Check if array is sorted and rotated leetcode 1752 python

okay, let's dive deep into the leetcode problem 1752, "check if array is sorted and rotated," with a comprehensive python tutorial, explanations, code examples, and optimization considerations.

**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 0). otherwise, return `false`.

**example:**

* **input:** `nums = [3, 4, 5, 1, 2]`
* **output:** `true`
* **explanation:** `[1, 2, 3, 4, 5]` is the original sorted array.
you can rotate this sorted array to get `[3, 4, 5, 1, 2]`.

* **input:** `nums = [2, 1, 3, 4]`
* **output:** `false`
* **explanation:** there's no sorted array that can be rotated to get `[2, 1, 3, 4]`.

* **input:** `nums = [1, 2, 3]`
* **output:** `true`
* **explanation:** `[1, 2, 3]` is a sorted array and it can be rotated by 0 positions.

**understanding the problem:**

the key is to recognize that a sorted and rotated array will have at most *one* "break" in the non-decreasing order. a break is a place where `nums[i] nums[i+1]`. if there are more than one such breaks, the array cannot be sorted and rotated.

**algorithm and code:**

here's a python solution with detailed explanations:

**explanation:**

1. **initialization:**
* `n = len(nums)`: gets the length of the input array.
* `breaks = 0`: initializes a counter to track how many times the non-decreasing order is violated.

2. **iteration and break detection:**
* `for i in range(n)`: iterates through the array elements.
* `nums[i] nums[(i + 1) % n]`: this is the core comparison.
* `nums[i]` is the current element.
* `nums[(i + 1) % n]` is the *next* element, but it uses the modulo operator (`%`) to handle the "rotation" aspect. wh ...

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check array sorted rotated
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