Leetcode #278: First Bad Version

In this guide, we solve Leetcode #278 First Bad Version in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

Difficulty: Easy
Premium: No
Tags: Binary Search, Interactive

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

Define the search bounds.
Check the mid point condition.
Narrow the bounds until convergence.

Example

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.

Python Solution

# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:


class Solution:
    def firstBadVersion(self, n: int) -> int:
        l, r = 1, n
        while l < r:
            mid = (l + r) >> 1
            if isBadVersion(mid):
                r = mid
            else:
                l = mid + 1
        return l

Complexity

The time complexity is $O(\log n)$ , and the space complexity is $O(1)$ . The space complexity is $O(1)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:


class Solution:
    def firstBadVersion(self, n: int) -> int:
        l, r = 1, n
        while l < r:
            mid = (l + r) >> 1
            if isBadVersion(mid):
                r = mid
            else:
                l = mid + 1
        return l

Leetcode #278: First Bad Version

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #278: First Bad Version

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview