# Solving Leetcode Interviews in Seconds with AI: Check if Array Is Sorted and Rotated


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1752" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false. There may be duplicates in the original array. Note: An array A rotated by x positions results in an array B of the same length such that B[i] == A[(i+x) % A.length] for every valid index i.   Example 1:  Input: nums = [3,4,5,1,2] Output: true Explanation: [1,2,3,4,5] is the original sorted array. You can rotate the array by x = 3 positions to begin on the element of value 3: [3,4,5,1,2].  Example 2:  Input: nums = [2,1,3,4] Output: false Explanation: There is no sorted array once rotated that can make nums.  Example 3:  Input: nums = [1,2,3] Output: true Explanation: [1,2,3] is the original sorted array. You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.                      Constraints:  1 <= nums.length <= 100 1 <= nums[i] <= 100  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Find the Potential Rotation Point:** Iterate through the array to find an index where `nums[i] > nums[i+1]`. This indicates a potential rotation point. If no such point exists, the array is already sorted (or has a single element) and thus a rotated sorted array.
*   **Verify Sorted Order Around Rotation Point:**  Once a potential rotation point is found, check if the array is sorted *before* the rotation point and *after* the rotation point.  Also ensure the last element is <= first element (after the potential rotation).
*   **Handle Edge Cases:**  Consider cases where the array is already sorted (no rotation required) or when the rotation point is at the end of the array.

*   **Time Complexity:** O(n), where n is the length of the input array.
*   **Space Complexity:** O(1)

	
	# Code
	```python
	def check(nums):
    """
    Checks if an array is a rotated sorted array.

    Args:
        nums: A list of integers.

    Returns:
        True if the array is a rotated sorted array, False otherwise.
    """
    n = len(nums)
    k = 0  # Count of decreases indicating rotation
    for i in range(n):
        if nums[i] > nums[(i + 1) % n]:
            k += 1

    return k <= 1
	```