Algorithm
Problem Name: 278. 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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
// Forward declaration of isBadVersion API.
bool isBadVersion(int version);
int firstBadVersion(int n) {
int left, right, mid;
left = 1; right = n;
while (left < right) {
mid = left + (right - left) / 2;
if (isBadVersion(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return right;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
bool isBadVersion(int version);
class Solution {
public:
int firstBadVersion(int n) {
int lower = 1, upper = n;
while(lower < upper)
if(isBadVersion(lower + (upper - lower)/2)) upper = lower + (upper - lower)/2;
else lower = lower + (upper - lower)/2 + 1;
return lower;
}
};
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
if (n == 1) {
return 1;
}
int start = 1;
int end = n;
int minIdx = Integer.MAX_VALUE;
while (start < = end) {
int mid = start + (end - start) / 2;
if (isBadVersion(mid)) {
minIdx = Math.min(minIdx, mid);
end = mid - 1;
}
else {
start = mid + 1;
}
}
return minIdx;
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const solution = function(isBadVersion) {
return function(n) {
let left = 1;
let right = n;
while (left < right) {
let mid = left + Math.floor( (right - left) / 2 );
if (isBadVersion(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};
};
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def firstBadVersion(self, n):
"""
:type n: int
:rtype: int
"""
l, r = 0, n
while l <= r:
mid = (l + r) // 2
if isBadVersion(mid):
r = mid - 1
else:
l = mid + 1
return r + 1
Input
Output
#6 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0278_FirstBadVersion : VersionControl
{
public int FirstBadVersion(int n)
{
int left = 1;
int right = n;
while (left < right)
{
int mid = left + (right - left) / 2;
if (IsBadVersion(mid))
right = mid;
else
left = mid + 1;
}
return left;
}
}
public class VersionControl
{
public bool IsBadVersion(int version)
{
return version > 3;
}
}
}
Input
Output