Sets-STL C++


Problem Statement :


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:

1 x: Add an element x to the set.
2 x: Delete an element x from the set. (If the number x is not present in the set, then do nothing).
3 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.


Solution :



title-img 


                            Solution in C :

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


int main() {
    int c,t,n;
    scanf("%d",&n);
    set<int> ns;
    while(n--) {
        scanf("%d %d",&c,&t);
        switch(c) {
            case 1:
                ns.insert(t);
                break;
            case 2:
                ns.erase(t);
                break;
            case 3:
                if (ns.find(t)!=ns.end())
                    cout << "Yes"<<endl;
                else
                    cout <<"No"<<endl;
                break;
            default:
                cout<<"invalid switch value: "<<c<<endl;
                
        }
    }
    
    
    return 0;
}








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