gordon college prof. brinton

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Associative Containers

Gordon College

Prof. Brinton

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STL - Assoc. Containers

• Set– The key is the dataset<int> intSet;set<string> keyword; set<time24> timeSet;

To support the STL container - set; the programmer mustoverload the == operator and < operator by comparingthe key field and giving a Boolean result.

record object

key field other fields

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STL - Assoc. Containers

• Map– Stores entries as key-value pair.– In a pair, the first component is the key; the second is the

value. Each component may have a different data type.

map<int, record> studentFile;

studentFile[2343554] = new studentFile;studentFile[2343554].addToBalance(112);

key value

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Set and Map

• Both containers do not allow duplicate keys.

• Multiset and multimap (also STL containers) allow duplicate keys)

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STL sets and maps

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Degenerate search tree red-black tree

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STL Set include <set>

• Constructors:set(); default - empty setset(T *first, T *last); use pointers

or iteratorsExamples:

set<int> first; // empty set of intsint myints[]= {10,20,30,40,50};set<int> second (myints,myints+5); // pointersset<int> third (second); // a copy of secondset<int> fourth (second.begin(), second.end()); // iterators into second

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• Operations:bool empty() const; void clear(); int size() const;int count(const T& key) const; //return either 1 or

0

iterator find(const T& key);const_iterator find(const T& key) const;// returns either an iterator or end()

NOTE: STL Associative Containers iterators do access the elements in the defined order.

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• Operations:pair<iterator, bool> insert(const T& key);// this return a pair object

Example:string t[]={"this","is","a","test"};set<string> s(t,t+4);

pair<set<string>::iterator,bool> result = s.insert("OK");(result.second)?cout << "TRUE":cout << "FALSE";

result = s.insert("OK");(result.second)?cout << "TRUE":cout << "FALSE";

Result:

TRUEFALSE

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• Operations:void erase ( iterator position );void erase ( iterator first, iterator last );size_type erase ( const key_type& x ); //returns either 1 or 0

void swap ( set<Key,Compare,Allocator>& st );

• Obtaining the bounds:pair<iterator,iterator> equal_range ( const key_type& x ) const;iterator lower_bound ( const key_type& x ) const; //iterator of first value not < x

iterator upper_bound ( const key_type& x ) const; //iterator of first value > x

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• Iterators:

iterator begin(); const_iterator begin(const);

iterator end(); const_iterator end(const);

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Set operators

• Union

setC = setA + setB;

setC = 1 2 3 4 5 6 7 8 11 15• Intersection A*B

setC = setA * setB;

setC = 3 5 7 8 11• Difference

setC = setA - setB;

setC = 1 4

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Map include <map>

• Operations are like set’s operations– swap– begin– end– size– empty– operator[] //unlike set– find– erase

See the following examples

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#include <string.h>#include <iostream>#include <map>using namespace std;int main(){ map<int, string> Employees; Employees[5234] = "Mike C."; Employees[3374] = "Charlie M."; Employees[1923] = "David D."; Employees[7582] = "John A."; Employees[5328] = "Peter Q.";

cout << "Employees[3374]=" << Employees[3374] << endl << endl; cout << "Map size: " << Employees.size() << endl; for( map<int,string>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

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#include <string.h>#include <iostream>#include <map>

using namespace std;int main(){ map<int, string> Employees; Employees[5234] = "Mike C."; Employees[3374] = "Charlie M."; Employees[1923] = "David D."; Employees[7582] = "John A."; Employees[5328] = "Peter Q.";

cout << "Employees[3374]=" << Employees[3374] << endl << endl; cout << "Map size: " << Employees.size() << endl; for( map<int,string>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

Employees[3374]=Charlie M.

Map size: 51923: David D.3374: Charlie M.5234: Mike C.5328: Peter Q.7582: John A.

