feb19 - Art of Balance

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!

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

you may go through following solution it may help you Solution

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