Alternating Groups II

Problem Statement

Consider a scenario where we have a circular arrangement of tiles, each of which can be either red or blue, represented by 0 for red and 1 for blue in an array colors. Alongside, we are given an integer k, representing the size of a group.

An "alternating" group is defined as k contiguous tiles in the configuration where each consecutive tile (except the first and the last within the group) alternates in color compared to its immediate neighbors. The goal is to determine and return the number of such alternating groups in this circular tile arrangement. It's critical to remember that in this arrangement, the first tile comes immediately after the last tile, forming a closed loop.

Examples

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

Approach and Intuition

To count the number of alternating groups in the circular array of tiles, we should consider the following steps:

1. Interpret the circular nature of the array:

   • This implies that after the last tile of the array, it loops back to the first tile. Hence, for any tile index i, the next tile is (i + 1) % n where n is the number of tiles (length of the colors array).

2. Analyze sections of k contiguous tiles:

   • Traverse through each tile, considering as the starting point of a potential k length segment.
   • Each segment will wrap around the end of the array if needed.

3. Verify alternating pattern within the segment:

   • For each potential starting point, check the colors of k consecutive tiles.
   • Ensure the color of each tile (except potentially the first and last in the segment) alternates with respect to its immediate neighbors.

4. Increment the count if a valid alternating segment is found:

   • If all the sequential checks from the previous step hold true, increment the count of valid groups.

Considerations based on constraints:

• Since the minimum length for colors is 3 and maximum is 105, a direct checking method should suffice without significant performance concerns.
• As k always satisfies 3 <= k <= colors.length, there will always be substantial segments to evaluate looping behavior.

Hence, the approach will efficiently handle the input constraints while properly counting all valid alternating segments.

Solutions

class Solution {
public:
    int countKGroups(vector<int>& paint, int threshold) {
        int total = paint.size();
        int count = 0;
        int sequenceLength = 1;
        int previousColor = paint[0];
    
        for (int j = 1; j < total + threshold - 1; j++) {
            int currentIndex = j % total;
    
            if (paint[currentIndex] == previousColor) {
                sequenceLength = 1;
                previousColor = paint[currentIndex];
                continue;
            }
    
            sequenceLength++;
    
            if (sequenceLength >= threshold) {
                count++;
            }
    
            previousColor = paint[currentIndex];
        }
    
        return count;
    }
};