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using namespace std;

int main(){ map<string, int> Employees; Employees["Mike C."] = 5234; Employees["Charlie M."] = 3374; Employees.insert(pair<string,int>("David D.",1923)); Employees.insert(map<string,int>::value_type("John A.",7582)); Employees.insert(make_pair("Peter Q.",5328)); cout << "Map size: " << Employees.size() << endl;

for( map<string, int>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

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using namespace std;

int main(){ map<string, int> Employees; Employees["Mike C."] = 5234; Employees["Charlie M."] = 3374; Employees.insert(pair<string,int>("David D.",1923)); Employees.insert(map<string,int>::value_type("John A.",7582)); Employees.insert(make_pair("Peter Q.",5328)); cout << "Map size: " << Employees.size() << endl;

for( map<string, int>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

Map size: 5Charlie M.: 3374David D.: 1923John A.: 7582Mike C.: 5234Peter Q.: 5328

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struct cmp_str { bool operator()(char const *a, char const *b) { return strcmp(a, b) < 0; }};

int main(){ map<char *, int, cmp_str> Employees; Employees["Mike C."] = 5234; Employees["Charlie M."] = 3374; Employees.insert(pair<char *,int>("David D.",1923)); Employees.insert(map<char *,int>::value_type("John A.",7582)); Employees.insert(make_pair((char *)"Peter Q.",5328)); cout << "Map size: " << Employees.size() << endl;

for( map<char *, int, cmp_str>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

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struct cmp_str { bool operator()(char const *a, char const *b) { return strcmp(a, b) < 0; }};

int main(){ map<char *, int, cmp_str> Employees; Employees["Mike C."] = 5234; Employees["Charlie M."] = 3374; Employees.insert(pair<char *,int>("David D.",1923)); Employees.insert(map<char *,int>::value_type("John A.",7582)); Employees.insert(make_pair((char *)"Peter Q.",5328)); cout << "Map size: " << Employees.size() << endl;

for( map<char *, int, cmp_str>::iterator ii=Employees.begin(); ii!=Employees.end(); ++ii) { cout << (*ii).first << ": " << (*ii).second << endl; }}

Map size: 5Charlie M.: 3374David D.: 1923John A.: 7582Mike C.: 5234Peter Q.: 5328

Could this be ordered differently?

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Map details

• Associative ArraysM[“Computer Science”] = 20;

Keys are unique therefore:

M[“Computer Science”] = 26;

replaces the value in the tree for CS.

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Map details

• Creating a map:

#include <map>

int main()

{

map<string, double> A;

…

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Map Details• [ ] operator

Steps

1. Search the for map for key

2. If key found, return a reference to the value object associated with key.

3. If key not found, create a new object of type value associated with key, and return a reference to that.

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int salary = Employee[“Jack Smith”];

Caution: this will create a new tree entry if it doesn’t exist and assign 0 to variable salary

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Better technique:

if (( itr = Employee.find(“Jack Smith”)) == Employee.end()) cout << “ the employee does not exist.” << endl;

Else cout << “Employee Salary: “ << itr->second << endl;

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Map Details• Pair Class

1. Contains two public members, first and second

2. Store key/value association

itr->first

itr->second

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Map Details• Pair Class

Easiest way to construct pair:

pair<string, int> a;a = make_pair(string(“Jack Smith”), int(100000));

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Map Details• Element requirements:

Each key and value must be assignable (an assignment operator which performs a “deep copy”

Deep copytarget of the assignment should be equal but independent of the source

a = b;

the value of a should match b - however if you change a it should not affect b (and vice versa)

Note: if we talking about a linked list then a shallow copy would have two pointers pointing to the same list.

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Also each key should adhere to “weak ordering”:

1. A<A is false2. Equality can be determined with (!(A<B) && !(B<A))

Note: only using the less than operator3. If A<B and B<C then A<C must be true

int, char, double, etc. (built-in types) are strict weak ordered

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value type used must have a defined default constructor.

map <int, bigNumber>

The built in types have a type of default constructor:

map<int, string>

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Map DetailsOperation Description Time

Complexity

swap Swaps elements with another map. This operation is performed in constant time.

Ex: map1.swap(map2);

O(1)

begin Returns an iterator to the first pair in the map.

Ex: itr = map1.begin();

O(1)

end Returns an iterator just beyond the end of the map.

