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:
call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true
Then 4 is the first bad version.
Example 2:
Input: n = 1, bad = 1
Output: 1
Constraints:
- 1 <= bad <= n <= 231 - 1
My Solution
We can solve this problem with binary search. Start will be 1 and end will be n at the beginning. We take the midpoint and if mid is bad, we know that the first bad version is in the left half of the range, so end will be mid-1. If mid is good, then we know the first bad version is in the right half of the range, so we set start to mid+1. We end when end is set at the last good version and start is at the first bad version. We just return start at the end.
The time complexity of binary search is O(log n). Space complexity is O(1) since we only require constant extra space.
/* 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 <= end) {
int mid = start + (end-start)/2;
if (isBadVersion(mid)) {
end = mid-1;
}
else {
start = mid+1;
}
}
return start;
}
}