How do you solve First Bad Version on LeetCode?

First Bad Version (LeetCode #278) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Using binary search significantly reduces the number of API calls.

What is the brute force solution for First Bad Version?

The brute force approach for First Bad Version is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves checking each version one by one until we find the first bad version. This is straightforward but inefficient, especially for large inputs. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve First Bad Version?

First Bad Version uses the following concepts: Binary Search, Interactive. The recommended patterns to study are: Binary Search, Interactive Problems.

What are the interview tips for First Bad Version?

Always consider the efficiency of your solution, especially when dealing with large inputs. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice coding binary search problems to get comfortable with the pattern.

What are common mistakes when solving First Bad Version?

Not using binary search when applicable, leading to unnecessary API calls. Confusing the conditions for updating left and right pointers.

#278

First Bad Version

Easy

Binary SearchInteractiveBinary SearchInteractive Problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

Using binary search allows us to efficiently narrow down the search space for the first bad version. This is much faster than checking each version sequentially.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, left = 1 and right = n.
2Step 2: While left < right, calculate mid = left + (right - left) // 2.
3Step 3: If isBadVersion(mid) is true, set right = mid (search left side). Otherwise, set left = mid + 1 (search right side).
4Step 4: When left equals right, return left as the first bad version.

solution.py11 lines

1# Full working Python code
2class Solution:
3    def firstBadVersion(self, n: int) -> int:
4        left, right = 1, n
5        while left < right:
6            mid = left + (right - left) // 2
7            if isBadVersion(mid):
8                right = mid
9            else:
10                left = mid + 1
11        return left

ℹ

Complexity note: The time complexity is O(log n) due to the binary search approach, which halves the search space with each iteration. The space complexity remains O(1) as we are using a constant amount of space.

1Using binary search significantly reduces the number of API calls.
2Understanding the relationship between versions helps in efficiently narrowing down the search.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.