Sets STL Hackerrank Solution in C++ | STL Hackerrank Solution

Sets-STL Hackerrank Solution in C++. Sets are a part of the C++ STL. Sets are containers that store unique elements following a specific order. Here are some of the frequently used member functions of sets:

Declaration:

set<int>s; //Creates a set of integers.

Size:

int length=s.size(); //Gives the size of the set.

Insert:

s.insert(x); //Inserts an integer x into the set s.

Erasing an element:

s.erase(val); //Erases an integer val from the set s.

Finding an element:

set<int>::iterator itr=s.find(val); //Gives the iterator to the element val if it is found otherwise returns s.end() .

Ex: set<int>::iterator itr=s.find(100); //If 100 is not present then it==s.end().

To know more about sets click Here. Coming to the problem, you will be given Q queries. Each query is of one of the following three types:

• :x Add an element x to the set.
• :x Delete an element x from the set. (If the number x is not present in the set, then do nothing).
• :x If the number x is present in the set, then print "Yes"(without quotes) else print "No"(without quotes).

Input Format

The first line of the input contains Q where Q is the number of queries. The next Q lines contain 1 query each. Each query consists of two integers y and x where y is the type of the query and x is an integer.

Constraints

1 <= Q <= 10^5

1 <= y <= 3

1 <= x <= 10^9

Output Format

For queries of type 3 print "Yes"(without quotes) if the number x is present in the set and if the number is not present, then print "No"(without quotes).

Each query of type 3 should be printed in a new line.

Sample Input

8

1 9

1 6

1 10

1 4

3 6

3 14

2 6

3 6

Sample Output

Yes

No

No

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Sets STL Hackerrank Solution in C++ 

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
using namespace std;

int main()
{
	/*Enter your code here. Read input from STDIN. Print output to STDOUT */
	int n;
	set<int> s;
	for (int i = 0; i < n; i++)
	{
		int a;
		cin >> a;
		s.insert(a);
	}

	cin >> n;
	for (int i = 0; i < n; i++)
	{
		int q, t;
		cin >> t >> q;
		switch (t)
		{
			case 1:
				{
					s.insert(q);
					break;
				}

			case 2:
				{
					s.erase(q);
					break;
				}

			case 3:
				{
					set<int>::iterator itr = s.find(q);
					if (itr == s.end())
						cout << "No" << endl;
					else
						cout << "Yes" << endl;
					break;
				}
		}
	}

	return 0;
}

The Output of Sets STL Hackerrank Solution

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