SOLUTION Of Remove for X
Data structures
A data structure is a way to store data in the memory of a computer. It is important to choose an appropriate data structure for a problem, because each data structure has its own advantages and disadvantages. The crucial question is: which operations are efficient in the chosen data structure?
This chapter introduces the most important data structures in the C++ standard library. It is a good idea to use the standard library whenever possible, because it will save a lot of time. Later in the book we will learn about more sophisticated data structures that are not available in the standard library.
Dynamic arrays
A dynamic array is an array whose size can be changed during the execution of the program. The most popular dynamic array in C++ is the vector structure, which can be used almost like an ordinary array.
The following code creates an empty vector and adds three elements to it:
vector<int> v;
v.push_back(3); // [3]
v.push_back(2); // [3,2]
v.push_back(5); // [3,2,5]
After this, the elements can be accessed like in an ordinary array:
cout << v[0] << "\n"; // 3 cout << v[1] << "\n"; // 2 cout << v[2] << "\n"; // 5
The function size returns the number of elements in the vector. The following code iterates through the vector and prints all elements in it:
for (int i = 0; i < v.size(); i++) { cout << v[i] << "\n";
}
A shorter way to iterate through a vector is as follows:
for (auto x : v) { cout << x << "\n";
}
The function back returns the last element in the vector, and the function pop_back removes the last element:
vector<int> v;
v.push_back(5);
v.push_back(2); cout << v.back() << "\n"; // 2
v.pop_back();
cout << v.back() << "\n"; // 5
The following code creates a vector with five elements:
vector<int> v = {2,4,2,5,1};
Another way to create a vector is to give the number of elements and the initial value for each element:
// size 10, | initial value 0 |
vector<int> | v(10); |
// size 10, | initial value 5 |
vector<int> | v(10, 5); |
The internal implementation of a vector uses an ordinary array. If the size of the vector increases and the array becomes too small, a new array is allocated and all the elements are moved to the new array. However, this does not happen often and the average time complexity of push_back is O(1).
The string structure is also a dynamic array that can be used almost like a vector. In addition, there is special syntax for strings that is not available in other data structures. Strings can be combined using the + symbol. The function substr(k, x) returns the substring that begins at position k and has length x, and the function find(t) finds the position of the first occurrence of a substring t.
The following code presents some string operations:
string a = "hatti"; string b = a+a; cout << b << "\n"; // hattihatti b[5] = ’v’;
cout << b << "\n"; // hattivatti string c = b.substr(3,4); cout << c << "\n"; // tiva
Set structures
A set is a data structure that maintains a collection of elements. The basic operations of sets are element insertion, search and removal.
The C++ standard library contains two set implementations: The structure set is based on a balanced binary tree and its operations work in O(logn) time. The structure unordered_set uses hashing, and its operations work in O(1) time on average.
The choice of which set implementation to use is often a matter of taste. The benefit of the set structure is that it maintains the order of the elements and provides functions that are not available in unordered_set. On the other hand, unordered_set can be more efficient.
The following code creates a set that contains integers, and shows some of the operations. The function insert adds an element to the set, the function count returns the number of occurrences of an element in the set, and the function erase removes an element from the set.
