Follow up for “Search in Rotated Sorted Array”: What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array.

public class Solution {
    /**
     * param A : an integer rotated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean
     */
    public boolean search(int[] A, int target) {
        // write your code here
        int len = A.length;
        if (A == null || len == 0) return false;

        int lo = 0, hi = len - 1;
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            if (A[mid] == target) return true;
            else if (A[lo] < A[mid]) {
                if (A[lo] <= target && target < A[mid]) {
                    hi = mid - 1;
                } else {
                    lo = mid + 1;
                }
            }else if (A[lo] > A[mid]) {
                if (A[mid] < target && target <= A[hi]) {
                    lo = mid + 1;
                } else {
                    hi = mid - 1;
                }
            // if A[start] == A[mid] we can only increment lo by 1 until we have found some element different from A[mid]
            } else {
                lo ++;
                // we can compare A[mid] with A[left] or A[right], both will work. However, if we compare with A[left] then we increment lo, if we compare with A[right], we decrement hi
            }
        }
        return false;
    }
}

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: bool
        """

        if not nums:
            return False

        lo, hi = 0, len(nums) - 1
        while lo <= hi:
            mid = (lo + hi) / 2
            if nums[mid] == target:
                return True
            elif nums[mid] < nums[hi]:
                if (nums[mid] < target and target <= nums[hi]):
                    lo = mid + 1
                else:
                    hi = mid - 1
            elif nums[mid] > nums[hi]:
                if (nums[lo] <= target and target < nums[mid]):
                    hi = mid - 1
                else:
                    lo = mid + 1
            else:
                hi -= 1
        return False