Hello Codeforces!

Today I wanted to gain contributions share my small invention during my upsolving. I don't know if there existed a similar idea yet, but as far as I can tell, the editorial for the problem does not use my idea. I think it would be nice to share with you guys and have your opinions on this idea.

Idea explanation

How do we build Merge-sort tree again?

Merge sort tree is a Segment tree, each node of whose contains a sorted set/vector of every element in its range. We build it in a bottom-up manner. To build the parent nodes, we merge 2 sets/vectors of the children the same as we do in the merge-sort algorithm in $$$O(n)$$$, hence the name Merge-sort tree.

But there is another way.

Let A be the array we are building the merge-sort tree on, it[i] be the vector that contains sorted elements in the range conquer by the i-th node. Let's create another array B that stores the following pairs: B[i].first = A[i], B[i].second = i. We sort the array B in the increasing order of first. Then we iterate each pair element value, position of B in that sorted order, push the value to the back of every it[i] where i is the node of the merge-sort tree that contains position.

We can see that I sort the array before inserting, so I decided to call this trick sort before insert. Do you guys have a better name?

Some advantages of "sort before insert" over the classical way

• We can handle the case where we have multiple element located in the same position.
• With efficient style we can build this tree iteratively.

Here is the small snippet for building the tree.

const int N = 1e5;  // limit for array size
int n;  // array size
vector<int> it[2 * N];

void build(const vector<int>& a) {
  vector<pair<int, int>> b;
  for (int i = 0; i < (int)a.size(); ++i) {
    b.emplace_back(a[i], i);
  }
  sort(b.begin(), b.end());
  for (auto [val, p]: b) {
    for (p += (int)a.size(); p > 0; p >>= 1)
      it[p].push_back(val);
  }
}

So now we can call merge-sort tree — quick-sort tree from now :))

Can we do with ranges?

As we can see, each node in the merge-sort tree contains only elements of its range, and in the sort before insert version, we used point-update to build the tree. Can we do the same with range-like update? Yes, yes we can with sort before insert!

But what does each node contains exactly? We already know that for every interval, we can break it into $$$O(\log n)$$$ sub-intervals, each of them will correspond to a node of the segment tree. So if we have some intervals with some associated value, we can add these values to every node corresponding to the sub-interval of the considering interval, so that the vectors of the nodes will still be sorted.

Here is the implementation of the idea.

const int N = 1e5;  // limit for array size
int n;  // array size
vector<int> it[2 * N];

struct Range {
  int l, r, value;  // [l, r)
};
void build(vector<Range> a) {
  sort(a.begin(), a.end(), [](const Range& u, const Range& v) { return u.value < v.value});
  int n = (int)a.size();
  for (auto [l, r, val]: b) {
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) it[l++].push_back(val);
      if (r & 1) it[--r].push_back(val);
    }
  }
}