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.
 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 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.

 call isBadVersion(3) -> false
 call isBadVersion(5) -> true
 call isBadVersion(4) -> true

 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

public class First_Bad_Version {

    /* 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;

            // @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;

                if (isBadVersion(mid)) {
                    end = mid;
                } else {
                    start = mid + 1;
                }
            }

            return start;
        }
    }
}

class VersionControl{
    // stub
    public boolean isBadVersion(int n) {
        // stub
        return false;
    }
}

// OJ: https://leetcode.com/problems/first-bad-version/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int firstBadVersion(int n) {
        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
# def isBadVersion(version):

class Solution(object):
  def firstBadVersion(self, n):
    """
    :type n: int
    :rtype: int
    """
    lo = 1
    hi = n
    while lo < hi:
      mid = lo + (hi - lo) / 2
      if isBadVersion(mid):
        hi = mid
      else:
        lo = mid + 1
    return lo

278 - First Bad Version

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