Introduction

In this blog post, we will explore how to solve the LeetCode problem "81" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values). Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4]. Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums. You must decrease the overall operation steps as much as possible. 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 Constraints: 1 <= nums.length <= 5000 -104 <= nums[i] <= 104 nums is guaranteed to be rotated at some pivot. -104 <= target <= 104 Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• Key Approach:
  • Adapt binary search to handle the rotated array. The core idea is to determine which half of the array (left or right) is sorted and then check if the target falls within that sorted half.
  • Handle duplicates carefully. Duplicates can make it difficult to determine which half is sorted. When nums[mid] == nums[low], increment low until nums[mid] != nums[low] to proceed with the binary search correctly.
• Complexity:
  • Runtime: O(log n) in the average and best cases, O(n) in the worst case (when many duplicates exist). Storage: O(1).

Code

    def search(nums: list[int], target: int) -> bool:
    """
    Searches for a target in a rotated sorted array (with duplicates).

    Args:
        nums: The rotated sorted array.
        target: The target value to search for.

    Returns:
        True if the target is found, False otherwise.
    """
    low = 0
    high = len(nums) - 1

    while low <= high:
        mid = (low + high) // 2

        if nums[mid] == target:
            return True

        # Handle duplicates: if nums[low] == nums[mid], we can't easily determine
        # which half is sorted, so increment low.
        if nums[low] == nums[mid]:
            low += 1
            continue  #Crucial for handling duplicates

        # Check if the left half is sorted
        if nums[low] <= nums[mid]:
            if nums[low] <= target < nums[mid]:
                high = mid - 1
            else:
                low = mid + 1
        # Otherwise, the right half is sorted
        else:
            if nums[mid] < target <= nums[high]:
                low = mid + 1
            else:
                high = mid - 1

    return False