leetcode278

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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.

Idea: binary search

template:

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public class Solution {
    /**
     * @param A an integer array sorted in ascending order
     * @param target an integer
     * @return an integer
     */
    public int findPosition(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0, end = nums.length - 1;
        // 要点1: start + 1 < end
        while (start + 1 < end) {
	    // 要点2：start + (end - start) / 2
            int mid = start + (end - start) / 2;
            // 要点3：=, <, > 分开讨论，mid 不+1也不-1
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        // 要点4: 循环结束后，单独处理start和end
        if (nums[start] == target) {
            return start;
        }
        if (nums[end] == target) {
            return end;
        }
        return -1;
    }
}

Code

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