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1752. Check if Array Is Sorted and Rotated

Description

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

Solutions

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

  • class Solution {
public:
    bool check(vector<int>& nums) {
        int cnt = 0;
        for (int i = 0, n = nums.size(); i < n; ++i) {
            cnt += nums[i] > (nums[(i + 1) % n]);
        }
        return cnt <= 1;
    }
};

  • class Solution:
    def check(self, nums: List[int]) -> bool:
        return sum(nums[i - 1] > v for i, v in enumerate(nums)) <= 1


  • func check(nums []int) bool {
	cnt := 0
	for i, v := range nums {
		if v > nums[(i+1)%len(nums)] {
			cnt++
		}
	}
	return cnt <= 1
}

  • function check(nums: number[]): boolean {
    const n = nums.length;
    return nums.reduce((r, v, i) => r + (v > nums[(i + 1) % n] ? 1 : 0), 0) <= 1;
}


  • impl Solution {
    pub fn check(nums: Vec<i32>) -> bool {
        let n = nums.len();
        let mut count = 0;
        for i in 0..n {
            if nums[i] > nums[(i + 1) % n] {
                count += 1;
            }
        }
        count <= 1
    }
}

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