Monday, February 15, 2021

Leetcode solution 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.

Example 4:

Input: nums = [1,1,1]
Output: true
Explanation: [1,1,1] is the original sorted array.
You can rotate any number of positions to make nums.

Example 5:

Input: nums = [2,1]
Output: true
Explanation: [1,2] is the original sorted array.
You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 100


 Problem link

Video Tutorial

You can find the detailed video tutorial here

Thought Process

Simple problem, just need to find a pattern
  • If always non-decreasing, true
  • If it has one decreasing pivot, the last element needs to be less or equal than the first element
  • If it has more than one decreasing pivot, false

Solutions


Time Complexity: O(N) since going through the array once
Space Complexity: O(1) no extra space is used

References

  • None