Easy

Alternating Groups IC++

Full explanation · Time O(n) · Space O(1)

// Time:  O(n)
// Space: O(1)

// sliding window, two pointers
class Solution {
public:
    int numberOfAlternatingGroups(vector<int>& colors) {
        static const int k = 3;

        int result = 0;
        for (int right = 0, left = 0, curr = 0; right < size(colors) + k - 1; ++right) {
            if (right - left + 1 == k) {
                result += curr == k - 1 ? 1 : 0;
                curr -= colors[left] != colors[(left + 1) % size(colors)] ? 1 : 0;
                ++left;
            }
            curr += colors[right % size(colors)] != colors[(right + 1) % size(colors)] ? 1 : 0;
        }
        return result;
    }
};

// Time:  O(n)
// Space: O(1)
// sliding window
class Solution2 {
public:
    int numberOfAlternatingGroups(vector<int>& colors) {
        int result = 0;
        for (int i = 0; i < size(colors); ++i) {
            if (colors[i] != colors[(i + 1) % size(colors)] && colors[(i + 1) % size(colors)] != colors[(i + 2) % size(colors)]) {
                ++result;
            }
        }
        return result;
    }
};