main.go

// Source: https://leetcode.com/problems/check-if-array-is-sorted-and-rotated
// Title: Check if Array Is Sorted and Rotated
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

// The array is originally sorted if and only if
// there exists only one `i` such that A[i] > A[(i+1) % A.length]
func check(nums []int) bool {
	n := len(nums)
	found := false
	for i := range n {
		if nums[i] > nums[(i+1)%n] {
			if found {
				return false
			} else {
				found = true
			}
		}
	}
	return true
}