#3206

Easy

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Alternating Groups I

Count all alternating groups in a circular array by tracking tiles with distinct neighbors efficiently using sliding window updates.

array sliding window

Problem Statement

You are given a circular array colors representing tiles colored either 0 or 1. An alternating group consists of three consecutive tiles where the middle tile has a different color from both neighbors. Compute how many such alternating groups exist in the circular arrangement.

Return an integer representing the total number of alternating groups in colors. The array length ranges from 3 to 100, and each element is either 0 or 1. Evaluate each triplet carefully, considering the circular wrap-around.

Examples

Example 1

Input: colors = [1,1,1]

Output: 0

Example 2

Input: colors = [0,1,0,0,1]

Output: 3

Alternating groups:

Constraints

3 <= colors.length <= 100
0 <= colors[i] <= 1

Solution Approach

Sliding Window Check

Use a sliding window of size three across the array. At each step, verify if the middle tile differs from its previous and next tile. Count it if it forms an alternating group.

Circular Handling

Treat the array as circular by connecting the end back to the start. Use modulo operations to correctly reference neighbors at the boundaries.

Efficient Counting

Instead of recalculating for each window, maintain a running count and only update when the window moves, leveraging previous comparisons to reduce redundant checks.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since each tile is checked once in a sliding window. Space complexity is O(1) as only counters and indices are maintained without extra arrays.

What Interviewers Usually Probe

Can you apply a sliding window instead of nested loops?
How do you handle the circular nature of the array?
Are you updating counts efficiently without redundant recalculations?

Common Pitfalls or Variants

Common pitfalls

Forgetting the circular wrap-around, causing missed groups at array ends.
Incorrectly comparing tiles without considering the middle tile distinction.
Recomputing full windows instead of using running updates, increasing time complexity.

Follow-up variants

Count alternating groups of size 4 instead of 3 in a circular array.
Use more than two colors and identify alternating patterns.
Return indices of all alternating groups instead of just the count.

FAQ

What is an alternating group in Alternating Groups I?

An alternating group is a sequence of three tiles where the middle tile differs from its immediate neighbors in a circular array.

How do I handle the circular array when checking groups?

Use modulo arithmetic to wrap around the array ends so the first and last tiles are considered neighbors.

Can I solve this without a sliding window?

Yes, but a sliding window with running updates reduces redundant checks and keeps time complexity O(n).

What is the time complexity for Alternating Groups I?

The sliding window approach checks each tile once, so the time complexity is O(n) with constant space.

Does GhostInterview detect common pitfalls for this problem?

Yes, it highlights issues like missed circular boundaries, miscounting middle tiles, and redundant recalculations.

terminal

Solution

Solution 1: Single Pass

We set $k = 3$, indicating that the length of the alternating group is $3$.

1

2

3

4

5

6

7

8

9

10

11

12

class Solution:
    def numberOfAlternatingGroups(self, colors: List[int]) -> int:
        k = 3
        n = len(colors)
        ans = cnt = 0
        for i in range(n << 1):
            if i and colors[i % n] == colors[(i - 1) % n]:
                cnt = 1
            else:
                cnt += 1
            ans += i >= n and cnt >= k
        return ans

Continue Practicing

#3208 alternating groups ii #3191 minimum operations to make binary array elements equal to one i #3254 find the power of k size subarrays i