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# More Operations on Sorted Sets

Authors: Darren Yao, Benjamin Qi, Andrew Wang

Finding the next element smaller or larger than a specified key in a set, and using iterators with sets.

### Prerequisites

C++

Resources
IUSACO

module is based off this

CP2

see decription of BSTs and heaps

Java

Resources
IUSACO

module is based off this

CP2

see decription of BSTs and heaps

In sets and maps where keys (or elements) are stored in sorted order, accessing or removing the next key higher or lower than some input key k is supported.

Keep in mind that insertion and deletion will take $\mathcal{O}(\log N)$ time for sorted sets, which is more than the average $\mathcal{O}(1)$ insertion and deletion for unordered sets, but less than the worst case $\mathcal{O}(N)$ insertion and deletion for unordered sets.

## Using Iterators

In Bronze, we avoided discussion of any set operations involving iterators.

C++

Resources
CPH

Java

In Java, iterators are helpful for looping through sets.

Iterators used with HashSet would yield the elements in random order:

Set<Integer> set = new HashSet<Integer>();set.add(1);set.add(3);set.add(0);set.add(-2);Iterator it = set.iterator();while (it.hasNext()) {	Integer i = (Integer)it.next();	System.out.print(i + " ");  // returns some random order}

But with TreeSet the elements are in sorted order:

Set<Integer> set = new TreeSet<Integer>();set.add(1);set.add(3);set.add(0);set.add(-2);Iterator it = set.iterator();while (it.hasNext()) {	Integer i = (Integer)it.next();	System.out.print(i + " ");  // returns -2 0 1 3}

Instead of creating an iterator and looping with it like in C++, Java provides a for-each loop which creates a hidden iterator and loops with it automatically:

Set<Integer> set = new TreeSet<Integer>();set.add(1);set.add(3);set.add(0);set.add(-2);for (int i : set) {	System.out.print(i + " ");  // returns -2 0 1 3}

### Warning!

You shouldn't modify sets when traversing it with set iterators like in any other iterators for Collections (this INCLUDES when using a for-each loop). The only modification possible is using the iterator remove() method which can only be used once before calling the next() method.

## Sorted Sets

C++

The sorted std::set also supports:

• lower_bound: returns an iterator to the least element greater than or equal to some element k.
• upper_bound: returns an iterator to the least element strictly greater than some element k.
set<int> s;s.insert(1);                        // s.insert(14);                       // [1, 14]s.insert(9);                        // [1, 9, 14]s.insert(2);                        // [1, 2, 9, 14]cout << *s.upper_bound(7) << '\n';  // 9cout << *s.upper_bound(9) << '\n';  // 14cout << *s.lower_bound(5) << '\n';  // 9cout << *s.lower_bound(9) << '\n';  // 9cout << *s.begin() << '\n';         // 1auto it = s.end();cout << *(--it) << '\n';    // 14s.erase(s.upper_bound(6));  // [1, 2, 14]

### Warning!

Suppose that we replace s.upper_bound(7) with upper_bound(begin(s),end(s),7), which was the syntax that we used for vectors in the prerequisite module. This will still output the expected results, but its time complexity is linear in the size of the set s rather than logarithmic, so make sure to avoid it!

Java

TreeSets in Java allow for a multitude of additional operations:

• first(): returns the lowest element in the set
• last(): returns the greatest element in the set
• lower(E v): returns the greatest element strictly less than v
• floor(E v): returns the greatest element less than or equal to v
• higher(E v): returns the least element strictly greater than v
• ceiling(E v): returns the least element greater than or equal to v
TreeSet<Integer> set = new TreeSet<Integer>();set.add(1);                          // set.add(14);                         // [1, 14]set.add(9);                          // [1, 9, 14]set.add(2);                          // [1, 2, 9, 14]System.out.println(set.higher(7));   // 9System.out.println(set.higher(9));   // 14System.out.println(set.lower(5));    // 2System.out.println(set.first());     // 1System.out.println(set.last());      // 14set.remove(set.higher(6));           // [1, 2, 14]System.out.println(set.higher(23));  // ERROR, no such element exists

One limitation of sorted sets is that we can't efficiently access the $k^{th}$ largest element in the set, or find the number of elements in the set greater than some arbitrary $x$. In C++, these operations can be handled using a data structure called an order statistic tree.