set<int> s;
s.insert(3);
s.insert(2);
s.insert(5); cout << s.count(3) << "\n"; // 1 cout << s.count(4) << "\n"; // 0
s.erase(3);
s.insert(4); cout << s.count(3) << "\n"; // 0 cout << s.count(4) << "\n"; // 1
A set can be used mostly like a vector, but it is not possible to access the elements using the [] notation. The following code creates a set, prints the number of elements in it, and then iterates through all the elements:
set<int> s = {2,5,6,8}; cout << s.size() << "\n"; // 4 for (auto x : s) { cout << x << "\n";
}
An important property of sets is that all their elements are distinct. Thus, the function count always returns either 0 (the element is not in the set) or 1 (the element is in the set), and the function insert never adds an element to the set if it is already there. The following code illustrates this:
set<int> s;
s.insert(5);
s.insert(5);
s.insert(5); cout << s.count(5) << "\n"; // 1
C++ also contains the structures multiset and unordered_multiset that otherwise work like set and unordered_set but they can contain multiple instances of an element. For example, in the following code all three instances of the number 5 are added to a multiset:
multiset<int> s; s.insert(5); s.insert(5); s.insert(5); cout << s.count(5) << "\n"; // 3 |
The function erase removes all instances of an element from a multiset:
s.erase(5);
cout << s.count(5) << "\n"; // 0
Often, only one instance should be removed, which can be done as follows:
s.erase(s.find(5)); cout << s.count(5) << "\n"; // 2
Map structures
A map is a generalized array that consists of key-value-pairs. While the keys in an ordinary array are always the consecutive integers 0,1,...,n−1, where n is the size of the array, the keys in a map can be of any data type and they do not have to be consecutive values.
The C++ standard library contains two map implementations that correspond to the set implementations: the structure map is based on a balanced binary tree and accessing elements takes O(logn) time, while the structure unordered_map uses hashing and accessing elements takes O(1) time on average.
The following code creates a map where the keys are strings and the values are integers:
map<string,int> m; m["monkey"] = 4; m["banana"] = 3; m["harpsichord"] = 9; cout << m["banana"] << "\n"; // 3
If the value of a key is requested but the map does not contain it, the key is automatically added to the map with a default value. For example, in the following code, the key ”aybabtu” with value 0 is added to the map.
map<string,int> m; cout << m["aybabtu"] << "\n"; // 0
The function count checks if a key exists in a map:
if (m.count("aybabtu")) {
// key exists
}
The following code prints all the keys and values in a map:
for (auto x : m) { cout << x.first << " " << x.second << "\n";
}
Iterators and ranges
Many functions in the C++ standard library operate with iterators. An iterator is a variable that points to an element in a data structure.
The often used iterators begin and end define a range that contains all elements in a data structure. The iterator begin points to the first element in the data structure, and the iterator end points to the position after the last element. The situation looks as follows:
{ 3, 4, 6, 8, 12, 13, 14, 17 }
↑ ↑
s.begin() s.end()
Note the asymmetry in the iterators: s.begin() points to an element in the data structure, while s.end() points outside the data structure. Thus, the range defined by the iterators is half-open.
Working with ranges
Iterators are used in C++ standard library functions that are given a range of elements in a data structure. Usually, we want to process all elements in a data structure, so the iterators begin and end are given for the function.
For example, the following code sorts a vector using the function sort, then reverses the order of the elements using the function reverse, and finally shuffles the order of the elements using the function random_shuffle.
sort(v.begin(), v.end()); reverse(v.begin(), v.end()); random_shuffle(v.begin(), v.end());
These functions can also be used with an ordinary array. In this case, the functions are given pointers to the array instead of iterators:
sort(a, a+n); reverse(a, a+n); random_shuffle(a, a+n);
Set iterators
Iterators are often used to access elements of a set. The following code creates an iterator it that points to the smallest element in a set:
set<int>::iterator it = s.begin();
A shorter way to write the code is as follows:
auto it = s.begin();
The element to which an iterator points can be accessed using the * symbol. For example, the following code prints the first element in the set:
auto it = s.begin(); cout << *it << "\n";
Iterators can be moved using the operators ++ (forward) and -- (backward), meaning that the iterator moves to the next or previous element in the set.
The following code prints all the elements in increasing order:
for (auto it = s.begin(); it != s.end(); it++) { cout << *it << "\n";
}
The following code prints the largest element in the set:
auto it = s.end(); it--; cout << *it << "\n";
The function find(x) returns an iterator that points to an element whose value is x. However, if the set does not contain x, the iterator will be end.
auto it = s.find(x); if (it == s.end()) { // x is not found } |
The function lower_bound(x) returns an iterator to the smallest element in the set whose value is at least x, and the function upper_bound(x) returns an iterator to the smallest element in the set whose value is larger than x. In both functions, if such an element does not exist, the return value is end. These functions are not supported by the unordered_set structure which does not maintain the order of the elements.
