How do you solve Alternating Groups I on LeetCode?

Alternating Groups I (LeetCode #3206) 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: The circular nature of the array means we need to handle the first and last elements as adjacent.

What is the brute force solution for Alternating Groups I?

The brute force approach for Alternating Groups I is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check every possible group of three contiguous tiles in the circle to see if they alternate. This is straightforward but inefficient because we will be checking overlapping groups multiple times. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Alternating Groups I?

Always clarify how to handle edge cases, like circular arrays. Explain your thought process clearly as you work through the problem. Consider both brute-force and optimized approaches, and discuss their trade-offs.

What are common mistakes when solving Alternating Groups I?

Forgetting to consider the circular nature of the array. Incorrectly counting groups by not checking all three tiles properly.

#3206

Alternating Groups I

Easy

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of checking every group of three tiles, we can use a single pass to identify groups of alternating tiles by leveraging the circular nature of the array. This reduces the number of checks significantly.

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Algorithm

4 steps

1Step 1: Initialize a count variable to 0.
2Step 2: Loop through the colors array, treating it as circular by using modulo operations.
3Step 3: For each tile, check if it forms an alternating group with its previous and next tiles.
4Step 4: Increment the count for each valid alternating group found.

solution.py7 lines

1def countAlternatingGroups(colors):
2    n = len(colors)
3    count = 0
4    for i in range(n):
5        if colors[i] != colors[(i - 1) % n] and colors[i] != colors[(i + 1) % n]:
6            count += 1
7    return count

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, checking each tile once. The space complexity is O(1) since we are using a constant amount of extra space.

1The circular nature of the array means we need to handle the first and last elements as adjacent.
2An alternating group is defined by the middle tile differing from its neighbors.

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