#3208

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Alternating Groups II

Solve the problem of counting alternating groups in a circle of tiles using a sliding window approach.

Problem Statement

You are given an array colors representing a circle of red and blue tiles. Each element in colors[i] is either 0 or 1, representing the color of the i-th tile. You are also given an integer k.

An alternating group consists of k consecutive tiles in the circle where each tile alternates in color (except for the first and last tiles, which must differ from their neighboring tiles). Your goal is to return the total number of alternating groups of size k that can be formed from the given tiles.

Examples

Example 1

Input: colors = [0,1,0,1,0], k = 3

Output: 3

Alternating groups:

Example 2

Input: colors = [0,1,0,0,1,0,1], k = 6

Output: 2

Alternating groups:

Example 3

Input: colors = [1,1,0,1], k = 4

Output: 0

Constraints

3 <= colors.length <= 105
0 <= colors[i] <= 1
3 <= k <= colors.length

Solution Approach

Sliding Window with Running State

Use a sliding window of size k to check for alternating groups. Move through the array by updating the state of the window each time. Keep track of the colors of the window's first and last tiles to determine if they alternate, checking each group efficiently.

Efficient Group Counting

By leveraging the sliding window approach, we can count alternating groups without recalculating the entire window each time. Instead, modify the state of the window incrementally, checking the next and previous tiles for alternation, which makes the solution more efficient.

Circle Handling

Since the tiles are arranged in a circle, the solution should correctly handle the transition from the end of the array back to the beginning. This can be done by treating the array as circular or using modular arithmetic to simulate circular behavior.

Complexity Analysis

Metric	Value
Time	O(n + k)
Space	O(1)

The time complexity is O(n + k), where n is the length of the colors array and k is the size of the group. The space complexity is O(1) since only a fixed number of variables are used during computation.

What Interviewers Usually Probe

The candidate should focus on a sliding window approach rather than brute-force checking of all possible groups.
They should demonstrate an understanding of the circular nature of the problem and how to handle the edge cases effectively.
Look for clear, incremental updates to the window state and a focus on efficiency rather than recalculating the entire window on each iteration.

Common Pitfalls or Variants

Common pitfalls

Failing to handle the circular nature of the problem, resulting in incorrect group counting when crossing the array's boundary.
Not updating the window state efficiently, leading to unnecessary recomputations.
Incorrectly assuming that the first and last tiles in the group do not need to alternate in color when checking for groups.

Follow-up variants

Adjusting the group size k dynamically based on input constraints.
Expanding the problem to handle groups with more than two colors.
Optimizing the approach further for arrays with larger sizes, using parallel processing techniques.

FAQ

What is the primary pattern to solve the Alternating Groups II problem?

The primary pattern is using a sliding window with running state updates to efficiently check for alternating groups of size k.

How does the circle of tiles impact the problem solution?

The circular arrangement requires special handling of the end and start of the array, ensuring the window can wrap around correctly.

Can I solve this problem with a brute-force approach?

While it's possible to use brute-force, the sliding window approach is more efficient and should be preferred for larger inputs.

How can GhostInterview help with this problem?

GhostInterview helps by guiding you through the sliding window approach, showing you how to optimize the solution and avoid common pitfalls.

What is the time complexity of the solution for Alternating Groups II?

The time complexity is O(n + k), where n is the number of tiles and k is the size of the alternating group.

terminal

Solution

Solution 1: Single Pass

We can unfold the ring into an array of length $2n$ and then traverse this array from left to right. We use a variable $\textit{cnt}$ to record the current length of the alternating group. If we encounter the same color, we reset $\textit{cnt}$ to $1$; otherwise, we increment $\textit{cnt}$. If $\textit{cnt} \ge k$ and the current position $i$ is greater than or equal to $n$, then we have found an alternating group, and we increment the answer by one.

1

2

3

4

5

6

7

8

9

10

11

class Solution:
    def numberOfAlternatingGroups(self, colors: List[int], k: int) -> int:
        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

#3206 alternating groups i #3191 minimum operations to make binary array elements equal to one i #3254 find the power of k size subarrays i