Problem Description

The problem requires us to determine if a given array nums can be the result of rotating a sorted array. A sorted array is considered non-decreasing, meaning that the elements are in ascending order or equal. The array may contain duplicate elements.

A rotation means taking any number of elements from one end of the array and moving them to the other end without changing their order. For example, [3, 4, 5, 1, 2] is a rotated version of [1, 2, 3, 4, 5]. The challenge is to figure out if the nums array could be made from such a rotation of a sorted array. If the array has not been rotated or has been rotated in a way that it still maintains non-decreasing order, our function should return true; otherwise, it should return false.

One important property of a rotated non-decreasing array is that it will have at most one point where the next number is less than the current number, which happens at the rotation point. In a purely sorted array, this does not happen at all.

Intuition

The intuition behind the solution approach is the property of the rotated sorted array mentioned earlier: if we look at each pair of consecutive elements in the array, going from the start to the end, there should be at most one occurrence where a number is greater than the number that follows it.

The method check in the provided Python code implements this logic. It uses list comprehension to compare each pair of elements in the array (handling the edge case of the first element by using modulo indexing), and counts how many times nums[i - 1] > nums[i] occurs. This would equal zero for an unrotated sorted array and exactly one for a correctly rotated sorted array. If this count is greater than one, then there must be at least two decreases, indicating that the array is neither sorted nor a rotated version of a sorted array, so the function will return false. Otherwise, it returns true.

The expression sum(nums[i - 1] > v for i, v in enumerate(nums)) calculates the count. The comparison nums[i - 1] > v evaluates to True (which is interpreted as 1) or False (0). Summing these up gives the total number of times there's a decrease from one element to the next. The condition <= 1 makes sure that there is at most one such decrease, accounting for the rotation point in a rotated array.

Solution Approach

The solution implements a simple but clever algorithm that leverages the nature of a rotated, sorted list to count how many times an element is followed by a smaller element. The crux is that if a list is sorted and then rotated, it will contain at most one such occurrence (the rotation pivot), and if that list had been rotated more than once, there should be two or more such occurrences.

No specific data structures are required for this approach as it works directly on the given list of numbers. The algorithm follows these steps:

1. Initialize a counter to zero. This will count the occurrences where a number is followed by a smaller number.
2. Iterate through the list of numbers using Python's enumerate function, which gives both the index i and the value v at each step.
3. Compare each number with the next one (but since it could wrap around, the previous to the first is the last, hence nums[i - 1] > v). If the previous number is greater, we have found an occurrence and the list comprehension turns it into a 1, otherwise a 0.
4. Sum the values from the list comprehension. This sum represents the number of decreases found in the list.
5. If the sum is greater than 1, return false, since this indicates the list was not just rotated from a sorted list (it was either unsorted before or tampered with after rotation).
6. Otherwise, return true, which implies it is either the original sorted list or a properly rotated version.

Here's how the code works:

class Solution:
    def check(self, nums: List[int]) -> bool:
        # Use list comprehension to sum the occurrences of `nums[i-1] > nums[i]`
        # Enclose this condition in parentheses to get a boolean (True/False) value which translates to (1/0)
        # when summed. This conditional will be True at most once in a rotated array.
        # Use `enumerate` to get both index and value for each element
        # Return True if there are 0 or 1 occurrences of a drop from `nums[i-1]` to `nums[i]`, otherwise False
        return sum(nums[i - 1] > v for i, v in enumerate(nums)) <= 1

The algorithm's complexity is O(n), where n is the number of elements in the list, because it involves one pass over the data. No sorting or nested loops are involved, making it efficient for this use case.

Example Walkthrough

Let's assume we have the following array: [4, 5, 1, 2, 3]. According to our problem description, we need to determine if this could be the result of a sorted array that has been rotated.

Following our solution approach:

1. We initialize our counter to zero. This will keep track of the number of times an element is followed by a smaller element.

2. We iterate through the array [4, 5, 1, 2, 3] using Python’s enumerate function:

   • Comparing 4 (index 0) with 5 (index 1) yields False (since 4 < 5, there is no decrease here).
   • Comparing 5 (index 1) with 1 (index 2) yields True (5 > 1, so we have a decrease).
   • Comparing 1 (index 2) with 2 (index 3) yields False.
   • Comparing 2 (index 3) with 3 (index 4) yields False.
   • Lastly, since the array might be rotated, we also compare the last and the first element: 3 (last) with 4 (first) yields False.

3. We count the number of True values from these comparisons using a list comprehension.

   • We have one True in our example which means the count is 1.

4. The sum of decreases is 1, which is not greater than 1, so according to our rule:

5. We return true, indicating that our array [4, 5, 1, 2, 3] could have been a sorted array that was rotated.

This example illustrates a case where the array is a rotated version of a sorted array as there is only one point of decrease, which aligns with the 'rotated sorted array' property.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n) where n is the length of the input array nums. This is because the code iterates through the list exactly once with the enumerate(nums) function and performs a constant-time operation of comparison and summation in each iteration.

The space complexity of the code is O(1) since it only uses a fixed amount of extra space. The extra space is independent of the input size, it only includes a few variables to keep track of the count of rotations which does not scale with the size of the input list.

Learn more about how to find time and space complexity quickly using problem constraints.