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
public class Solution {
public bool Check(int[] nums) {
bool isRotationFound = false;
bool isSorted = true;
for(int i=1;i<nums.Length;i++){
if(nums[i-1]>nums[i]){
if(isRotationFound){
return false;
}
else{
isRotationFound = true;
}
}
}
if(isRotationFound){
if(nums[nums.Length-1] > nums[0]){
return false;
}
}
return true;
}
}Time Complexity: O(n) Space Complexity: O(1)
