Introduction to Sets & Maps

C++

In sorted sets, the elements are sorted in order of element. Insertions, deletions, and searches are all $\mathcal{O}(\log N)$, where $N$ is the number of elements in the set.

std::set implements sorted sets in C++. Some operations on an std::set named s include:

• s.insert(x), which adds the element x to s if not already present.
• s.erase(x), which removes the element x from s if present.
• s.count(x), which returns 1 if s contains x and 0 if it doesn't.

You can iterate through a set in sorted order using a for-each loop.

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set<int> s;
s.insert(1);  // [1]
s.insert(4);  // [1, 4]
s.insert(2);  // [1, 2, 4]
s.insert(1);  // [1, 2, 4]
// the insert method did nothing because 1 was already in the set
cout << s.count(1) << endl;  // 1
s.erase(1);  // [2, 4]
cout << s.count(5) << endl;  // 0
s.erase(0);  // [2, 4]

Java

In sorted sets, the pairs are sorted in order of key. Insertions, deletions, and searches are all $\mathcal{O}(\log N)$, where $N$ is the number of elements in the set.

Some operations on a TreeSet named set include:

• set.add(x), which adds the element x to set if not already present.
• set.remove(x), which removes the element x from set if present.
• set.contains(x), which checks whether set contains the element x.

You can iterate through a set in sorted order using a for-each loop.

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Set<Integer> set = new TreeSet<>();
set.add(1); // [1]
set.add(4); // [1, 4]
set.add(2); // [1, 2, 4]
set.add(1); // [1, 2, 4]
// the add method did nothing because 1 was already in the set
System.out.println(set.contains(1)); // true
set.remove(1); // [2, 4]
System.out.println(set.contains(5)); // false
set.remove(0); // [2, 4]

Python

Python's built-in set uses hashing to support $\mathcal{O}(1)$ insertion, deletion, and searches. Some operations on a Python set named s include:

• s.add(x): adds element x to s if not already present
• s.remove(x): removes an element x from set if present
• x in s: checks whether s contains the element x

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s = set()
s.add(1) # {1}
s.add(4) # {1, 4}
s.add(2) # {1, 4, 2}
s.add(1) # {1, 4, 2}
# the add method did nothing because 1 was already in the set
print(1 in s) # True
s.remove(1) # {4, 2}
print(5 in s) # False
s.remove(0) # {4, 2}
# if the element to be removed does not exist, nothing happens

C++

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#include <bits/stdc++.h>

using namespace std;

int main() {
	int n;
	cin >> n;

	set<int> distinctNumbers;
	for (int i = 0; i < n; i++) {

Java

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// Source: Daniel

import java.io.*;
import java.util.*;

public class DistinctNumbers {
	public static void main(String[] args) throws IOException {
		Kattio io = new Kattio();
		int n = io.nextInt();
		Set<Integer> set = new HashSet<>();

Python

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n = int(input())  # unused
nums = [int(x) for x in input().split()]
distinct_nums = set(nums)
print(len(distinct_nums))

We can do this more efficiently by skipping the creation of the list, and use a set comprehension directly:

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n = int(input())  # unused
distinct_nums = {int(x) for x in input().split()}
print(len(distinct_nums))

C++

In sorted maps, the pairs are sorted in order of key. Insertions, deletions, and searches are all $\mathcal{O}(\log N)$, where $N$ is the number of pairs in the map.

std::map implements sorted maps in C++. Some operations on an std::map named m include:

• m[key], which returns a reference to the value associated with the key key.

  • If key is not present in the map, then the value associated with key is constructed using the default constructor of the value type. For example, if the value type is int, then calling m[key] for a key not within the map sets the value associated with that key to 0. As another example, if the value type is std::string, then calling m[key] for a key not within the map sets the value associated with that key to the empty string. More discussion regarding what happens in this case can be found here.
  • Alternatively, m.at(key) behaves the same as m[key] if key is contained within m but throws an exception otherwise.
  • m[key] = value will assign the value value to the key key.
• m.count(key), which returns the number of times the key is in the map (either one or zero), and therefore checks whether a key exists in the map.
• m.erase(key), which removes the map entry associated with the specified key if the key was present in the map.

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map<int, int> m;
m[1] = 5;  // [(1, 5)]
m[3] = 14;  // [(1, 5); (3, 14)]
m[2] = 7;  // [(1, 5); (2, 7); (3, 14)]
m[0] = -1;  // [(0, -1); (1, 5); (2, 7); (3, 14)]
m.erase(2);  // [(0, -1); (1, 5); (3, 14)]
cout << m[1] << endl;  // 5
cout << m.count(7) << endl;  // 0
cout << m.count(1) << endl;  // 1
cout << m[2] << endl;  // 0

Java

In sorted maps, the pairs are sorted in order of key. Insertions, deletions, and searches are all $\mathcal{O}(\log N)$, where $N$ is the number of pairs in the map.

In a TreeMap, the put(key, value) method assigns a value to a key and places the key and value pair into the map. The get(key) method returns the value associated with the key. The containsKey(key) method checks whether a key exists in the map. Lastly, remove(key) removes the map entry associated with the specified key.

