< Back 278. First Bad Version We're given an API called isBadVersion() -> bool, and we're asked to find the first bad version from a range of n versions. Every version after the first bad version, from versions[bad:n + 1] will return True. We use binary search and decrease our search space to the left if the version isn't bad, and vice versa if the version is bad. Eventually, our left pointer will be at the first bad version. The solution is as follows: class Solution: def firstBadVersion(self, n: int) -> int: l, r = 1, n while l <= r: m = (l + r) // 2 if not isBadVersion(m): l = m + 1 else: r = m - 1 return l _ Time Complexity: O(logn) - Standard binary search time complexity. _ Space Complexity: O(1) - We store a left and right pointer.