Min window O(n)

    deque<int> d;
    for (int i = 0; i < n; ++i) {
        while (!d.empty() && a[d.back()] >= a[i]) {
            d.pop_back();
        }
        d.push_back(i);
        while (d.front() <= i - k) {
            d.pop_front();
        }
        if (i >= k - 1) {
            cout << a[d.front()] << '\n';
        }
    }