Find Minimum in Rotated Sorted Array II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

C++ Find Minimum in Rotated Sorted Array II LeetCode Solution

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

Java Find Minimum in Rotated Sorted Array II LeetCode Solution

public int findMin(int[] nums) {
	 int l = 0, r = nums.length-1;
	 while (l < r) {
		 int mid = (l + r) / 2;
		 if (nums[mid] < nums[r]) {
			 r = mid;
		 } else if (nums[mid] > nums[r]){
			 l = mid + 1;
		 } else {  
			 r--;  //nums[mid]=nums[r] no idea, but we can eliminate nums[r];
		 }
	 }
	 return nums[l];
}
 

Python 3 Find Minimum in Rotated Sorted Array II LeetCode Solution

class Solution:
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return min(nums)

Array-1180
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