# 1752. Check if Array Is Sorted and Rotated

## Description

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 A[i] == B[(i+x) % A.length], where % is the modulo operation.

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 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

## Solutions

• class Solution {
public boolean check(int[] nums) {
int cnt = 0;
for (int i = 0, n = nums.length; i < n; ++i) {
if (nums[i] > nums[(i + 1) % n]) {
++cnt;
}
}
return cnt <= 1;
}
}

• class Solution {
public:
bool check(vector<int>& nums) {
int cnt = 0;
for (int i = 0, n = nums.size(); i < n; ++i) {
cnt += nums[i] > (nums[(i + 1) % n]);
}
return cnt <= 1;
}
};

• class Solution:
def check(self, nums: List[int]) -> bool:
return sum(nums[i - 1] > v for i, v in enumerate(nums)) <= 1


• func check(nums []int) bool {
cnt := 0
for i, v := range nums {
if v > nums[(i+1)%len(nums)] {
cnt++
}
}
return cnt <= 1
}

• function check(nums: number[]): boolean {
const n = nums.length;
return nums.reduce((r, v, i) => r + (v > nums[(i + 1) % n] ? 1 : 0), 0) <= 1;
}


• impl Solution {
pub fn check(nums: Vec<i32>) -> bool {
let n = nums.len();
let mut count = 0;
for i in 0..n {
if nums[i] > nums[(i + 1) % n] {
count += 1;
}
}
count <= 1
}
}