How do you solve Alternating Groups II on LeetCode?

Alternating Groups II (LeetCode #3208) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the circular nature of the problem is crucial for finding valid groups.

What is the brute force solution for Alternating Groups II?

The brute force approach for Alternating Groups II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible starting point in the circular array and counts how many groups of size k have alternating colors. This method is straightforward but inefficient due to the nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Alternating Groups II?

Alternating Groups II uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Alternating Groups II?

Always clarify the problem constraints and edge cases before diving into coding. Discuss your thought process and potential approaches with the interviewer. Test your solution with various cases, including edge cases, to ensure its correctness.

What are common mistakes when solving Alternating Groups II?

Failing to account for the circular nature when checking groups. Not properly validating the alternating condition for all tiles in the group.

#3208

Alternating Groups II

Medium

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently count valid alternating groups. By leveraging the circular nature of the array, we can avoid redundant checks and reduce the time complexity significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a new array that simulates the circular nature by duplicating the original array.
2Step 2: Use a sliding window of size k to check for alternating colors in the new array.
3Step 3: Count valid groups as you slide the window across the array.

solution.py12 lines

1# Full working Python code
2
3def countAlternatingGroups(colors, k):
4    n = len(colors)
5    colors = colors + colors[:k-1]  # Duplicate for circular effect
6    count = 0
7    for i in range(n):
8        if i + k - 1 < len(colors):
9            if all(colors[j] != colors[j + 1] for j in range(i, i + k - 1)):
10                count += 1
11    return count
12

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a limited number of times. The space complexity is O(n) due to the duplication of the array for circular behavior.

1Understanding the circular nature of the problem is crucial for finding valid groups.
2Using a sliding window approach can significantly reduce the number of checks needed.

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