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
Solution:
class Solution {
public boolean check(int[] n) {
int l= n.length; // Calculating length of given array.
int nRotation=0;// Declaring a variable to keep a count of number of rotations.
//Running a for loop to identify any rotation in given array.
for(int i=0;i<l;i++){
if(n[i]>n[(i+1)%l]){
nRotation++;
}
}
//If number of rotations are greater than 1 return false.
if(nRotation>1)
return false;
//Else return True.
return true;
}
}
Top comments (0)