---
id: harder-ordered
title: "Harder Problems with Ordered Sets"
author: Benjamin Qi
prerequisites: 
 - Silver - Introduction to Ordered Sets
 - Silver - Introduction to Stacks & Queues
description: ""
frequency: 2
---

import { Problem } from "../models";

export const metadata = {
  problems: {
    sample: [
      new Problem("CSES", "Bit Inversions", "1188", "Normal", false, []),
    ],
    general: [
      new Problem("Silver", "Milk Measurement", "763", "Normal", false, []),
      new Problem("Silver", "Convention II", "859", "Normal", false, []),
      new Problem("Gold", "Snow Boots", "813", "Normal", false, []),
      new Problem("CF", "Jury Marks", "contest/831/problem/C", "Hard", false, [], "Hard, would recommend skipping (1700 on CF)"),
      new Problem("CF", "Mahmoud & Ehab & Function", "contest/862/problem/E", "Hard", false, [], "Hard, do not attempt until Gold/Plat (2100 on CF)"),
      new Problem("CF", "Tournament", "contest/878/problem/C", "Very Hard", false, [], "First solve problem for $n$-th tournament only. Extremely hard, do not attempt (2700 on CF)"),
    ]
  }
};


## Example: Using Iterators

<problems-list problems={metadata.problems.sample} />

<spoiler title="Solution">

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

#define sz(x) (x).size()

string s;
int m;
multiset<int> ans, ret;
 
void modify(int x) {
  if (x == 0 || x == sz(s)) return;
  auto it = ans.find(x);
  if (it != end(ans)) {
    int a = *prev(it), b = *next(it);
    ret.erase(ret.find(x-a)), ret.erase(ret.find(b-x));
    ret.insert(b-a);
    ans.erase(it);
  } else {
    it = ans.insert(x);
    int a = *prev(it), b = *next(it);
    ret.erase(ret.find(b-a));
    ret.insert(x-a), ret.insert(b-x);
  }
}
 
int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);
  cin >> s >> m;
  ans.insert(0); ans.insert(sz(s));
  for (int i = 0; i < sz(s)-1; ++i) 
    if (s[i] != s[i+1]) ans.insert(i+1);
  for (auto it = ans.begin(); next(it) != ans.end(); it ++)
    ret.insert(*next(it)-*it);
  for (int i = 0; i < m; ++i) {
    int x; cin >> x;
    modify(x-1); modify(x);
    cout << *ret.rbegin() << " ";
  }
}
```

<br />

Note that `multiset` has a high constant factor, so replacing `ret` with an array and a `priority_queue` reduces the runtime by a factor of 2.

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

#define sz(x) (x).size()

string s;
int m;
set<int> ans;
priority_queue<int> ret;
int cnt[200005];
 
void ad(int x) { cnt[x] ++; ret.push(x);  }
void modify(int x) {
  if (x == 0 || x == sz(s)) return;
  auto it = ans.find(x);
  if (it != end(ans)) {
    int a = *prev(it), b = *next(it); ans.erase(it);
    cnt[x-a] --, cnt[b-x] --; ad(b-a);
  } else {
    it = ans.insert(x).first;
    int a = *prev(it), b = *next(it);
    cnt[b-a] --, ad(x-a), ad(b-x);
  }
}
 
int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);
  cin >> s >> m;
  ans.insert(0); ans.insert(sz(s));
  for (int i = 0; i < sz(s)-1; ++i) 
    if (s[i] != s[i+1]) ans.insert(i+1);
  for (auto it = ans.begin(); next(it) != ans.end(); it ++) 
    ad(*next(it)-*it);
  for (int i = 0; i < m; ++i) {
    int x; cin >> x;
    modify(x-1); modify(x);
    while (!cnt[ret.top()]) ret.pop();
    cout << ret.top() << " ";
  }
}
```

</spoiler>

## Problems

<problems-list problems={metadata.problems.general} />