Formatted question description: https://leetcode.ca/all/1752.html

# 1752. Check if Array Is Sorted and Rotated

Easy

## 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.

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

## Solution

Loop over nums and find the minimum value. Then store the indices of the minimum value, and check whether one of the indices can be the start index of a sorted array. Skip the indices that are right after another index with the minimum value.

class Solution {
public boolean check(int[] nums) {
int min = Integer.MAX_VALUE;
for (int num : nums)
min = Math.min(min, num);
List<Integer> minIndices = new ArrayList<Integer>();
int length = nums.length;
for (int i = 0; i < length; i++) {
if (nums[i] == min)
}
int size = minIndices.size();
for (int i = 0; i < size; i++) {
if (i > 0 && minIndices.get(i) - minIndices.get(i - 1) == 1)
continue;
int minIndex = minIndices.get(i);
int[] newNums = new int[length];
for (int j = 0; j < length; j++)
newNums[j] = nums[(minIndex + j) % length];
boolean flag = true;
for (int j = 1; j < length; j++) {
if (newNums[j] < newNums[j - 1]) {
flag = false;
break;
}
}
if (flag)
return true;
}
return false;
}
}