How do you solve Alternating Groups III on LeetCode?

Alternating Groups III (LeetCode #3245) 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 array is crucial.

What is the time complexity of Alternating Groups III?

The optimal solution for Alternating Groups III runs in O(n) time and uses O(n) space. The complexity is O(n) for preprocessing the groups and O(1) for each query, making it efficient for large inputs.

What is the brute force solution for Alternating Groups III?

The brute force approach for Alternating Groups III is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible group of tiles to see if they alternate colors. This is straightforward but inefficient because we might have to check many overlapping groups. The optimal Optimal Solution approach improves this to O(n).

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

Alternating Groups III uses the following concepts: Array, Binary Indexed Tree, Ordered Set. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Alternating Groups III?

Clarify the circular nature of the problem with the interviewer. Discuss potential optimizations early on to show your thought process. Practice edge cases, like all tiles being the same color.

What are common mistakes when solving Alternating Groups III?

Not considering the circular nature of the array when checking groups. Failing to update the group sizes after a color change.

#3245

Alternating Groups III

Hard

ArrayBinary Indexed TreeOrdered SetHash MapArray

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)

We can preprocess the colors to find maximal alternating groups and store their sizes. This allows us to quickly answer queries about group sizes without rechecking the entire array.

⚙️

Algorithm

3 steps

1Step 1: Preprocess the colors array to find all maximal alternating groups and their sizes.
2Step 2: Store the sizes of these groups in a frequency map for quick lookup.
3Step 3: For type 1 queries, simply return the count from the frequency map. For type 2 queries, update the colors and adjust the groups as necessary.

solution.py26 lines

1from collections import defaultdict
2
3def count_alternating_groups(colors, queries):
4    n = len(colors)
5    group_sizes = defaultdict(int)
6    def update_groups():
7        count = 0
8        size = 1
9        for i in range(n):
10            if colors[i] != colors[(i + 1) % n]:
11                size += 1
12            else:
13                if size > 1:
14                    group_sizes[size] += 1
15                size = 1
16        if size > 1:
17            group_sizes[size] += 1
18    update_groups()
19    results = []
20    for query in queries:
21        if query[0] == 1:
22            results.append(group_sizes[query[1]])
23        elif query[0] == 2:
24            colors[query[1]] = query[2]
25            update_groups()
26    return results

ℹ

Complexity note: The complexity is O(n) for preprocessing the groups and O(1) for each query, making it efficient for large inputs.

1Understanding the circular nature of the array is crucial.
2Efficiently managing updates to the colors is key to optimizing the solution.

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