HackerEarth's basic programming solutions( Ali and Helping innocent people, Duration, Book of Potion making ) :

Problem 21: Ali and Helping innocent people

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()
{
    int count;
    string s;
    cin>>s;
if((((s[0]-'0')+(s[1]-'0'))%2==0)&&(((s[3]-'0')+(s[4]-'0'))%2==0)&&(((s[4]-'0')+
(s[5]-'0'))%2==0)&&(((s[7]-'0')+(s[8]-'0'))%2==0)&&(s[2]!='A') && 
  (s[2]!='E') && (s[2]!='I') && (s[2]!='O') && (s[2]!='U') && (s[2]!='Y') )
    {
        cout<<"valid"<<endl;
    }
    else
    {
        cout<<"invalid"<<endl;
    }
}

Since the string structure is same in each output , we can simply write all the valid conditions in the if parenthesis.

AND is used , therefore all conditions must be true in order to print valid.





Problem 22: Duration

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()
{
    int t,p,q,r,s,md,hd;
    cin>>t;
    while(t--)
    {
cin>>p>>q>>r>>s; //p=SH , q=SM , r=EH , s=EM
     if(q>s)
     {
     q=60-q;
     md=q+s;
     hd=r-(p+1);
     cout<<hd<<" "<<md<<endl;
     }
     else if(q==s)
     {
         md=0;
         hd=abs(p-r);
         cout<<hd<<" "<<md<<endl;
     }
     else{
        md=s-q;
        hd=(r-p);
        cout<<hd<<" "<<md<<endl;
     }
    }
}

You can use the example to understand the code.

input                 output
2                     0 30   
1 44 2 14             5 41 
2 42 8 23





Problem 23: Book of Potion making

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 int sum;
    string s;
    cin>>s;
    sum=(s[0]-'0')*1+(s[1]-'0')*2+(s[2]-'0')*3+(s[3]-'0')*4+(s[4]-'0')*5+(s[5]-'0')*6
  +(s[6]-'0')*7+(s[7]-'0')*8+(s[8]-'0')*9+(s[9]-'0')*10;
    if(sum%11==0)
    {
        cout<<"Legal ISBN"<<endl;
    }
    else
    {
        cout<<"Illegal ISBN"<<endl;
    }
}

Since the string structure is same in each output , we can simply write all the sum and then check whether it's divisible by 11 or not.

 

 

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

 

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