Question

Formatted question description: https://leetcode.ca/all/278.html

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

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

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;
    }
}

############

/* 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 left = 1, right = n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// 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 an integer
# def isBadVersion(version):


class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        left, right = 1, n
        while left < right:
            mid = (left + right) >> 1
            if isBadVersion(mid):
                right = mid
            else:
                left = mid + 1
        return left

############

# 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

/**
 * Forward declaration of isBadVersion API.
 * @param   version   your guess about first bad version
 * @return 	 	      true if current version is bad
 *			          false if current version is good
 * func isBadVersion(version int) bool;
 */

func firstBadVersion(n int) int {
	left, right := 1, n
	for left < right {
		mid := (left + right) >> 1
		if isBadVersion(mid) {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

/**
 * Definition for isBadVersion()
 *
 * @param {integer} version number
 * @return {boolean} whether the version is bad
 * isBadVersion = function(version) {
 *     ...
 * };
 */

/**
 * @param {function} isBadVersion()
 * @return {function}
 */
var solution = function (isBadVersion) {
    /**
     * @param {integer} n Total versions
     * @return {integer} The first bad version
     */
    return function (n) {
        let left = 1;
        let right = n;
        while (left < right) {
            const mid = (left + right) >>> 1;
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    };
};

// The API isBadVersion is defined for you.
// isBadVersion(version:i32)-> bool;
// to call it use self.isBadVersion(version)

impl Solution {
    pub fn first_bad_version(&self, n: i32) -> i32 {
        let mut left = 1;
        let mut right = n;
        while left < right {
            let mid = left + (right - left) / 2;
            if self.isBadVersion(mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        left
    }
}

278 - First Bad Version

Question

Algorithm

Code

All Problems

All Solutions

Welcome to Subscribe On Youtube

Question

Algorithm

Code

All Problems

All Solutions