LeetCode 1752. Check if Array Is Sorted and Rotated Solution in Java, C++, Python & More | Explanation + Code

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 B[i] == A[(i+x) % A.length] for every valid index i.

 

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 = 2 positions to begin on 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

Solution 1

PythonJavaC++GoTypeScriptRustC

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

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; } }(code-box)

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; } };(code-box)

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

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; }(code-box)

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 } }(code-box)

bool check(int* nums, int numsSize) { int count = 0; for (int i = 0; i < numsSize; i++) { if (nums[i] > nums[(i + 1) % numsSize]) { count++; } } return count <= 1; }(code-box)