Sponsored
Sponsored
This approach uses the concept of binary search to efficiently find the first bad version by minimizing the number of calls made to the given isBadVersion API. By leveraging binary search, we can reduce the problem set size by half with each iteration, therefore achieving O(log n) time complexity.
Time Complexity: O(log n)
Space Complexity: O(1)
1#include <stdbool.h>
2
3// Forward declaration of isBadVersion API.
4bool isBadVersion(int version);
5
6int firstBadVersion(int n) {
7 int left = 1, right = n;
8 while (left < right) {
9 int mid = left + (right - left) / 2;
10 if (isBadVersion(mid)) right = mid;
11 else left = mid + 1;
12 }
13 return left;
14}The code above implements the binary search algorithm. It initializes two pointers, left and right, to perform the search. If isBadVersion(mid) is true, the first bad version must be at mid or to its left, hence right is adjusted. Otherwise, the bad version must be to the right of mid, hence left is adjusted. The loop continues until left meets right, at which point the first bad version is identified.
This approach explores a recursive implementation of the binary search technique to identify the first bad version. Employing recursion allows the continual division of the search problem into smaller subproblems until the solution is found.
Time Complexity: O(log n)
Space Complexity: O(log n) - due to recursion stack
1class
This Python solution embeds a binarySearch helper function to recursively zero in on the first bad version. Each call analyzes the status of the mid-point version, narrowing the searcher's focus until the exact location of the first bad version is discerned.