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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;
}
}

############

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

############

# 1752. Check if Array Is Sorted and Rotated
# https://leetcode.com/problems/check-if-array-is-sorted-and-rotated

class Solution:
def check(self, nums: List[int]) -> bool:
n = len(nums)
k = 0

for i in range(n):
if nums[i] > nums[(i+1)%n]:
k += 1

if k > 1: return False

return True


• 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;
}
};

• 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
}
}