Introduction

In this blog post, we will explore how to solve the LeetCode problem "33" 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 ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= 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, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2]. Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 Example 2: Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1 Example 3: Input: nums = [1], target = 0 Output: -1 Constraints: 1 <= nums.length <= 5000 -104 <= nums[i] <= 104 All values of nums are unique. nums is an ascending array that is possibly rotated. -104 <= target <= 104

Explanation

Here's the solution to the rotated sorted array search problem, along with explanations and code:

• High-Level Approach:

  • Utilize Binary Search: The core idea is to adapt binary search to work on the rotated sorted array.
  • Identify Sorted Portion: In each iteration, determine which half of the array (left or right) is sorted.
  • Search in Sorted Portion: Check if the target lies within the sorted portion. If so, narrow down the search to that portion; otherwise, search in the other half.

• Complexity:

  • Runtime: O(log n), Storage: O(1)

Code

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

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

    Returns:
        The index of the target if found, otherwise -1.
    """
    left, right = 0, len(nums) - 1

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

        if nums[mid] == target:
            return mid

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

    return -1