Exactly K (Sliding Window Sub-pattern)

// Count subarrays satisfying an "at most" rule.
// For a fixed right end, every start in [left..right] is valid.
long long countAtMost(vector<int>& a, int k) {
    if (k < 0) return 0;                  // "at most -1" is impossible (use k <= 0 for distinct counts)
    long long count = 0;
    int left = 0;
    for (int right = 0; right < (int)a.size(); right++) {
        add(a[right]);
        while (!valid(k)) { remove(a[left]); left++; }
        count += right - left + 1;        // subarrays ending at right
    }
    return count;
}

// "Exactly K" is not monotonic, so use two "at most" counts:
long long countExactly(vector<int>& a, int k) {
    return countAtMost(a, k) - countAtMost(a, k - 1);
}