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Map<Integer, Integer> map = new TreeMap<Integer, Integer>();
map.put(1, 5);  // [(1, 5)]
map.put(3, 14);  // [(1, 5); (3, 14)]
map.put(2, 7);  // [(1, 5); (2, 7); (3, 14)]
map.remove(2);  // [(1, 5); (3, 14)]
System.out.println(map.get(1));  // 5
System.out.println(map.containsKey(7));  // false
System.out.println(map.containsKey(1));  // true

Python

Colloquially, maps are referred to as dicts in python. They act as hash maps, so they have $\mathcal{O}(1)$ insertion, deletion, and searches.

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d = {}
d[1] = 5  # {1: 5}
d[3] = 14  # {1: 5, 3: 14}
d[2] = 7  # {1: 5, 2: 7, 3: 14}
del d[2]  # {1: 5, 3: 14}
print(d[1])  # 5
print(7 in d)  # False
print(1 in d)  # True

C++

An std::map stores entries as pairs in the form {key, value}. To iterate over maps, you can use a for loop. The auto keyword suffices to iterate over any type of pair (here, auto substitutes for pair<const int, int>).

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for (const auto& x : m) {
	cout << x.first << " " << x.second << endl;
}
for (auto x : m) {
	cout << x.first << " " << x.second << endl;
}

/* both output the following:
0 -1
1 5
3 14
*/

The first method (iterating over const references) is generally preferred over the second because the second will make a copy of each element that it iterates over. Additionally, you can pass by reference when iterating over a map, allowing you to modify the values (but not the keys) of the pairs stored in the map:

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for (auto& x : m) {
	x.second = 3;
}
for (pair<int,int> x : m) {
	cout << x.first << " " << x.second << endl;
}

/*
0 3
1 3
3 3
*/

Java

To iterate over maps, you can use a for loop over the keys.

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for (int k : m.keySet()) {
	System.out.println(k + " " + m.get(k));
}

Python

To iterate over dicts, there are three options. Dicts will be returned in the same order of insertion in Python 3.6+. You can iterate over the keys:

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for key in d:
	print(key)

You can iterate over the values:

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for value in d.values():
	print(value)

You can iterate over the key-value pairs:

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for key, value in d.items():
	print(key, value)

Python

For example, the following code attempts to remove every entry from a map, but results in a runtime error.

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d = {i: i for i in range(10)}

for i in d:
	del d[i]

Traceback (most recent call last):
  File "test.py", line 3, in <module>
    for i in d:
RuntimeError: dictionary changed size during iteration

One way is to get around this is to create a new map.

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d = {i: i for i in range(10)}

# only includes every third element
d_new = dict(item for i, item in enumerate(d.items()) if i % 3 == 0)

print("new dict:", d_new)  # new dict: {0: 0, 3: 3, 6: 6, 9: 9}

Another is to maintain a list of all the keys you want to remove and remove them after the iteration finishes:

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d = {i: i for i in range(10)}

# removes every third element
to_remove = {key for i, key in enumerate(d) if i % 3 == 0}

for key in to_remove:
	del d[key]

print("new dict:", d)  # new dict: {1: 1, 2: 2, 4: 4, 5: 5, 7: 7, 8: 8}

C++

For example, the following code attempts to remove every entry from a map, but results in a segmentation fault.

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map<int, int> m;

for (int i = 0; i < 10; ++i)
	m[i] = i;

for (auto &it : m) {
	cout << "Current Key: " << it.first << endl;

	m.erase(it.first);
}

The reason is due to "iterators, pointers and references referring to elements removed by the function [being] invalidated" (as stated in the documentation for erase), though iterators are beyond the scope of this module.

One way to get around this is to just create a new map instead of removing from the old one.

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map<int, int> m, M;

for (int i = 0; i < 10; ++i)
	m[i] = i;

int current_iteration = 0;

for (const auto &it : m) {
	// only includes every third element
	if (current_iteration % 3 == 0) {

Another is to maintain a list of all the keys you want to erase and erase them after the iteration finishes.

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map<int, int> m;
for (int i = 0; i < 10; ++i) {
	m[i] = i;
}

vector<int> to_erase;
int current_iteration = 0;

for (const auto &it : m) {
	// removes every third element	

Java

Modifying a Collection (Set, Map, etc.) in the middle of a for-each loop will cause a ConcurrentModificationException. See the following snippet for an example:

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Map<Integer, Integer> m = new TreeMap<>();

for (int i = 0; i < 10; i++) {
	m.put(i, i);
}
for (int key : m.keySet()) {
	m.remove(key);  // ConcurrentModificationException thrown!!
}

One work-around is to use Iterator and the .remove() method to remove elements while looping over them, like in the next code snippet:

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Set<Integer> s = new TreeSet<>();
// s starts as {0, 1, 2}
s.add(0); s.add(1); s.add(2);

Iterator<Integer> iter = s.iterator();
while (iter.hasNext()) {
	int key = iter.next();
	if (key == 0 || key == 2) {
		iter.remove();
	}

However, Iterator is outside the scope of this module.

The easiest option (in most cases) if you want to remove/insert mutiple elements at once is to use your Container's .addAll(c) or .removeAll(c) methods. That means that you should put all the elements you want to remove (or add) in a new Collection, and then use that new Collection as the parameter of the .addAll(c) or .removeAll(c) method that you call on your original Collection. See the following code snippet for an example (it works equivalently to the code above):

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Set<Integer> s = new TreeSet<>();
// s starts as {0, 1, 2}
s.add(0); s.add(1); s.add(2);

Set<Integer> toRemove = new TreeSet<>();
for (Integer a : s) {
	if (a == 0 || a == 2) {
		toRemove.add(a);
	}
}

C++