In this **Leetcode First Bad Version problem solution**, 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 the version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

## Problem solution in Python.

class Solution: def firstBadVersion(self, n): """ :type n: int :rtype: int """ first = 0 last = n while first <= last: mid = (first+last)//2 if isBadVersion(mid): if not isBadVersion(mid-1): return mid else: last = mid - 1 else: first = mid + 1 return -1

## Problem solution in Java.

public class Solution extends VersionControl { public int firstBadVersion(int n) { if(n==1) return 1; int low = 1; int high = n; while(low<=high){ int mid = low + (high-low)/2; if(isBadVersion(mid)){ // move towards good version high = mid-1; }else{ // if good version // move towards more recent good version low = mid+1; } } return low; } }

## Problem solution in C++.

class Solution { public: int find(int start, int end) { if(start>end) { return -1; } int mid = start+(end-start)/2; bool isBad = isBadVersion(mid); if(isBad && (mid == 0 || !isBadVersion(mid-1))) { return mid; } if(isBad) { return find(start, mid-1); } else { return find(mid+1, end); } } int firstBadVersion(int n) { return find(1, n); } };

## Problem solution in C.

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; }

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