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

• Youtube
• B站

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

public boolean check(int[] nums) {
    if (nums == null || nums.length == 0) {
        return false;
    }
    if (nums.length == 1 || nums.length == 2) {
        return true;
    }

    int decreasingCount = 0;
    for (int i = 1; i < nums.length; i++) {
        if (nums[i] < nums[i - 1]) {
            if (decreasingCount != 0) {
                return false;
            }
            decreasingCount++;
        }
    }
    return decreasingCount == 0 ? true : nums[nums.length - 1] <= nums[0];
} 

 

Time Complexity: O(N) since going through the array once

Space Complexity: O(1) no extra space is used

References

• None