https://leetcode.com/problems/first-bad-version/description/

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

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        //i am a manager and has a team developing a product.
        //the latest product fails. Because the new version is based on previous version. So i need to find the earliest version that fails.
        //i can use isBadVersion to check if it is bad.
        //i can't use API for too many times.
        //our product versions is like: [1,2,3....n], is in ascending order
        //check if [i] is bad and [i-1] is good then return [i]
            //if [i-1] is bad also then check left part
        //check if [i] is good and [i+1] is bad then return [i+1]
            //if [i+1] is good also then check right part
        int low=0,high=n;
        for(int mid=low+((high-low)>>1);low<=high;){
            if(isBadVersion(mid)){
                if(mid-1>=0&&!isBadVersion(mid-1)){
                    return mid;
                }else{
                    //check left
                    high=mid-1;
                    mid=low+((high-low)>>1);
                }
            }else {
                if(mid+1<=n-1&&isBadVersion(mid+1)){
                    return mid+1;
                }else{
                    //check right
                    low=mid+1;
                    mid=low+((high-low)>>1);
                }
            }
        }
        return n;
    }
};

It can be more clear.

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int left=1, right=n;
        for(;left<=right;){
            int mid=left+((right-left)>>1);
            if(isBadVersion(mid)){
                //check left. It may cause empty array.
                //[Bad1,Bad2]->[]. return left
                //[bad]->[]. return left
                right=mid-1;
            }else{
                //check right
                //[good,bad]->[bad]
                left=mid+1;
            }
        }
      	// when break, all exit is to return left.
        return left;
    }
};

Another way to think return value is to consider the break condition.

When not found, right=mid-1 < left. Of course we need the one in the back.