For example, the following code finds the element nearest to x:
auto it = s.lower_bound(x); if (it == s.begin()) { cout << *it << "\n"; } else if (it == s.end()) { it--; cout << *it << "\n"; } else { int a = *it; it--; int b = *it; if (x-b < a-x) cout << b << "\n"; else cout << a << "\n"; } |
The code assumes that the set is not empty, and goes through all possible cases using an iterator it. First, the iterator points to the smallest element whose value is at least x. If it equals begin, the corresponding element is nearest to x. If it equals end, the largest element in the set is nearest to x. If none of the previous cases hold, the element nearest to x is either the element that corresponds to it or the previous element.
Other structures
Bitset
A bitset is an array whose each value is either 0 or 1. For example, the following code creates a bitset that contains 10 elements:
bitset<10> s; s[1] = 1; s[3] = 1; s[4] = 1; s[7] = 1; cout << s[4] << "\n"; // 1 cout << s[5] << "\n"; // 0
The benefit of using bitsets is that they require less memory than ordinary arrays, because each element in a bitset only uses one bit of memory. For example, if n bits are stored in an int array, 32n bits of memory will be used, but a corresponding bitset only requires n bits of memory. In addition, the values of a bitset can be efficiently manipulated using bit operators, which makes it possible to optimize algorithms using bit sets.
The following code shows another way to create the above bitset:
bitset<10> s(string("0010011010")); // from right to left cout << s[4] << "\n"; // 1 cout << s[5] << "\n"; // 0
The function count returns the number of ones in the bitset:
bitset<10> s(string("0010011010")); cout << s.count() << "\n"; // 4
The following code shows examples of using bit operations:
bitset<10> a(string("0010110110")); bitset<10> b(string("1011011000")); cout << (a&b) << "\n"; // 0010010000 cout << (a|b) << "\n"; // 1011111110 cout << (a^b) << "\n"; // 1001101110
Deque
A deque is a dynamic array whose size can be efficiently changed at both ends of the array. Like a vector, a deque provides the functions push_back and pop_back, but it also includes the functions push_front and pop_front which are not available in a vector.
A deque can be used as follows:
deque<int> d;
d.push_back(5); // [5]
d.push_back(2); // [5,2]
d.push_front(3); // [3,5,2]
d.pop_back(); // [3,5]
d.pop_front(); // [5]
The internal implementation of a deque is more complex than that of a vector, and for this reason, a deque is slower than a vector. Still, both adding and removing elements take O(1) time on average at both ends.
Stack
A stack is a data structure that provides two O(1) time operations: adding an element to the top, and removing an element from the top. It is only possible to access the top element of a stack.
The following code shows how a stack can be used:
stack<int> s;
s.push(3);
s.push(2);
s.push(5); cout << s.top(); // 5
s.pop(); cout << s.top(); // 2
Queue
A queue also provides two O(1) time operations: adding an element to the end of the queue, and removing the first element in the queue. It is only possible to access the first and last element of a queue.
The following code shows how a queue can be used:
queue<int> q;
q.push(3);
q.push(2);
q.push(5); cout << q.front(); // 3
q.pop(); cout << q.front(); // 2
Priority queue
A priority queue maintains a set of elements. The supported operations are insertion and, depending on the type of the queue, retrieval and removal of either the minimum or maximum element. Insertion and removal take O(logn) time, and retrieval takes O(1) time.
While an ordered set efficiently supports all the operations of a priority queue, the benefit of using a priority queue is that it has smaller constant factors. A priority queue is usually implemented using a heap structure that is much simpler than a balanced binary tree used in an ordered set.