O(1)

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Complexity

size Returns the number of elements contained by the map. You should use empty() to test whether or not size equals 0, because empty() is faster.

cout << map1.size();

O(n)

empty Returns true if the map contains zero elements.

O(1)

Operator [ ] Accepts a key and returns a reference to the object associated with key. Note that this operator always creates an element associated with key if it does not exist.

O(log n)

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Complexity

find Accepts a key and returns an iterator to the pair element if found. If the key is not found in the map, this method returns end().

if ( (itr = map1.find("Bob")) == map1.end())

cerr << "Bob was not found in the map." << endl;

else

cout << "Bob was found in the map." << endl;

O(log n)

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Complexity

erase There are three overloaded forms of this method.

The first accepts a single iterator, and removes the element implied by the iterator from the map. The map should be non-empty. O(1).

map1.erase(map1.begin()); // remove the first element

O(1)

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Complexity


The second accepts two iterators, that specify a range of values within the map to be removed. The time complexity is logarithmic plus linear time for the length of the sequence being removed. The two iterators must be valid iterators within the sequence contained by the map. Formally, the range of values [starting, ending) are removed. At worst, this is an O(n) operation.

Ex: map1.erase(map1.begin(), map1.end()); // erase the entire map

O(n)

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Complexity


The last accepts an object of the key type, and removes all occurrences of the key from map. Note that since map elements all have unique keys, this will erase at most 1 element. O(log n).

Ex: map1.erase(string("Bob")); // removes all occurrences of "Bob"

O(log n)

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Multimap/Multisetbegin endrbegin rendclear emptycountequal_range lower_bound upper_bound insert erasefindmax_sizeoperator=sizeswap

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Multimap/Multiset DetailsOperation Description

constructor creates an empty multiset

multiset<int> S

size returns the number of elements

S.size()

insert S.insert( const T& ) : adds an item to the set. If the item is already present, there will be multiple copies in the multiset/multimap

S.insert(6);

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count S.count( const T& t ) : returns the number of times that the item t is found in the set. This will be 0 if the item is not found

erase S.erase( const T& t ) : removes the all items equal to t from the multiset.

erase S.erase( multiset<T>::iterator i ) : removes the item pointed to by i.

S.erase( v ); // ALL: removes all of items == v from the multiset

S.erase( S.find( v ) ); // ONE: removes the first item == v from the multiset

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reverse reverse( iterator begin, iterator end ) : reverses the order of the items:

find S.find( const T& t ) : returns an iterator that points to the first item == t, or S.end() if the item could not be found.

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equal_range S.equal_range(*iter) - returns a pair of iterators such that all occurrences of the item are in the iterator range

pair<multiset <int>::iterator, multiset <int>::iterator > p;

p = S.equal_range(5);

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Multimap/Multiset Detailsequal_range

multiset<int>::iterator it; pair<multiset<int>::iterator,multiset<int>::iterator>

ret;

int myints[]= {77,30,16,2,30,30};multiset<int> mymultiset (myints, myints+6); // 2 16 30

30 30 77

ret = mymultiset.equal_range(30); for (it=ret.first; it!=ret.second; ++it) ++(*it); // 2 16 31 31 31 77

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Multimap/Multiset Details

• How is multiset implemented?

Multiset:

BST - with integer value representing each occurrence of the element

insert {find - if not present create node else increment count}erase {remove node and report back count}

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7 20

9 304

(2)

(1)(5)

(2) (3) (1)

HOW WOULD THESE BE HANDLED?clear emptycountfindsize

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Multimap/Multiset Details

• How is multimap implemented?

Multimap:

BST - with list or map as the node’s payloadinsert {find - if not present create node else add value to list/set}erase {remove node and report back count}

HOW WOULD THESE BE HANDLED?clear emptycountfindsize

There is not guarantee as to order of duplicates

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7 20

9 304

(“Tom” -> “Jack” -> “Joe”)

(“Tom”)

(“Jack” -> “Joe”)(“Jack”)

(“Jack” -> “Joe”)

(“Emma”)

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