81. Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.

• Would this affect the run-time complexity? How and why?

// Binary Search
bool search(vector<int>& nums, int target) { // time: O(logn); space: O(1)
    int left = 0, right = nums.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return true;
        if (nums[left] < nums[mid]) { // 1st half is in order
            if (nums[left] <= target && target < nums[mid]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else if (nums[left] > nums[mid]) { // 2nd half is in order
            if (nums[mid] < target && target <= nums[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        } else { // handle the case like: [3, 1, 2, 3, 3, 3]
            ++left;
        }
    }
    return false;
}

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        left = 0
        right = len(nums) - 1
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] == target:
                return True
            if nums[left] < nums[mid]:
                if nums[left] <= target and target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            elif nums[left] > nums[mid]:
                if nums[mid] < target and target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
            else:
                left = left + 1
        return False

33. Search in Rotated Sorted Array