HackerEarth's basic programming solutions( Minimize Cost, Magical Word, Best Index ) :

Problem 42: Minimize Cost

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)

"Still working on this problem....."
 

 

 

 

Problem 43: Magical Word

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;

   cin>>t;

   while(t--)

   {

      int l;

      string s;

      cin>>l>>s;

      for(int i=0;i<s.length();i++)

       {

         if(s[i]<=69 && s[i]>=65)

         s[i]='C';

         else if(s[i]<=72 && s[i]>=70)

         s[i]='G';

         else if(s[i]<=76 && s[i]>=73)

         s[i]='I';

         else if(s[i]<=81 && s[i]>=77)

         s[i]='O';

         else if(s[i]<=86 && s[i]>=82)

         s[i]='S';

         else if(s[i]<=93 && s[i]>=87)

         s[i]='Y';

         else if(s[i]<=99 && s[i]>=94)

         s[i]='a';

         else if(s[i]<=102 && s[i]>=100)

         s[i]='e';

         else if(s[i]<=105 && s[i]>=103)

         s[i]='g';

         else if(s[i]<=108 && s[i]>=106)

         s[i]='k';

         else if(s[i]<=111 && s[i]>=109)

         s[i]='m';

         else if(s[i]<=127 && s[i]>=112)

         s[i]='q';

else

         s[i]='C';

         }

         cout<<s<<endl;

     }  

}

This code is simple. There's no need for any explanation. Take a look at the ASCII table and try to understand what i did.
 

 

 

 

Problem 44: Best Index

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 i,j,l,k,n;

cin>>n;

long long a[n+1];

long double sum,val=-9999999999;

a[0]=0;

for(i=1;i<n+1;i++)

{

//same approach used as in play with numbers

cin>>a[i];

a[i]+=a[i-1];

}

for(i=1;i<n+1;i++)

{

sum=0;

k=i;

l=1;

while(k+l<n)

{

l++;

k+=l;

}

sum=a[k]-a[i-1];

if(sum>val)

val=sum;

}

cout<<fixed<<setprecision(0)<<val<<endl; //see note below

}

If you will use only setprecision(0) , the result will come in scientific notation , so use fixed with it , if you don't want that. The answer will not be accepted if you use alone setprecision(0).

By the way you can also use directly this ---> printf("%.0Lf\n",val);

 

 

 

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