Easy
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
class Solution {
fun check(nums: IntArray): Boolean {
var checker = 0
for (i in 1 until nums.size) {
if (nums[i - 1] > nums[i]) {
checker++
}
}
// checking if the last element is greater than the first
if (nums[nums.size - 1] > nums[0]) {
checker++
}
return checker <= 1
}
}