First Bad Version

Finding the first bad version among the ( n ) versions is a classic binary search problem, where we have to identify the exact point where the versions start to become bad.

Since we want to minimize the number of calls to the API, we can use a binary search to identify the boundary between the good and bad versions. Here’s the Python code to achieve this:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

class Solution:
    def firstBadVersion(self, n: int) -> int:
        left, right = 1, n

        while left < right:
            mid = left + (right - left) // 2
            if isBadVersion(mid):
                right = mid # If the middle version is bad, the first bad version must be at or to the left of mid
            else:
                left = mid + 1 # If the middle version is good, the first bad version must be to the right of mid

        return left

Explanation

  • We initialize left and right pointers to cover the entire range of versions.
  • We perform binary search by dividing the current range in half at each step.
  • If the mid version is bad, the first bad version must be at or to the left of mid, so we set right = mid.
  • If the mid version is good, the first bad version must be to the right of mid, so we set left = mid + 1.
  • When left and right converge, left will be at the first bad version.

Key Takeaways

  • Binary search is a perfect choice for this problem since it minimizes the number of calls to the API and finds the boundary efficiently.
  • The time complexity of this algorithm is (O(\log n)), where (n) is the number of versions.
  • The space complexity is (O(1)).
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution:
    def firstBadVersion(self, n: int) -> int:
        i = 1
        j = n
        while (i < j):
            pivot = (i+j) // 2
            if (isBadVersion(pivot)):
                j = pivot
            else:
                i = pivot + 1
        return i

Problem Classification

This problem can be classified in the following categories:

  1. Binary Search: Since the task is to find the first bad version with minimum API calls, and given that all versions after a bad one are also bad, this indicates a clear binary search pattern. With binary search, we divide our search space in half with each comparison, which reduces the number of calls we make to the API.

  2. Version Control: This is related to version control since it’s about finding the first bad version of a product. The principle is similar to the versioning of software in development processes where different versions are maintained and checked for bugs or problems.

  3. Error/Exception Handling and Debugging: Since the problem involves identifying a ‘bad version’ that could be analogous to a software bug or error, it can also fall under this category. It’s about tracing the point of failure in a sequence of versions or operations.

  4. Software Testing: The problem deals with the concept of checking different versions of a product for a failure or ‘bad’ condition. This is similar to different types of software testing methodologies where individual units or components of a software product are tested to ensure they are working correctly.

Language Agnostic Coding Drills

  1. Understanding Binary Search: The core of this problem is understanding how to use binary search to quickly narrow down the potential solutions. The concept of binary search can be practiced with an array of sorted numbers where one has to find a specific number. This technique can be extended to this problem where instead of searching for a specific number, we’re searching for a version where the property of being bad changes.

  2. Understanding APIs and Mock Functions: In the problem, a function isBadVersion() is provided which acts as an API. This function can return True or False based on the input. A simple drill can involve creating a mock function that returns True or False based on some condition and then using this function in code.

  3. Handling Conditions with Binary Search: This drill involves understanding how to adapt the binary search process based on conditions. In the case of this problem, if the pivot version is bad, then all future versions will also be bad, so the search space should be reduced to the left (i.e., versions less than the pivot). Otherwise, if the pivot version is not bad, all previous versions will also not be bad, so the search space should be reduced to the right (i.e., versions greater than the pivot). This can be practiced by modifying the condition in a simple binary search problem.

  4. Loop Structure and Termination Conditions: This problem involves a while loop where the conditions for the loop are not as straightforward as iterating over an array or a fixed number of iterations. This can be practiced by creating problems that involve custom loop conditions and ensuring that the loop terminates correctly.

Approach:

The approach to solve this problem is to implement a binary search algorithm where the midpoint is checked for being a bad version. If the midpoint is a bad version, then the search space is reduced to everything before the midpoint. If the midpoint is not a bad version, then the search space is reduced to everything after the midpoint. This process continues until the first bad version is found.

This involves:

  1. Initializing two pointers at the start and end of the versions list (1 and n in this case).

  2. Calculate the midpoint and check if it’s a bad version using the given API function.

  3. If the midpoint is a bad version, move the end pointer to the midpoint.

  4. If the midpoint is not a bad version, move the start pointer to one position after the midpoint.

  5. Repeat the process until the start pointer is not less than the end pointer.

  6. Return the start pointer, which will be the first bad version.

Each of the drills listed above contributes to understanding a part of this problem-solving approach. By mastering each drill, you will be well-equipped to tackle this problem.

Targeted Drills in Python

  1. Understanding Binary Search:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

# Drill: Implement a function that uses binary search to find a number in a sorted list

def binary_search(numbers, target):
    left, right = 0, len(numbers) - 1
    
    while left <= right:
        mid = (left + right) // 2
        if numbers[mid] == target:
            return mid
        elif numbers[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
            
    return -1 # not found
  1. Understanding APIs and Mock Functions:
1
2
3
4
5
6
7
8
9

# Drill: Implement a mock function that checks if a number is even (instead of a version being bad), and use it in a function

def is_even(n):
    return n % 2 == 0

def print_even(numbers):
    for num in numbers:
        if is_even(num):
            print(num)
  1. Handling Conditions with Binary Search:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

# Drill: Modify the binary search function to find the first number in a sorted array that is greater than a given target

def binary_search_first_greater(numbers, target):
    left, right = 0, len(numbers) - 1
    
    while left < right:
        mid = (left + right) // 2
        if numbers[mid] <= target:
            left = mid + 1
        else:
            right = mid
            
    if numbers[left] > target:
        return left
    return -1 # no number greater than target
  1. Loop Structure and Termination Conditions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
10

# Drill: Implement a function that iterates over an array until it finds a number that is not even

def find_first_odd(numbers):
    i = 0
    while i < len(numbers) and is_even(numbers[i]):
        i += 1
        
    if i < len(numbers):
        return numbers[i]
    return -1 # no odd number found

Each of these drills can be combined into the final solution, where you implement a binary search over the versions, use a provided API to check if a version is bad, handle the conditions based on whether the midpoint version is bad or not, and ensure that the loop terminates correctly to find the first bad version.