3208. Alternating Groups II LeetCode Solution

In this guide, you will get 3208. Alternating Groups II LeetCode Solution with the best time and space complexity. The solution to Alternating Groups II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Alternating Groups II solution in C++
4. Alternating Groups II solution in Java
5. Alternating Groups II solution in Python
6. Additional Resources

Problem Statement of Alternating Groups II

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

3208. Alternating Groups II LeetCode Solution in C++

class Solution {
 public:
  int numberOfAlternatingGroups(vector<int>& colors, int k) {
    const int n = colors.size();
    int ans = 0;
    int alternating = 1;

    for (int i = 0; i < n + k - 2; ++i) {
      alternating =
          colors[i % n] == colors[(i - 1 + n) % n] ? 1 : alternating + 1;
      if (alternating >= k)
        ++ans;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

3208. Alternating Groups II LeetCode Solution in Java

class Solution {
  public int numberOfAlternatingGroups(int[] colors, int k) {
    final int n = colors.length;
    int ans = 0;
    int alternating = 1;

    for (int i = 0; i < n + k - 2; ++i) {
      alternating = colors[i % n] == colors[(i - 1 + n) % n] ? 1 : alternating + 1;
      if (alternating >= k)
        ++ans;
    }

    return ans;
  }
}
// code provided by PROGIEZ

3208. Alternating Groups II LeetCode Solution in Python

class Solution:
  def numberOfAlternatingGroups(self, colors: list[int], k: int) -> int:
    n = len(colors)
    ans = 0
    alternating = 1

    for i in range(n + k - 2):
      alternating = (1 if colors[i % n] == colors[(i - 1) % n]
                     else alternating + 1)
      if alternating >= k:
        ans += 1

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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