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Binary Search - Easy Level - Question 1

Binary Search - Easy Level - Question 1


Leetcode 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 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:

1 <= bad <= n <= 2^31 - 1


Analysis:

The n value could be very large, so the naïve way to go back one by one is too slow.

One observation is that: if the xth version is good, then all the previous versions before x are good as well. So it is suitable for the binary search.

So we can make a guess first, then can shrink the search range by half.


See the code below:

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

class Solution {
public:
    int firstBadVersion(int n) {
        int left = 1, right = n;
        while(left < right) {
            int mid = left + (right - left) / 2;
            if(isBadVersion(mid)) right = mid;
            else left = mid + 1;
        }
        return left;
    }
};




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