Introduction

In this blog post, we will explore how to solve the LeetCode problem "154" 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

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. Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Explanation

• Binary Search with Handling Duplicates: Employ binary search to efficiently narrow down the search space. The key is to handle duplicate elements effectively, as they can disrupt the standard binary search logic. If nums[mid] == nums[left], increment left to shrink the search space.
  • Identify Sorted Subarrays: In each iteration, determine whether the left or right subarray is sorted. If the left part is sorted, the minimum element must be in the right part, and vice-versa.
  • Update Search Space: Based on the sorted subarrays, adjust the left and right pointers to continue the binary search in the appropriate direction, ensuring that the minimum element is always within the search space.

• Runtime Complexity: O(log n) in the average case, O(n) in the worst case (when most elements are duplicates). Storage Complexity: O(1).

Code

    def findMin(nums):
    """
    Finds the minimum element in a rotated sorted array that may contain duplicates.

    Args:
        nums: A list of integers representing the rotated sorted array.

    Returns:
        The minimum element in the array.
    """

    left, right = 0, len(nums) - 1

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

        if nums[mid] > nums[right]:
            left = mid + 1
        elif nums[mid] < nums[right]:
            right = mid
        else:
            # Handle duplicate case: nums[mid] == nums[right]
            right -= 1

    return nums[left]