Max Suffix Query with Insertions Only

Note: This was originally in Gold, but the practice problems were too hard ...

To solve this problem, we need a data structure that supports operations similar to the following:

1. Add a pair $(a,b)$.
2. For any $x$, query the maximum value of $b$ over all pairs satisfying $a\ge x$.

This can be solved with a segment tree, but a simpler option is to use a map. We rely on the fact that if there exist pairs $(a,b)$ and $(c,d)$ in the map such that $a\le c$ and $b\le d$, we can simply ignore $(a,b)$ (and the answers to future queries will not be affected). So at every point in time, the pairs $(a_1,b_1),(a_2,b_2),\dots ,(a_k,b_k)$ that we store in the map will satisfy $a_1<a_2<\cdots <a_k$ and $b_1>b_2>\cdots >b_k$.

• Querying for a certain $x$ can be done with a single lower_bound operation, as we just want the minimum $i$ such that $a_i\ge x$.
• When adding a pair $(a^{\prime },b^{\prime })$, first check if there exists $(a,b)$ already in the map such that $a\ge a^{\prime },b\ge b^{\prime }$.
  • If so, then do nothing.
  • Otherwise, insert $(a^{\prime },b^{\prime })$ into the map and repeatedly delete pairs $(a,b)$ such that $a\le a^{\prime },b\le b^{\prime }$ from the map until none remain.

If there are $N$ insertions, then each query takes $\mathcal{O}(\log N)$ time and adding a pair takes $\mathcal{O}(\log N)$ time amortized.

Implementation

1#define lb lower_bound
2
3map<int,ll> m;
4void ins(int a, ll b) {
5    auto it = m.lb(a); if (it != end(m) && it->s >= b) return;
6    it = m.insert(it,{a,b}); it->s = b;
7    while (it != begin(m) && prev(it)->s <= b) m.erase(prev(it));
8}
9ll query(int x) { auto it = m.lb(x);
10    return it == end(m) ? 0 : it->s; }

You can check the analysis for the full solution to the original problem.

Problems