# Leetcode – First Bad Version Solution in Python

The First Bad Version is a popular problem on LeetCode, often asked in coding interviews. This article will provide an in-depth guide on how to approach and solve this problem in Python. We will explore various aspects of the problem, understand its context, analyze different strategies to solve it, and evaluate their complexities.

## Introduction

First, let’s understand the problem statement:

Suppose you are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad. You are given an API bool isBadVersion(version) which returns whether version number ‘version’ is bad. Implement a function to find the first bad version, i.e., the version where the downfall started. 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.

## Understanding the Problem

In this problem, we need to find the first bad version from a series of versions numbered from 1 to n. We can only use the given API isBadVersion(version) to check whether a particular version is bad or not. The key point is to minimize the number of calls to this API.

## Naive Approach: Linear Search

The simplest approach would be to start checking from version 1 to n and return the version where isBadVersion returns true for the first time. This approach is known as a Linear Search.

def firstBadVersion(n):
for i in range(1, n+1):
return i

### Time Complexity:

O(n) – We may need to scan through all n elements in the worst case.

### Space Complexity:

O(1) – No additional space is being used.

Though this approach is simple, it is inefficient, especially when the number of versions is large.

## Optimal Approach: Binary Search

Since the versions are sorted and once a version is bad, all subsequent versions are bad as well, we can leverage the Binary Search algorithm to optimize the search process.

1. Initialize two pointers: left at 1 and right at n.
2. While left is less than or equal to right: a. Compute the middle point mid. b. If isBadVersion(mid) is True, then it means the first bad version is to the left of mid (including mid itself). Hence, set right to mid - 1. c. Otherwise, the first bad version must be to the right of mid. Hence, set left to mid + 1.
3. When the loop ends, left will be pointing at the first bad version. Return left.
def firstBadVersion(n):
left, right = 1, n
while left <= right:
mid = left + (right - left) // 2
right = mid - 1
else:
left = mid + 1
return left

### Time Complexity:

O(log n) – As we are dividing the search space by half in every step.

### Space Complexity:

O(1) – No additional space is being used.

## Analysis:

The binary search approach is considerably faster than the naive approach for large n. It significantly reduces the number of calls to isBadVersion, which is particularly useful if such calls are expensive in terms of time or cost.

## Potential Optimization: Exponential Backoff

In scenarios where the first bad version is close to the beginning, an exponential backoff approach can sometimes be more efficient than binary search. The idea is to start with a small range and exponentially increase it until you find a bad version, then perform a binary search within that range.

def firstBadVersion(n):
return 1

# Exponential search to find a bad version
right = 2
while right < n and not isBadVersion(right):
right *= 2

# Binary search in the range [right/2, min(n, right)]
left = right // 2
right = min(n, right)
while left <= right:
mid = left + (right - left) // 2
return left