First Bad Version

                                            
// 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;
    }
};

                                            
public class Solution extends VersionControl {
    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;
    }
}

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

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

                                            
/* The isBadVersion API is defined in the parent class 
VersionControl.
      bool IsBadVersion(int version); */

public class Solution : VersionControl {
    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;
    }
}

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

                                            
/* The isBadVersion API is defined in the parent 
class VersionControl.
      public function isBadVersion($version){} */
class Solution extends VersionControl {
    /**
     * @param Integer $n
     * @return Integer
     */
    function firstBadVersion($n) {
        $left = 1;
        $right = $n;

        while ($left < $right) {
            $mid = $left + intval(($right - $left) / 2);
            
            if ($this->isBadVersion($mid)) {
                $right = $mid;
            } else {
                $left = $mid + 1;
            }
        }
        
        return $left;
    }
}

                                            
/**
 * The knows API is defined in the parent class VersionControl.
 *     func isBadVersion(_ version: Int) -> Bool{}
 */
 
class Solution : VersionControl {
    func firstBadVersion(_ n: Int) -> Int {
        var left = 1
        var right = n
        
        while left < right {
            let mid = left + (right - left) / 2
            if isBadVersion(mid) {
                right = mid
            } else {
                left = mid + 1
            }
        }
        
        return left
    }
}

                                            
# The is_bad_version API is already defined for you.
# @param {Integer} version
# @return {boolean} whether the version is bad
# def is_bad_version(version):

def first_bad_version(n)
    left = 1
    right = n

    while left < right
        mid = left + (right - left) / 2
        if is_bad_version(mid)
            right = mid
        else
            left = mid + 1
        end
    end

    return left
end

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

To solve this coding challenge, we need to find the first bad version of a product efficiently, given the constraint that a version is bad if and only if all subsequent versions are also bad. This problem is a classic example of utilizing a binary search algorithm to minimize the number of API calls to check whether a version is bad.

Explanation

The problem itself is well-suited for a binary search approach due to the ordered nature of the versions from 1 to n. With binary search, we can halve the search space after each API call, which ensures that we are efficiently narrowing down to the first bad version. To break down the approach step-by-step:

Initialization : Start with two pointers:
left

set to 1 (the first version) and
right

set to
n

(the last version).
Binary Search Loop :

Calculate the middle point:
mid = (left + right) // 2

.
Use the
isBadVersion(mid)

API to check if
mid

is a bad version.
If
mid

is a bad version, it means that the first bad version could be
mid

itself or another version before
mid

. Hence, adjust the
right

pointer to
mid

.
If
mid

is not a bad version, it means that the first bad version must be after
mid

. Therefore, adjust the
left

pointer to
mid + 1

.

Termination : The binary search loop continues until
left

is equal to
right

. This point,
left

(or
right

as both will be equal), will be the first bad version.

Here is the detailed pseudocode to achieve the solution:

                                            
# Initialize the binary search variables
left = 1            # Starting from the first version
right = n           # Ending at the last version

# Perform binary search
while left < right:
mid = left + (right - left) // 2   # Compute the middle point to avoid overflow

if isBadVersion(mid):  # Check if mid version is bad
right = mid       # If mid is bad, the first bad version is at mid or before it
else:
left = mid + 1    # If mid is not bad, the first bad version is after mid

# At the end of the loop, left will be the first bad version
return left

Step-by-Step Explanation

Initialization :

We set two pointers,
left

and
right

, to represent the current search range.
left

starts at 1 (the beginning of the versions), and
right

starts at
n

(the end of the versions).

Binary Search Loop :

Calculate the midpoint to split the search range into two halves.
Use the
isBadVersion(mid)

to determine if the midpoint is a bad version.
If
isBadVersion(mid)

returns
True

, adjust the
right

pointer to
mid

, effectively narrowing the search range to the left half including
mid

.
If
isBadVersion(mid)

returns
False

, adjust the
left

pointer to
mid + 1

, narrowing the search range to the right half excluding
mid

.

Termination :

The loop ends when
left

equals
right

, which ensures that the first bad version is found. This will be the point where
left

and
right

meet.

This approach ensures that we find the first bad version with the minimum number of API calls, in O(log n) time complexity, making it very efficient for large values of

n

. Feel free to ask if you need any additional clarification or further breakdown of any part of the solution!