feb19 - Art of Balance

general
1

Can Someone review my solution for the problem Art of Balance:
int offset(ll t[],ll goal,ll rows,ll excluded){
float off=0,set=0;
for(int i=0;i<rows;i++)
{
off=off+abs(goal - t[i]);
}
if(excluded==0){
return (off/2);}
else{

    for(int i=rows;i<=(rows+excluded);i++)
    {
        set=set+t[i];
    }
    return ((off+set)/2);
}

}
bool nonzero(int i)
{return (i!=0);}

int count(string s, char c)
{

int res = 0; 

for (int i=0;i<s.length();i++) 


	if (s[i] == c) 
		res++; 

return res;

}

int main()
{
ll t,occ[26],stdc,numofcounts,ans=0,modifiedlength,slength,y;
float special;
cin>>t;
while(t–)
{
string str;
cin>>str;

 slength=str.length();
    for(int i=0;i<=25;i++)
    {
	char c = 'A'+i; 
	occ[i]=count(str, c);
    }
    
     sort(occ,occ+26,greater<ll>());
       

    numofcounts=count_if(begin(occ),end(occ),nonzero);
    
    
     if(slength%numofcounts==0)
      { ans=offset(occ,slength/numofcounts,numofcounts,0);  }
    
      else{
          for(int i=0;;i++)
          {
              if(slength%(numofcounts-i)==0){
                  
                  
                  if((i)>(y=((slength/i)%numofcounts)))
                  {
                      modifiedlength=y+numofcounts;
                  }
                  else
                  {
                      modifiedlength=numofcounts-i;
                  }  
                  break;}
          }
          ans=offset(occ,slength/modifiedlength,modifiedlength,numofcounts-modifiedlength);
      }
      cout<<ans<<endl;
}
return 0;

}

thanking you!

2

for minimum operations, the balance string can have more than the number of distinct characters in the original string

3

you may go through following solution it may help you Solution