Naive Solution

The most straightforward approach is to iterate through the entire array and check if the target value exists. This is a brute-force method.

Code (Python)

def search_brute_force(nums, target):
    for num in nums:
        if num == target:
            return True
    return False

Time Complexity

O(n), where n is the number of elements in the array.

Space Complexity

O(1), as we're using constant extra space.

Optimal Solution: Modified Binary Search

Because the array is rotated and initially sorted, we can use a modified binary search to find the target more efficiently. The main challenge arises due to the presence of duplicates, which can make it difficult to determine which half of the array is sorted.

Algorithm

1. Initialize left and right pointers to the start and end of the array, respectively.
2. While left <= right:
   • Calculate the mid index.
   • If nums[mid] == target, return True.
   • Handle the duplicate case: If nums[left] == nums[mid], increment left to shrink the search space. Duplicates obscure the sorted nature, so we must linearly advance the left bound until the sorted nature is revealed.
   • Determine if the left half is sorted: If nums[left] <= nums[mid]:
     • If the target lies within the sorted left half (nums[left] <= target < nums[mid]), update right = mid - 1.
     • Otherwise, update left = mid + 1.
   • Otherwise, the right half is sorted (or unsorted because of duplicates):
     • If the target lies within the sorted right half (nums[mid] < target <= nums[right]), update left = mid + 1.
     • Otherwise, update right = mid - 1.
3. If the target is not found, return False.

Code (Python)

def search_rotated_sorted_array(nums, target):
    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2

        if nums[mid] == target:
            return True

        if nums[left] == nums[mid]:
            left += 1  # Handle duplicate case
            continue

        if nums[left] <= nums[mid]:  # Left half is sorted
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:  # Right half is sorted
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1

    return False

Time Complexity

In the worst-case scenario, where there are many duplicates (e.g., [1,1,1,1,1,1,1,1,1,1,1,1,1,2]), the algorithm might degenerate to O(n) because the nums[left] == nums[mid] condition forces the left pointer to increment linearly. In the best and average cases (where there are fewer duplicates), the time complexity will be closer to O(log n).

Space Complexity

O(1), as we're using constant extra space for pointers.

Edge Cases

• Empty array: The code handles empty arrays gracefully as the while loop will not execute.
• Single-element array: The code works correctly for single-element arrays.
• Target at the beginning or end: The code handles targets at the beginning or end of the array.
• Array with all duplicates: The code handles arrays with all duplicate elements, albeit with O(n) time complexity.