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Sets-STL

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333 Discussions

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  • dangis_bakutis
     + 0 comments
    #include <iostream>
#include <set>

using namespace std;

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

    set<int> set;

    while (q > 0)
    {
        int y, x;
        cin >> y >> x;

        switch (y)
        {
        case 1:
        {
            set.insert(x);
            break;
        }
        case 2:
        {
            set.erase(x);
            break;
        }
        default:
            if (set.end() != set.find(x))
                cout << "Yes" << endl;
            else
                cout << "No" << endl;
            break;
        }

        q--;
    }
}

  • akshaykandasamy1
     + 0 comments

    int main() { set s; int q; int a; int y; cin>>q; while(q>0){ cin>>a; if(a==1){ cin>>y;

            s.insert(y);
    }
    else if(a==2){
        cin>>y;
        auto it=s.find(y);
        if(it!=s.end()){
            s.erase(*it);
        }
        else{
            continue;
        }
    }
    else if(a==3){
        cin>>y;
        auto itr=s.find(y);
        if(itr!=s.end()){
            cout<<"Yes"<<endl;
        }
        else{
            cout<<"No"<<endl;
        }

    }
    q--;
}  
return 0;

    }

  • yashdeora98294
     + 0 comments

    Here is Sets-STL problem solution in C++ - https://programmingoneonone.com/hackerrank-sets-stl-solution-in-cpp.html

  • buivanchung22109
     + 0 comments

    //M.Kohaku

    include

    using namespace std;

    int main(){ sets; int n; cin >> n; for(int i=0;i p; cin >> p.first >> p.second; int x = p.first; int y = p.second; if(x==1){ s.insert(y); } else if(x==2){ s.erase(y); } else if(x==3){ if(s.count(y)) cout << "Yes" << endl; else cout << "No" << endl; } } }

  • kubalak_2k18
     + 0 comments

    My solution:

    int main() {
    set<int> values;
    int queries, qType, qVal;
    cin >> queries; 
    while(queries) {
        cin >> qType >> qVal;
        switch (qType) {
            case 1:
                values.insert(qVal);
                break;
            case 2:
                values.erase(qVal);
                break;
            case 3:
                if(values.find(qVal) != values.end())
                    cout << "Yes\n";
                else
                    cout << "No\n";
                break;
        }
        queries--;
    }
    return 0;
}