LeetCode #3208 — MEDIUM

Alternating Groups II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented by colors[i]:

colors[i] == 0 means that tile i is red.
colors[i] == 1 means that tile i is blue.

An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles).

Return the number of alternating groups.

Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.

Example 1:

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

Output: 3

Explanation:

Alternating groups:

Example 2:

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

Output: 2

Explanation:

Alternating groups:

Example 3:

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

Output: 0

Explanation:

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: There is a circle of red and blue tiles. You are given an array of integers colors and an integer k. The color of tile i is represented by colors[i]: colors[i] == 0 means that tile i is red. colors[i] == 1 means that tile i is blue. An alternating group is every k contiguous tiles in the circle with alternating colors (each tile in the group except the first and last one has a different color from its left and right tiles). Return the number of alternating groups. Note that since colors represents a circle, the first and the last tiles are considered to be next to each other.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window

Example 1

[0,1,0,1,0]
3

Example 2

[0,1,0,0,1,0,1]
6

Example 3

[1,1,0,1]
4

Step 02

Core Insight

What unlocks the optimal approach

Try to find a tile that has the same color as its next tile (if it exists).
Then try to find maximal alternating groups by starting a single for loop from that tile.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3208: Alternating Groups II
class Solution {
    public int numberOfAlternatingGroups(int[] colors, int k) {
        int n = colors.length;
        int ans = 0, cnt = 0;
        for (int i = 0; i < n << 1; ++i) {
            if (i > 0 && colors[i % n] == colors[(i - 1) % n]) {
                cnt = 1;
            } else {
                ++cnt;
            }
            ans += i >= n && cnt >= k ? 1 : 0;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3208: Alternating Groups II
func numberOfAlternatingGroups(colors []int, k int) (ans int) {
	n := len(colors)
	cnt := 0
	for i := 0; i < n<<1; i++ {
		if i > 0 && colors[i%n] == colors[(i-1)%n] {
			cnt = 1
		} else {
			cnt++
		}
		if i >= n && cnt >= k {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3208: Alternating Groups II
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

// Accepted solution for LeetCode #3208: Alternating Groups II
/**
 * [3208] Alternating Groups II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn number_of_alternating_groups(colors: Vec<i32>, k: i32) -> i32 {
        let n = colors.len();
        let k = k as usize - 1;

        let mut result = 0;
        let mut count = 0;

        for i in n..(2 * n + k - 1) {
            if colors[i % n] != colors[(i + 1) % n] {
                count += 1;
            } else {
                count = 0;
            }

            if count >= k {
                result += 1;
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3208() {
        assert_eq!(1, Solution::number_of_alternating_groups(vec![0, 0, 1], 3));
        assert_eq!(1, Solution::number_of_alternating_groups(vec![0, 1, 1], 3));
        assert_eq!(
            3,
            Solution::number_of_alternating_groups(vec![0, 1, 0, 1, 0], 3)
        );
        assert_eq!(
            2,
            Solution::number_of_alternating_groups(vec![0, 1, 0, 0, 1, 0, 1], 6)
        );
    }
}

// Accepted solution for LeetCode #3208: Alternating Groups II
function numberOfAlternatingGroups(colors: number[], k: number): number {
    const n = colors.length;
    let [ans, cnt] = [0, 0];
    for (let i = 0; i < n << 1; ++i) {
        if (i && colors[i % n] === colors[(i - 1) % n]) {
            cnt = 1;
        } else {
            ++cnt;
        }
        ans += i >= n && cnt >= k ? 1 : 0;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.