(Analysis by Benjamin Qi)

This is a harder version of a problem from last month's Silver contest.

Let $A_a$ denote the color of fence segment $a$ for each $a\in [1,N]$.

Approach 1:

The minimum number of strokes required to paint a subrange $[a,b]$ is equal to $b-a+1$ minus the number of indices $(l,r)$ such that $A_l=A_r<\min_{l<i<r}{A_i}$ (similarly as Modern Art from the Gold contest).

In the sample case, the pairs of indices are $(1,4)$, $(2,3)$, $(4,5)$, and $(6,8)$.

To generate all such pairs of indices (there are $\mathcal O(N)$ of them), we can use a monotonic stack (as alluded to by the analysis for the silver problem). The diagram below displays which elements are in the stack at what times for the sample case ($1$ and $2$ remain in the stack at the end):

            3
2-2     2---2- ...
1-----1-1------- ...


Computing the number of intervals contained within some query interval can be done offline in $\mathcal{O}(\log N)$ per query; answer queries in increasing order of right endpoint and store a BIT for the left endpoints.

Time Complexity: $\mathcal{O}((N+Q)\log N)$

#include <bits/stdc++.h>
using namespace std;

#define f first
#define s second

const int MX = 2e5+5;

int N,Q;
vector<pair<int,int>> query[MX];

struct {
int bit[MX];
int sum(int i) {
int sum = 0;
for (;i;i-=i&-i) sum += bit[i];
return sum;
}
int query(int l, int r) { return sum(r)-sum(l-1); }
void inc(int i) { for (;i<MX;i+=i&-i) ++bit[i]; }
} BIT;

int main() {
cin >> N >> Q;
vector<int> A(N), ans(Q);
for (int& t: A) cin >> t;
for (int i = 0; i < Q; ++i) {
int l,r; cin >> l >> r; --l,--r;
query[r].push_back({l,i});
}
vector<int> stk;
for (int i = 0; i < N; ++i) {
while (!stk.empty() && A[stk.back()] > A[i]) stk.pop_back();
if (!stk.empty() && A[stk.back()] == A[i]) { // consider pair (stk.back(),i)
BIT.inc(1+stk.back());
stk.back() = i;
} else stk.push_back(i);
for (pair<int,int> q: query[i])
ans[q.s] = i-q.f+1-BIT.query(q.f+1,i+1);
}
for (int t: ans) cout << t << "\n";
}


Approach 2 (courtesy of Spencer Compton):

The pairs of indices above join the segments into several connected components. For example, the connected components in the sample case are $[1,4,5],[2,3],[6,8],[7]$. Define $\texttt{is_last}[i]$ to be true if $i$ is the last number in its group and there exists some $j>i$ such that $A_j<A_i$. So for the sample case, $\texttt{is_last}[3]$ and $\texttt{is_last}[7]$ are both true. As in the above solution, we can generate all such indices $i$ with a monotonic stack.

The answer for a query $[l,r]$ is equal to the number of $i\in [l,r]$ such that $\texttt{is_last}[i]$ is true (which we can compute in $\mathcal{O}(1)$ using prefix sums), plus some additional contribution by connected components that continue past $r$, if the greatest index included in such a component that is at most $r$ is at least $l$.

For example, the answer to the query $(3,6)$ in the sample case is $3$ because

• $\texttt{is_last}[3]$ is true
• The component starting at index $1$ continues past index $6$, greatest index at most $6$ is $5\ge 3$.
• The component starting at index $6$ continues past index $6$, greatest index at most $6$ is $6\ge 3$.

We can maintain these indices in a stack $\texttt{last_found}$. For every query, binary search on the stack to count the number of indices at least $l$. The time complexity remains the same.

#include <bits/stdc++.h>
using namespace std;

const int MX = 2e5+5;

#define f first
#define s second
#define sz(x) (int)(x).size()

int N,Q;
vector<pair<int,int>> query[MX];

int main() {
cin >> N >> Q;
vector<int> A(N);
for (int& t: A) cin >> t;
for (int i = 0; i < Q; ++i) {
int l,r; cin >> l >> r; --l,--r;
query[r].push_back({l,i});
}
vector<bool> is_last(N);
vector<pair<int,int>> stk;
for (int i = 0; i < N; ++i) {
while (!stk.empty() && stk.back().f > A[i]) {
is_last[stk.back().s] = true;
stk.pop_back();
}
if (!stk.empty() && stk.back().f == A[i]) stk.back().s = i;
else stk.push_back({A[i],i});
}
vector<int> cum_last{0}, last_found;
vector<int> ans(Q);
for (int r = 0; r < N; ++r) {
cum_last.push_back(cum_last.back());
if (!last_found.empty() && A[r] == A[last_found.back()])
last_found.pop_back();
last_found.push_back(r);
if (is_last[r]) {
++cum_last.back();
last_found.pop_back();
}
for (pair<int,int> p: query[r]) {
int lo = 0, hi = sz(last_found);
while (lo < hi) {
int mid = (lo+hi)/2;
if (p.f <= last_found[mid]) hi = mid;
else lo = mid+1;
}
ans[p.s] = cum_last[r+1]-cum_last[p.f]+sz(last_found)-lo;
}
}
for (int t: ans) cout << t << "\n";
}