Suppose in a company, one product manager is leading a team who develops a new product. Suppose latest version fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version will be bad. So we have an array A with n elements [1, 2, … n] and we have to find the first bad version from this array.
Consider we have a function isBadVersion(version_id), this will return whether the version is bad or not. For an example, suppose n = 5, and version = 4 is first bad version. So if the isBadVersion(3) returns false isBadVersion(5) returns true, and isBadVersion(4) also returns true, then the first bad version is 4
To solve this, we will follow these steps −
Let us see the following implementation to get better understanding −
first_bad = 0 def isBadVersion(version): if version >= first_bad: return True return False class Solution: def firstBadVersion(self, n): if n <2: return n start = 1 end = n while(start<=end): mid = (start+end)//2 if isBadVersion(mid) and not isBadVersion(mid-1): return mid elif isBadVersion(mid-1): end = mid-1 else: start = mid+1 ob1 = Solution() first_bad = 4 op = ob1.firstBadVersion(5) print(op)