Solution For First Bad Version

Problem Statement:

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.

Solution:

The problem statement is essentially asking us to perform a binary search on a sorted list of n elements. We know that all versions after a bad version are also bad. Therefore, if version mid is bad, we know that the first bad version must be in the range [1, mid]. Similarly, if version mid is not bad, we know that the first bad version must be in the range [mid + 1, n]. We can use this observation to perform a binary search.

Let’s start by initializing the left and right pointers to the first and last versions, respectively. We’ll use a while loop to keep dividing the range in half until we find the first bad version.

class Solution:
def firstBadVersion(self, n):
left = 1
right = n
while left < right:
mid = left + (right - left) // 2
if isBadVersion(mid):
right = mid
else:
left = mid + 1
return left

Initially, the range is [1, n]. We calculate the midpoint mid as shown above.

If mid is bad, we know that the first bad version must be in the range [1, mid], so we set right = mid. Otherwise, we know that the first bad version must be in the range [mid + 1, n], so we set left = mid + 1.

We repeat this process until the range is reduced to a single version. At this point, we have found the first bad version.

Note that we use the expression left + (right - left) // 2 to calculate the midpoint. This is done to prevent integer overflow. If we use the expression (left + right) // 2 and left + right is too large, we may get an incorrect result.

Step by Step Implementation For First Bad Version

/**
 * public class SVNRepo {
 *     public static boolean isBadVersion(int k);
 * }
 * you can use SVNRepo.isBadVersion(k) to judge whether 
 * the kth code version is bad or not.
*/

public class Solution {
    /**
     * @param n: An integer
     * @return: An integer which is the first bad version.
     */
    public int findFirstBadVersion(int n) {
        // write your code here
        int start = 1, end = n;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (SVNRepo.isBadVersion(mid)) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (SVNRepo.isBadVersion(start)) {
            return start;
        }
        return end;
    }
}

# The isBadVersion API is already defined for you.
# @param version, an integer
# @return a bool
# def isBadVersion(version):

class Solution:
    def firstBadVersion(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 1:
            return 1
        left = 1
        right = n
        while left < right:
            mid = left + (right - left) // 2
            if isBadVersion(mid) == False:
                left = mid + 1
            else:
                right = mid
        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){
            let mid = Math.floor(left + (right - left)/2);
            if(isBadVersion(mid)){
                right = mid;
            }
            else{
                left = mid + 1;
            }
        }
        
        return left;
    };
};

You can use the following pseudo-code to solve the problem:

function isBadVersion(version):
    /*
    Returns true if the version passed in is a bad version, false otherwise.
    */
    
function firstBadVersion(n):
    /*
    Returns the first bad version.
    */
    
    // Set left and right pointers.
    left = 1
    right = n
    
    // Loop until left and right meet.
    while left <= right:
        
        // Get the middle version.
        mid = left + (right - left) // 2
        
        // Check if the middle version is a bad version.
        if isBadVersion(mid):
            
            // If it is, the first bad version is somewhere to the left
            // of the middle version.
            right = mid - 1
            
        else:
            
            // If it's not, the first bad version is somewhere to the right
            // of the middle version.
            left = mid + 1
            
    // At this point, left > right, so the first bad version is the
    // version pointed to by the left pointer.
    return left

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