1752: Check-if-Array-Is-Sorted-and-Rotated
Easy

Simple algorithm; just have to keep track of how many times nums[i] < nums[i-1] and make sure that the number of occurences is less than or equal to 1 (also remember to compare nums[0] < nums[nums.size()-1])

Code:

class Solution {
public:
    bool check(vector<int>& nums) {
        int count = 1;
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i] < nums[i-1]) {
                --count;
                if (count < 0) {
                    return false;
                }
            }
        }
        if (nums[0] < nums[nums.size()-1]) {
            --count;
            if (count < 0) {
                return false;
            }
        }
        return true;
    }
};

Complexity: