notes/OJ notes/pages/Leetcode First-Bad-Version.md

Leetcode First-Bad-Version

2022-07-09 09:52

Algorithms:

#algorithm #binary_search

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#binary_search

Links:

• Link to problem

---

Problem

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.

Examples

Example 1:

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.

Example 2:

Input: n = 1, bad = 1 Output: 1

Constraints

• 1 <= bad <= n <= 231 - 1

Thoughts

[!summary] This is a #binary_search problem

Note that Leetcode First-Bad-Version#Constraints, n can be 2**31, which means there might be integer overflow.

To address that, according to Binary Search Algorithm#How to implement Binary search, use mid = l + (r - l) / 2

In my first iteration, I use a first variable to keep track of the first bad version.

Later I realized that by the definition of Bi-search, left boundary will converge to the first one.

Solution

Second version, 0ms

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
  int firstBadVersion(int n) {
    // variant of BS
    // 1-indexed
    
    int r = n;
    int l = 1;
    int mid;
    
    do {
      mid = l + (r - l) / 2;
        
      if (isBadVersion(mid)) {
        // Search left
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    } while (l <= r);
      
    return l;
  }
};

First iteration, 4ms

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
  int firstBadVersion(int n) {
    // variant of BS
    // 1-indexed
    
    int r = n;
    int l = 1;
    int mid;
    int first = n;
    
    do {
      mid = l + (r - l) / 2;
        
      if (isBadVersion(mid)) {
        first = min(n, mid);
        // Search left
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    } while (l <= r);
      
    return first;
  }
};