278. First Bad Version (Leetcode|| Java || Easy || Binary Search)

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

SOLUTION:

/* 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;//We have to search in a range as per the question, so we use binary search
int end = n;

while(start <= end){
int mid = start + (end — start)/2;

boolean isBad = isBadVersion(mid);//We will check if the version or element we are at is bad or not and if it is bad we are going to move end toward mid and if it is not bad we will move our start from good versions toward the bad

if(start == end){
return end;
}
else if(isBad){
end = mid;
}else{
start = mid + 1;
}
}
return -1;//if no version is bad return -1
}
}

TIME COMPLEXITY :- O(log(n))

SPACE COMLEXITY :- O(1)

Thank you for reading. If you have any queries please let me know in the comment section.