278. First Bad Version (Easy)

https://leetcode.com/problems/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. 

Solutions

class Solution extends VersionControl {

    public int firstBadVersion(int n) {
        if (n == 1 || isBadVersion(1)) {
            return 1;
        }

        if (!isBadVersion(1) && isBadVersion(2)) {
            return 2;
        }

        return bisect(1, n);
    }

    public int bisect(long left, long right) {
        // be careful of the overflow issue
        int mid = (int) ((right + left) / 2);

        // outlet of recursion
        if (mid > 1 && isBadVersion(mid) && !isBadVersion(mid - 1)) {
            return mid;
        }

        // if version mid is contaminated the first bad version must be earlier but
        // not equal to version mid. If the mid is the first bad, execution should never
        // reach here and already exit from the above outlet
        if (isBadVersion(mid)) {
            return bisect(left, mid - 1);
        }

        // if version mid is a fine version, search the consequent releases
        return bisect(mid + 1, right);
    }
}

class Solution extends VersionControl {

    public int firstBadVersion(int n) {
        return helper(1, n);
    }

    public int helper(long i, long j) {
        // be careful of the overflow issue
        int m = (int) ((j + i) / 2);

        // actually, i > j will never happens since it already returned when i == j
        if (i >= j) {
            return (int) i;
        }

        // if version m is contaminate the first bad version must be earlier or equal to version m
        if (isBadVersion(m)) {
            return helper(i, m);
        }

        // if version m is a fine version, search the consequent releases
        return helper(m + 1, j);
    }
}

Incorrect Solutions

References