1752. Check if Array Is Sorted and Rotated

Problem Statement

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

Intuition

Intuition - 

Only one cut point 

3 4 5 1 2

Only one possible point where 5>1 not else 

and also index[0] > index[last] eg 3 >2 else not possible


        // check for the cut in the array eg
        // 3 4 5 1 2

        // 5 to 1  incr count 

        // also second check for such testcase

        // 2 1 3 4

        // Here count is incr but last should be smaller than first 
        // to filter such , below statement

Links

https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/

Video Links

Approach 1:

Iterate

C++

class Solution {
public:
    bool check(vector<int>& nums) {
        int count = 0;

        for(int i=1; i<nums.size(); i++){
            if(nums[i] < nums[i-1])
                count++;
        }
        if(nums[0] < nums[nums.size()-1])
            count++;

        return count<=1;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++