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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 <= 1001 <= nums[i] <= 100
Solutions
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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 } }