First Bad Version | Leetcode | Solution

Companies:

Hi, In this post we will be discussing the first bad version problem present on LeetCode

Video Solution

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. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

Sample Test Cases

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.

Problem Solution

Brute Force Approach

To solve this problem using brute force method, we can just start iterating from version 1 to version N and stop as soon as we find a version for which the function isBadVersion returns true. In the worst case, this method will run from 1 to N and will have a worst case time complexity of \(O(N)\)

Optimized Approach

To optimize the above brute force solution, we should make an observation:

If for any number X, isBadVersion(X) = True then X >= ans
If for any number X, isBadVersion(X) = False then X < ans

This is true, because according to the problem description isBadVersion function gives true for any version >= ans and gives false for any version < ans

Therefore, here we can use binary search approach to solve the problem as shown in the solution below :

C++

class Solution {
public:
    int firstBadVersion(long long n) {
      long long ans = n;
      long long low = 1, hi = n;
      while (low <= hi) {
        long long mid = (low + hi)/2;
        if (isBadVersion(mid) == true) {
          ans = min(ans, mid);
          hi = mid - 1;
        }
        else {
          low = mid + 1;
        }
      }
      return (int) ans;
    }
};