**
* @param {integer} n Total versions
* @return {integer} The first bad version
*/
return function (n) {
let left = 1;
let right = n;
while (left < right) {
const mid = (left + right) >>> 1;
right = mid;
} else {
left = mid + 1;
}
}
return left;
};
};


• // The API isBadVersion is defined for you.
// to call it use self.isBadVersion(version)

impl Solution {
pub fn first_bad_version(&self, n: i32) -> i32 {
let mut left = 1;
let mut right = n;
while left < right {
let mid = left + (right - left) / 2;
right = mid;
} else {
left = mid + 1;
}
}
left
}
}