0278 - First Bad Version (Easy)

Problem Link

https://leetcode.com/problems/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 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 <= 2^31 - 1

Approach 1: Binary Search

Prerequisite

• Binary Search

C++

Written by @wingkwong

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

class Solution {
public:
    int firstBadVersion(int n) {
        // init possible range
        int l = 1, r = n;
        while (l < r) {
            int m = l + (r - l) / 2;
            // [F,F,..,F,T,T,T,..,T]
            // exclude m
            if (!isBadVersion(m)) l = m + 1;
            // include m
            else r = m;
        }
        return l;
    }
};

JavaScript

Written by @radojicic23

/**
 * Definition for isBadVersion()
 * 
 * @param {integer} version number
 * @return {boolean} whether the version is bad
 * isBadVersion = function(version) {
 *     ...
 * };
 */

/**
 * @param {function} isBadVersion()
 * @return {function}
 */
var solution = function(isBadVersion) {
    /**
     * @param {integer} n Total versions
     * @return {integer} The first bad version
     */
    return function(n) {
        let l = 1;
        let r = n;
        while (l < r) {
            let mid = Math.floor((l + r) / 2);
            if (!isBadVersion(mid)) l = mid + 1;
            else r = mid;
        }
        return l;
    };
};

Python

Written by @radojicic23

# The isBadVersion API is already defined for you.
# def isBadVersion(version: int) -> bool:

class Solution:
    def firstBadVersion(self, n: int) -> int:
        l, r = 1, n
        while l < r:
            mid = (l + r) // 2
            if (isBadVersion(mid)):
                r = mid
            else:
                l = mid + 1
        return l