README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0200-0299/0278.First%20Bad%20Version/README_EN.md	Binary Search Interactive

278. First Bad Version

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Description

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

Solutions

Solution 1: Binary Search

We define the left boundary of the binary search as $l = 1$ and the right boundary as $r = n$.

While $l < r$, we calculate the middle position $\textit{mid} = \left\lfloor \frac{l + r}{2} \right\rfloor$, then call the isBadVersion(mid) API. If it returns $\textit{true}$, it means the first bad version is between $[l, \textit{mid}]$, so we set $r = \textit{mid}$; otherwise, the first bad version is between $[\textit{mid} + 1, r]$, so we set $l = \textit{mid} + 1$.

Finally, we return $l$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$.

Python3

# 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) >> 1
            if isBadVersion(mid):
                r = mid
            else:
                l = mid + 1
        return l

Java

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

C++

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

class Solution {
public:
    int firstBadVersion(int n) {
        int l = 1, r = n;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

Go

/**
 * Forward declaration of isBadVersion API.
 * @param   version   your guess about first bad version
 * @return 	 	      true if current version is bad
 *			          false if current version is good
 * func isBadVersion(version int) bool;
 */

func firstBadVersion(n int) int {
	l, r := 1, n
	for l < r {
		mid := (l + r) >> 1
		if isBadVersion(mid) {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

TypeScript

/**
 * The knows API is defined in the parent class Relation.
 * isBadVersion(version: number): boolean {
 *     ...
 * };
 */

var solution = function (isBadVersion: any) {
    return function (n: number): number {
        let [l, r] = [1, n];
        while (l < r) {
            const mid = (l + r) >>> 1;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
};

Rust

// The API isBadVersion is defined for you.
// isBadVersion(version:i32)-> bool;
// to call it use self.isBadVersion(version)

impl Solution {
    pub fn first_bad_version(&self, n: i32) -> i32 {
		let (mut l, mut r) = (1, n);
        while l < r {
            let mid = l + (r - l) / 2;
            if self.isBadVersion(mid) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l
    }
}

JavaScript

/**
 * 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, r] = [1, n];
        while (l < r) {
            const mid = (l + r) >>> 1;
            if (isBadVersion(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
};