## Sorted Maps

C++

The ordered map also allows:

• lower_bound: returns the iterator pointing to the lowest entry not less than the specified key
• upper_bound: returns the iterator pointing to the lowest entry strictly greater than the specified key respectively.
map<int, int> m;m = 5;     // [(3, 5)]m = 4;    // [(3, 5); (11, 4)]m = 491;  // [(3, 5); (10, 491); (11, 4)]cout << m.lower_bound(10)->first << " " << m.lower_bound(10)->second     << '\n';  // 10 491cout << m.upper_bound(10)->first << " " << m.upper_bound(10)->second     << '\n';  // 11 4m.erase(11);   // [(3, 5); (10, 491)]if (m.upper_bound(10) == m.end()) {	cout << "end" << endl;  // Prints end}

Java

The ordered map additionally supports firstKey / firstEntry and lastKey / lastEntry, returning the lowest key/entry and the highest key/entry, as well as higherKey / higherEntry and lowerKey / lowerEntry, returning the lowest key/entry strictly higher than the specified key, or the highest key/entry strictly lower than the specified key.

TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>();map.put(3, 5);                           // [(3, 5)]map.put(11, 4);                          // [(3, 5); (11, 4)]map.put(10, 491);                        // [(3, 5); (10, 491); (11, 4)]System.out.println(map.firstKey());      // 3System.out.println(map.firstEntry());    // (3, 5)System.out.println(map.lastEntry());     // (11, 4)System.out.println(map.higherEntry(4));  // (10, 491)map.remove(11);                          // [(3, 5); (10, 491)]System.out.println(map.lowerKey(4));     // 3System.out.println(map.lowerKey(3));     // ERROR

## Multisets

A multiset is a sorted set that allows multiple copies of the same element.

C++

In addition to all of the regular set operations,

• the count() method returns the number of times an element is present in the multiset. However, this method takes time linear in the number of matches so you shouldn't use it in a contest.
• To remove a value once, use ms.erase(ms.find(val)).
• To remove all occurrences of a value, use ms.erase(val).

### Warning!

Using ms.erase(val) erases all instances of val from the multiset. To remove one instance of val, use ms.erase(ms.find(val))!

multiset<int> ms;ms.insert(1);                  // ms.insert(14);                 // [1, 14]ms.insert(9);                  // [1, 9, 14]ms.insert(2);                  // [1, 2, 9, 14]ms.insert(9);                  // [1, 2, 9, 9, 14]ms.insert(9);                  // [1, 2, 9, 9, 9, 14]cout << ms.count(4) << '\n';   // 0cout << ms.count(9) << '\n';   // 3cout << ms.count(14) << '\n';  // 1ms.erase(ms.find(9));cout << ms.count(9) << '\n';  // 2ms.erase(9);cout << ms.count(9) << '\n';  // 0

Java

While there is no multiset in Java, we can implement one using the TreeMap from values to their respective frequencies. We declare the TreeMap implementation globally so that we can write functions for adding and removing elements from it. The first, last, higher, and lower operations still function as intended; just use firstKey, lastKey, higherKey, and lowerKey respectively.

static TreeMap<Integer, Integer> multiset = new TreeMap<Integer, Integer>();
public static void main(String[] args) { ... }
static void add(int x) {	if (multiset.containsKey(x)) {		multiset.put(x, multiset.get(x) + 1);	} else {		multiset.put(x, 1);	}

## Priority Queues

### Warning!

Priority queues are not implemented in the same way as sets and multisets, but they are included in this section because the operations that they perform can also be performed with sets.

Resources
CSA

A priority queue (or heap) supports the following operations: insertion of elements, deletion of the element considered highest priority, and retrieval of the highest priority element, all in $\mathcal{O}(\log N)$ time according to the number of elements in the priority queue. Priority queues are simpler and faster than sets, so you should use them instead whenever possible.

