java - Algorithm to find the simplest combination of integers that has not yet been used -

i looking algorithm finding simplest combination of integers 0 5 (that 1 consists of fewest number of integers) has not yet been used (the used combinations in list).

the order matter , combinations should returned in list.

for example, list used numbers this:

{{0},{1},{2},{3},{4},{0,0},{0,1},{0,2},...,{2,1},{2,2},...,{1,5,4},...}

in case, algorithm should return list {5}, {5} combination consists of fewest integers.

if list looks this:

{{0},{1},{2},{3},{4},{5},{0,0},{0,1},{0,2},{0,3},{0,5},...}

the algorithm should return list 0 , 4 ({0,4}).

as used in java, java answer preferable pseudo-code or other programming languages usable too.

thank in advance!

i guess example 2 wrong: {{0},{1},{2},{3},{4},{5},{0,1},{0,2},{0,3},{0,5},...} smallest solution {0,0}, not {0,4}

complete solutions here:

import java.util.*;  public class algorithm {      static list<list<integer>> getchildren(list<integer> node){         list<list<integer>> children = new arraylist<list<integer>>();         for(int = 0; < 6; i++){             list<integer> child = new arraylist<integer>(node);             child.add(i);             children.add(child);         }         return children;     }      static list<integer> find(queue<list<integer>> queue, set<list<integer>> set){          for(;;){             list<integer> head = queue.poll();             if(!set.contains(head)){                 return head;             } else {                 for(list<integer> child : getchildren(head)){                     queue.add(child);                 }             }         }     }      public static void main(string[] arg) {         queue<list<integer>> queue = new linkedlist<list<integer>>();         for(int = 0; < 6; i++){             queue.add(collections.singletonlist(i));         }         // example {{0},{1},{2},{3},{4},{5},{0,1},{0,2},{0,3},{0,5},...}         set<list<integer>> task = new hashset<list<integer>>();         task.add(arrays.aslist(0));         task.add(arrays.aslist(1));         task.add(arrays.aslist(2));         task.add(arrays.aslist(3));         task.add(arrays.aslist(4));         task.add(arrays.aslist(5));         task.add(arrays.aslist(0, 1));         task.add(arrays.aslist(0, 2));         task.add(arrays.aslist(0, 3));         task.add(arrays.aslist(0, 5));          system.out.println(find(queue, task));     }  }