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Leetcode 278. First Bad Version

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 returns whether version is bad. Implement a function to find the first bad version.

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.

Function Signature

int firstBadVersion(int n);

Clarifying Questions

  1. What are the constraints on n?
    • Typically, 1 <= n <= 2^31 - 1.
  2. Is there any guarantee that there is at least one bad version?
    • Yes, you can assume there is at least one bad version.
  3. Can we use a binary search approach?
    • Yes, binary search is a common approach to minimize the number of API calls.

Strategy

  1. Initialize the binary search:
    • Use two pointers, left and right, initialized to 1 and n respectively.
  2. Binary Search Loop:
    • While left is less than right:
      • Compute the middle point mid.
      • If isBadVersion(mid) returns true, then the bad version is at mid or before. Move the right pointer to mid.
      • Otherwise, move the left pointer to mid + 1.
  3. Termination:
    • When the loop ends, left will point to the first bad version.

Time Complexity

Code

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int left = 1;
        int right = n;
        
        while (left < right) {
            int mid = left + (right - left) / 2; // To avoid potential overflow
            
            if (isBadVersion(mid)) {
                right = mid; // Potential first bad version
            } else {
                left = mid + 1; // First bad version must be after `mid`
            }
        }
        
        return left; // At the end of the loop, `left` should point to the first bad version.
    }
};

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