C++

### C++

priority_queue<int> pq;pq.push(7);                // pq.push(2);                // [2, 7]pq.push(1);                // [1, 2, 7]pq.push(5);                // [1, 2, 5, 7]cout << pq.top() << endl;  // 7pq.pop();                  // [1, 2, 5]pq.pop();                  // [1, 2]pq.push(6);                // [1, 2, 6]

Java

### Java

In Java (unlike in C++), we delete and retrieve the lowest element.

PriorityQueue<Integer> pq = new PriorityQueue<Integer>();pq.add(7);                      // pq.add(2);                      // [7, 2]pq.add(1);                      // [7, 2, 1]pq.add(5);                      // [7, 5, 2, 1]System.out.println(pq.peek());  // 1pq.poll();                      // [7, 5, 2]pq.poll();                      // [7, 5]pq.add(6);                      // [7, 6, 5]

Python

### Python

In Python (unlike in C++), we delete and retrieve the lowest element.

Note that Python's priority queue is not encapsulated; heapq operates on a provided list directly by turning it into a heap, then doing operations on the heap.

### Warning!

Because of a heap's structure, printing out pq will not print out the elements in sorted order in Python; instead, it will print out the list. The comments below are a representation of what the heap contains, not what the contents of pq actually are.

import heapq
pq = []
"""The following line is not necessary since pq starts as an empty list. However,for lists of length greater than 1, heapify is required to turn the list into aheap."""heapq.heapify(pq)

## Introductory Problems

StatusSourceProblem NameDifficultyTags
CSESEasy
Show TagsSorted Set
CFEasy
Show TagsSorted Set
CSESNormal
Show TagsSorted Set
CSESNormal
Show TagsSorted Set

## Harder Example - Bit Inversions

### Warning!

Problems marked as "Hard" or beyond in this module would likely be too difficult to appear on a USACO Silver contest.

Focus Problem – try your best to solve this problem before continuing!

View Internal Solution

### Solution

We'll use iterators extensively.

Let the bit string be $s=s_0s_1s_2\ldots,s_{n-1}$. In the set dif, we store all indices $i$ such that $s_i\neq s_{i-1}$ (including $i=0$ and $i=n$). If the elements of dif are $0=dif_1, then the longest length is equal to

$\max(dif_2-dif_1,dif_3-dif_2,\ldots,dif_k-dif_{k-1}).$

We can store each of these differences in a multiset ret; after each inversion, we'll need to output the maximum element of ret.

Inverting a bit at a 0-indexed position x corresponds to inserting x into dif if it not currently present or removing x if it is, and then doing the same with x+1. Whenever we insert or remove an element of dif, we should update ret accordingly.

C++

#include <bits/stdc++.h>using namespace std;
#define sz(x) (x).size()
string s;int m;set<int> dif;multiset<int> ret;


Java

### Warning!

Java solutions are too slow for the CSES. Use C++ instead to get AC.

import java.io.*;import java.util.*;
class BitInversion {	public static TreeMap<Integer, Integer> ret =	    new TreeMap<Integer, Integer>();	public static String s;	public static int m;	public static Set<Integer> dif = new TreeSet<Integer>();


Note that multiset has a high constant factor, so replacing ret with a priority queue and an array that stores the number of times each integer occurs in the priority queue reduces the runtime by a factor of 2.

C++

#include <bits/stdc++.h>using namespace std;
#define sz(x) (int)(x).size()
string s;int m;set<int> dif;priority_queue<int> ret;int cnt;

Java

import java.io.*;import java.util.*;
class BitInversion {	public static PriorityQueue<Integer> pq =	    new PriorityQueue<Integer>(Collections.reverseOrder());	public static String s;	public static int m;	public static TreeSet<Integer> dif = new TreeSet<Integer>();	public static int cnt[];

## Harder Problems

StatusSourceProblem NameDifficultyTags
SilverNormal
Show TagsSorted Set
SilverNormal
Show TagsSorted Set
CFNormal
Show TagsGreedy, Sorted Set, Sorting
CFNormal
Show TagsGreedy, Multiset, Sorting
CFNormal
Show TagsCoordinate Compression, Prefix Sums, Sorted Set, Sorting
GoldHard