README.md

1752. Check if Array Is Sorted and Rotated

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.

Example 4:

Input: nums = [1,1,1]
Output: true
Explanation: [1,1,1] is the original sorted array.
You can rotate any number of positions to make nums.

Example 5:

Input: nums = [2,1]
Output: true
Explanation: [1,2] is the original sorted array.
You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

Solutions (Ruby)

1. Solution

# @param {Integer[]} nums
# @return {Boolean}
def check(nums)
  count = (1...nums.size).count { |i| nums[i] < nums[i - 1] }
  count += 1 if nums[0] < nums[-1]

  count < 2
end

Solutions (Rust)

1. Solution

impl Solution {
    pub fn check(nums: Vec<i32>) -> bool {
        match (1..nums.len()).filter(|&i| nums[i] < nums[i - 1]).count() {
            0 => true,
            1 => nums[0] >= nums[nums.len() - 1],
            _ => false,
        }
    }
}