In this **Leetcode Find Minimum in Rotated Sorted Array II problem solution** Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

- [4,5,6,7,0,1,4] if it was rotated 4 times.
- [0,1,4,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array. You must decrease the overall operation steps as much as possible.

## Problem solution in Python.

class Solution: def findMin(self, nums: List[int]) -> int: i=0 if not nums:return None while(i<len(nums)-1 and nums[i+1]>=nums[i]): i+=1 if i==len(nums)-1:i=-1 return nums[i+1]

## Problem solution in Java.

int low = 0, high = nums.length-1; while(low < high) { int mid = (low + high) / 2; if(nums[low] > nums[mid]) { high = mid; }else if(nums[mid] > nums[high]) { low = mid + 1; }else if(nums[low] < nums[mid]){ return nums[low]; }else if(nums[mid] < nums[high]) { return nums[mid]; }else { low++; } } return nums[low];

## Problem solution in C++.

int findMin(vector<int>& nums) { int left = 0, right = nums.size()-1; while (left < right) { int mid = left+(right-left)/2; if (nums[left] > nums[mid]){ left++; right=mid; } else if (nums[mid] > nums[right]){ left=mid+1; } else { if(nums[left] < nums[right]) return nums[left]; else right--; } } return nums[left]; }

## Problem solution in C.

int findMin(int* nums, int numsSize) { if (numsSize == 1 || nums[0] < nums[numsSize-1]) { return nums[0]; } for (int i = numsSize - 1; i > 0; i--) { if (nums[i] < nums[i-1]) { return nums[i]; } } }

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