-
Notifications
You must be signed in to change notification settings - Fork 0
/
Check if Array Is Sorted and Rotated.txt
56 lines (37 loc) · 1.28 KB
/
Check if Array Is Sorted and Rotated.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
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:
bool check(vector<int>& nums)
{
int count = 0;
int n = nums.size();
for(int i = 1 ; i < n ; i++)
{
if(nums[i-1] > nums[i])
count++;
}
if(nums[n-1] > nums[0])
count++;
return count <= 1;
}
};