void heapify(vector<int>& vec) {
    int n = vec.size();
    for (int a = 0, b = 1; b - 1 < n; a = b, b <<= 1) {
        nth_element(vec.begin() + a, vec.begin() + b, vec.end());
    }
}

You can implement a "heapify" by only four lines of code. And the time complexity is O(n).

Fancy, right? :)


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