1878. Check If Array Is Sorted And Rotated¶

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1878. Check if Array Is Sorted and Rotated

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

Solution¶

class Solution {
    public boolean check(int[] nums) {
        int n = nums.length, count = 0;
        for (int i = 1; i < n; i++) {
            if (nums[i - 1] > nums[i]) count++;
        }
        if (nums[n - 1] > nums[0]) count++;
        return count <= 1;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here