0278. First Bad Version

Easy | Binary Search | 24 ms (94.52%), 14.2 MB (73.95%)

❓ Problem Statement

Source: LeetCode - First Bad Version GitHub: Solution / Performance

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.

✍🏻 Constraints & Example

Constraints:

• 1 <= bad <= n <= 2^31 - 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.

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

💡 Ideas

Binary Search Problem Sorted array to find a minimum number that is the first bad version.

Boundary / Search Space

Left (Minimum) = 1 (first version) Right (Maximum) = n (last element could be the answer)

Condition

While Loop = Left < Right Return = isBadVersion(mid) == True

Return Value

Left (as index)

🤖 Python3

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return an integer
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        # (base case)
        if n == 1: return 1
        if n == 2: return 1 if isBadVersion(1) else 2
        
        # ==================================================
        #  Binary Search                                   =
        # ==================================================
        # time  : O(log(n))
        # space : O(1)

        l, r = 1, n
        while l < r:
            mid = (l + r) // 2
            
            if isBadVersion(mid): r = mid
            else: l = mid + 1
                
        return l

🤖 Java

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    /**
     * @time  : O(log(n))
     * @space : O(1)
     */
    
    public int firstBadVersion(int n) {
        /* base case */
        if (n == 1) return 1;
        
        int l = 1, r = n;
        while(l < r) {
            int mid = l + (r - l) / 2;
            
            if(isBadVersion(mid)) r = mid;
            else l = mid + 1;
        }
        
        return l;
    }
}