How do you solve Check if Array Is Sorted and Rotated on LeetCode?

Check if Array Is Sorted and Rotated (LeetCode #1752) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: An array can only have one drop point to be considered a rotated sorted array.

What is the time complexity of Check if Array Is Sorted and Rotated?

The optimal solution for Check if Array Is Sorted and Rotated runs in O(n) time and uses O(1) space. We only loop through the array once, making it O(n) time complexity, and we use a constant amount of space.

What is the brute force solution for Check if Array Is Sorted and Rotated?

The brute force approach for Check if Array Is Sorted and Rotated is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible rotation of the array to see if it is sorted. This is simple but inefficient because we may have to check every rotation one by one. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if Array Is Sorted and Rotated?

Check if Array Is Sorted and Rotated uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Check if Array Is Sorted and Rotated?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, especially with sorted arrays and duplicates. Explain your thought process clearly while coding; it shows your understanding.

What are common mistakes when solving Check if Array Is Sorted and Rotated?

Not considering edge cases with duplicates. Assuming the array is sorted without checking all elements.

#1752

Check if Array Is Sorted and Rotated

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking every rotation, we can find the number of times the array 'drops' from a higher to a lower value. If it drops more than once, it can't be a rotated sorted array.

⚙️

Algorithm

3 steps

1Step 1: Initialize a count variable to track the number of drops.
2Step 2: Loop through the array and compare each element with the next one, counting how many times the current element is greater than the next.
3Step 3: If the count is more than 1, return false. Otherwise, return true.

solution.py7 lines

1def check(nums):
2    count = 0
3    n = len(nums)
4    for i in range(n):
5        if nums[i] > nums[(i + 1) % n]:
6            count += 1
7    return count <= 1

ℹ

Complexity note: We only loop through the array once, making it O(n) time complexity, and we use a constant amount of space.

1An array can only have one drop point to be considered a rotated sorted array.
2Duplicates do not affect the sorted nature but can complicate the drop counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.