CareerCup

fun PrintChar(string A , string B){
     bool isPresentInB[128]=false ; // initialize all elements with false
     for(int i=0;i<B.size();i++)isPresentInB[B[i]]=true;
     bool isPresentInAbutNotInB[128]=false;
     for(int i=0;i<A.size();i++)if(!isPresentInB[A[i]]) isPresentInAbutNotInB[A[i]]=true;
  
     for(int i=0;i<128;i++)if(isPresentInAbutNotInB[i]) printf("%c\n",i);
}

temp = String A;
foreach elem in String B {
  if(elem is present in temp) {
    delete elem from temp;
  }
}
print temp;

you can insert all the characters in string B into a hash table, than check every character in string A. It takes linear time.
If you are not conformable with hashing, you can sort A and B respectively and than identify the characters in A but not in B by a scan in a similar way as merge sort.

+1 for the anonymous solution posted above. I thought of typing the same... Sigh...