First Bad Version LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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First Bad Version 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 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

First Bad Version Solutions

Time: O(log n)
Space: O(1)

C++

bool isBadVersion(int version);

class Solution {
 public:
  int firstBadVersion(int n) {
    int l = 1;
    int r = n;

    while (l < r) {
      const int m = l + (r - l) / 2;
      if (isBadVersion(m))
        r = m;
      else
        l = m + 1;
    }

    return l;
  }
};

Java

 
public class Solution extends VersionControl {
  public int firstBadVersion(int n) {
    int l = 1;
    int r = n;

    while (l < r) {
      final int m = l + (r - l) / 2;
      if (isBadVersion(m))
        r = m;
      else
        l = m + 1;
    }

    return l;
  }
}

Python


class Solution:
  def firstBadVersion(self, n: int) -> int:
    l = 1
    r = n

    while l < r:
      m = (l + r) >> 1
      if isBadVersion(m):
        r = m
      else:
        l = m + 1

    return l

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