Hackearth's Data Structures solutions( SnackDown Contest, Mark The Answer, Most Frequent ) :

Problem 16: SnackDown Contest

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder) 

#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long t;

    cin>>t;

    while(t--)

    {

        long long n,p,q,i;

        cin>>n;

        cin>>p;

        vector<long long>v1(p,0);

        for(i=0;i<p;i++)

        {

            cin>>v1[i];

        }

        cin>>q;

        vector<long long>v2(q,0);

        for(i=0;i<q;i++)

        {

            cin>>v2[i];

        }

        vector<long long>v3(p+q);

        merge(v1.begin(),v1.end(),v2.begin(),v2.end(),v3.begin()); //vector v3=v1+v2

        set<long long>s(v3.begin(),v3.end()); //converting vector v3 into set

        long long test=1;

        for(auto &e:s) //set traversal , see note below

        {

            if(e==test)

            {

                test++;

            }

        }

        if(test==n+1)

        {

            cout<<"YES"<<endl;

        }

        else

        {

            cout<<"NO"<<endl;

        }

    }

}

The advantage of converting merged vector v3 into set s is that it removes all the duplicate elements (helps when there's a question which can be solved by both) and also sort the elements.

So now we have to check only if the elements of the set forms a proper counting up to n , if they do that means all questions can be solved by them and if there's a discontinuity that means there's a question which can't be solved by both of them.

 

 

 

 

Problem 17: Mark The Answer

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder) 

#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long n,x,e,count=0,ct2=0; //e=element , n=no. of elements , ct2=count2

    cin>>n>>x;

    while(n--)

    {

        cin>>e;

        if(e<=x)

        {

            count++;

        }

        else

        {

            ct2++;

            if(ct2==2)

            {

                break;

            }

        }

    }

cout<<count;

}

This code is simple. There's no need for any explanation.

 

 

 

 

 

 

Problem 18: Most Frequent

Solution: (in c++)

( please guys before moving to the solution try it yourself at least 3-4 times , if you really wanna become a good coder)

#include<bits/stdc++.h>

using namespace std;

int main()

{

map<int,int>m;

int n,i,p=0,q;

cin>>n;

int a[n];

for(i=0;i<n;i++)

{

cin>>a[i];

m[a[i]]++; //see problem no. 3 explanation

}

for(auto &e:m) //map traversal

{

if(e.second>p)

{

p=e.second;

q=e.first;

}

}

cout<<q;

}

This code is simple. There's no need for any explanation.

 

 

 

 

Guys , if you have any queries related to a question or need more explanation, comment down below!