package com.freetymekiyan.algorithms.level.easy;
/**
* 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.
* <p>
* 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.
* <p>
* You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to
* find the first bad version. You should minimize the number of calls to the API.
* <p>
* Company Tags: Facebook
* Tags: Binary Search
* Similar Problems: (M) Search for a Range, (M) Search Insert Position, (E) Guess Number Higher or Lower
*/
public class FirstBadVersion {
/**
* Binary Search.
* Initial search range:
* n versions from 1 to n.
* Get the middle version m.
* If m is bad, then m+1 to n are all bad. The first bad version can be m or previous versions.
* If m is good, the first bad version can only be larger than m.
* Stop when there is only one version left.
*/
public int firstBadVersion(int n) {
int l = 1;
int r = n;
while (l < r) {
int m = l + (r - l) / 2; // Avoid overflow, since l >= 1, n <= Integer.MAX_VALUE
if (!isBadVersion(m)) {
l = m + 1;
} else {
r = m;
}
}
return l;
}
/**
* Dummy function, just for compilation.
*/
private boolean isBadVersion(int m) {
return m > 1;
}
/**
* Binary Search. Pruning.
* If we know m is a bad version, we can check m - 1 right after that.
* If m - 1 is 1, which means that's the only available good version, m would be the first bad version.
* If m - 1 is good, then m is the first bad version.
*/
public int firstBadVersionB(int n) {
int l = 1;
int r = n;
while (l < r) {
int m = l + (r - l) / 2;
if (!isBadVersion(m)) {
l = m + 1;
} else if (m == 1 || !isBadVersion(m - 1)) {
// Early return. If current version is bad and the previous one is good.
return m;
} else {
r = m;
}
}
return l;
}
}