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:
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 mid = \left\lfloor l + r⁄2 \right\rfloor, then call the isBadVersion(mid) API. If it returns true, it means the first bad version is between [l, mid], so we set r = mid; otherwise, the first bad version is between [mid + 1, r], so we set l = mid + 1.
Finally, we return l.
The time complexity is O(log n), and the space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScript
# 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(code-box)
/* 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;
}
}(code-box)
// 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;
}
};(code-box)
/**
* 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
}(code-box)
/**
* 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;
};
};(code-box)
// 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
}
}(code-box)
/**
* 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;
};
};(code-box)
