The powerset of {1, 2, 3}
is:
{{}, {2}, {3}, {2, 3}, {1, 2}, {1, 3}, {1, 2, 3}, {1}}
Let\'s say I have a Set
in Java:<
Some of the solutions above suffer when the size of the set is large because they are creating a lot of object garbage to be collected and require copying data. How can we avoid that? We can take advantage of the fact that we know how big the result set size will be (2^n), preallocate an array that big, and just append to the end of it, never copying.
The speedup grows quickly with n. I compared it to João Silva's solution above. On my machine (all measurements approximate), n=13 is 5x faster, n=14 is 7x, n=15 is 12x, n=16 is 25x, n=17 is 75x, n=18 is 140x. So that garbage creation/collection and copying is dominating in what otherwise seem to be similar big-O solutions.
Preallocating the array at the beginning appears to be a win compared to letting it grow dynamically. With n=18, dynamic growing takes about twice as long overall.
public static List> powerSet(List originalSet) {
// result size will be 2^n, where n=size(originalset)
// good to initialize the array size to avoid dynamic growing
int resultSize = (int) Math.pow(2, originalSet.size());
// resultPowerSet is what we will return
List> resultPowerSet = new ArrayList>(resultSize);
// Initialize result with the empty set, which powersets contain by definition
resultPowerSet.add(new ArrayList(0));
// for every item in the original list
for (T itemFromOriginalSet : originalSet) {
// iterate through the existing powerset result
// loop through subset and append to the resultPowerset as we go
// must remember size at the beginning, before we append new elements
int startingResultSize = resultPowerSet.size();
for (int i=0; i oldSubset = resultPowerSet.get(i);
// create a new element by adding a new item from the original list
List newSubset = new ArrayList(oldSubset);
newSubset.add(itemFromOriginalSet);
// add this element to the result powerset (past startingResultSize)
resultPowerSet.add(newSubset);
}
}
return resultPowerSet;
}