Find Minimum in Rotated Sorted Array II


Problem Statement :


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.

 

Example 1:

Input: nums = [1,3,5]
Output: 1
Example 2:

Input: nums = [2,2,2,0,1]
Output: 0
 

Constraints:

n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
nums is sorted and rotated between 1 and n times.


Solution :



title-img 


                            Solution in C :

int findMin(int* nums, int numsSize){
    int l=0, r=numsSize-1, mid;
    
    while (l<r){
        mid = l + (r-l)/2;
        if (nums[mid] < nums[r]) r=mid;
        else if (nums[mid] > nums[r]) l=mid+1;
        else --r;
    }
    
    return nums[l];
}
                        


                        Solution in C++ :

class Solution {
public:
    int findMin(vector<int>& nums) {
        int low = 0, high = nums.size() - 1;
        while(low < high){
            int mid = low + (high - low) / 2;
            if(nums[mid] == nums[high]) high--;
            else if(nums[high] > nums[mid]) high = mid;
            else low = mid + 1;
        }
        return nums[low];
        
    }
};
                    


                        Solution in Java :

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


                        Solution in Python : 
                            
def findMin(self, nums: List[int]) -> int:
    l,r=0,len(nums)-1
    while l<r:
        mid=(l+r)//2
        if nums[mid]>nums[r]:
            l=mid+1
        elif nums[mid]<nums[r]:
            r=mid
        else:
            r-=1
    return nums[l]


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