By default, the elements in a C++ priority queue are sorted in decreasing order, and it is possible to find and remove the largest element in the queue. The following code illustrates this:
priority_queue<int> q;
q.push(3);
q.push(5);
q.push(7);
q.push(2); cout << q.top() << "\n"; // 7
q.pop();
cout << q.top() << "\n"; // 5
q.pop();
q.push(6); cout << q.top() << "\n"; // 6
q.pop();
If we want to create a priority queue that supports finding and removing the smallest element, we can do it as follows:
priority_queue<int,vector<int>,greater<int>> q;
Policy-based data structures
The g++ compiler also supports some data structures that are not part of the C++ standard library. Such structures are called policy-based data structures. To use these structures, the following lines must be added to the code:
#include <ext/pb_ds/assoc_container.hpp> using namespace __gnu_pbds;
After this, we can define a data structure indexed_set that is like set but can be indexed like an array. The definition for int values is as follows:
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_set;
Now we can create a set as follows:
indexed_set s;
s.insert(2);
s.insert(3);
s.insert(7);
s.insert(9);
The speciality of this set is that we have access to the indices that the elements would have in a sorted array. The function find_by_order returns an iterator to the element at a given position:
auto x = s.find_by_order(2); cout << *x << "\n"; // 7
And the function order_of_key returns the position of a given element:
cout << s.order_of_key(7) << "\n"; // 2
If the element does not appear in the set, we get the position that the element would have in the set:
cout << s.order_of_key(6) << "\n"; // 2 cout << s.order_of_key(8) << "\n"; // 3
Both the functions work in logarithmic time.
Comparison to sorting
It is often possible to solve a problem using either data structures or sorting. Sometimes there are remarkable differences in the actual efficiency of these approaches, which may be hidden in their time complexities.
Let us consider a problem where we are given two lists A and B that both contain n elements. Our task is to calculate the number of elements that belong to both of the lists. For example, for the lists
A = [5,2,8,9,4] and B = [3,2,9,5],
the answer is 3 because the numbers 2, 5 and 9 belong to both of the lists.
A straightforward solution to the problem is to go through all pairs of elements in O(n2) time, but next we will focus on more efficient algorithms.
Algorithm 1
We construct a set of the elements that appear in A, and after this, we iterate through the elements of B and check for each elements if it also belongs to A. This is efficient because the elements of A are in a set. Using the set structure, the time complexity of the algorithm is O(nlogn).
Algorithm 2
It is not necessary to maintain an ordered set, so instead of the set structure we can also use the unordered_set structure. This is an easy way to make the algorithm more efficient, because we only have to change the underlying data structure. The time complexity of the new algorithm is O(n).
Algorithm 3
Instead of data structures, we can use sorting. First, we sort both lists A and
B. After this, we iterate through both the lists at the same time and find the common elements. The time complexity of sorting is O(nlogn), and the rest of the algorithm works in O(n) time, so the total time complexity is O(nlogn).
Efficiency comparison
The following table shows how efficient the above algorithms are when n varies and the elements of the lists are random integers between 1...109:
n Algorithm 1 Algorithm 2 Algorithm 3
106 1.5 s 0.3 s 0.2 s 2·106 3.7 s 0.8 s 0.3 s
3·106 5.7 s 1.3 s 0.5 s
4·106 7.7 s 1.7 s 0.7 s
5·106 10.0 s 2.3 s 0.9 s
Algorithms 1 and 2 are equal except that they use different set structures. In this problem, this choice has an important effect on the running time, because Algorithm 2 is 4–5 times faster than Algorithm 1.
However, the most efficient algorithm is Algorithm 3 which uses sorting. It only uses half the time compared to Algorithm 2. Interestingly, the time complexity of both Algorithm 1 and Algorithm 3 is O(nlogn), but despite this, Algorithm 3 is ten times faster. This can be explained by the fact that sorting is a simple procedure and it is done only once at the beginning of Algorithm 3, and the rest of the algorithm works in linear time. On the other hand, Algorithm 1 maintains a complex balanced binary tree during the whole algorithm.
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