How to generate a power set of a given set? - algorithm

Question

I am studying for an interview and I stumbled upon this question online under the "Math" category.
Generate power set of given set:
int A[] = {1,2,3,4,5};
int N = 5;
int Total = 1 << N;
for ( int i = 0; i < Total; i++ ) {
for ( int j = 0; j < N; j++) {
if ( (i >> j) & 1 )
cout << A[j];
}
cout <<endl;
}
Please I do not want an explicit answer. I just want clarifications and hints on how to approach this problem.
I checked power set algorithm on google and I still do not understand how to address this problem.
Also, could someone reiterate what the question is asking for.
Thank you.

Answer 1

Power set of a set A is the set of all of the subsets of A.
Not the most friendly definition in the world, but an example will help :
Eg. for {1, 2}, the subsets are : {}, {1}, {2}, {1, 2}
Thus, the power set is {{}, {1}, {2}, {1, 2}}
To generate the power set, observe how you create a subset : you go to each element one by one, and then either retain it or ignore it.
Let this decision be indicated by a bit (1/0).
Thus, to generate {1}, you will pick 1 and drop 2 (10).
On similar lines, you can write a bit vector for all the subsets :
{} -> 00
{1} -> 10
{2} -> 01
{1,2} -> 11
To reiterate : A subset if formed by including some or all of the elements of the original set. Thus, to create a subset, you go to each element, and then decide whether to keep it or drop it. This means that for each element, you have 2 decisions. Thus, for a set, you can end up with 2^N different decisions, corresponding to 2^N different subsets.
See if you can pick it up from here.

Answer 2

Create a power-set of: {"A", "B", "C"}.
Pseudo-code:
val set = {"A", "B", "C"}
val sets = {}
for item in set:
for set in sets:
sets.add(set + item)
sets.add({item})
sets.add({})
Algorithm explanation:
1) Initialise sets to an empty set: {}.
2) Iterate over each item in {"A", "B", "C"}
3) Iterate over each set in your sets.
3.1) Create a new set which is a copy of set.
3.2) Append the item to the new set.
3.3) Append the new set to sets.
4) Add the item to your sets.
4) Iteration is complete. Add the empty set to your resultSets.
Walkthrough:
Let's look at the contents of sets after each iteration:
Iteration 1, item = "A":
sets = {{"A"}}
Iteration 2, item = "B":
sets = {{"A"}, {"A", "B"}, {"B"}}
Iteration 3, item = "C":
sets = {{"A"}, {"A", "B"}, {"B"}, {"A", "C"}, {"A", "B", "C"}, {"B", "C"}, {"C"}}
Iteration complete, add empty set:
sets = {{"A"}, {"A", "B"}, {"B"}, {"A", "C"}, {"A", "B", "C"}, {"B", "C"}, {"C"}, {}}
The size of the sets is 2^|set| = 2^3 = 8 which is correct.
Example implementation in Java:
public static <T> List<List<T>> powerSet(List<T> input) {
List<List<T>> sets = new ArrayList<>();
for (T element : input) {
for (ListIterator<List<T>> setsIterator = sets.listIterator(); setsIterator.hasNext(); ) {
List<T> newSet = new ArrayList<>(setsIterator.next());
newSet.add(element);
setsIterator.add(newSet);
}
sets.add(new ArrayList<>(Arrays.asList(element)));
}
sets.add(new ArrayList<>());
return sets;
}
Input: [A, B, C]
Output: [[A], [A, C], [A, B], [A, B, C], [B], [B, C], [C], []]

Answer 3

Power set is just set of all subsets for given set. It includes all subsets (with empty set). It's well-known that there are 2N elements in this set, where N is count of elements in original set.
To build power set, following thing can be used:
Create a loop, which iterates all integers from 0 till 2N-1
Proceed to binary representation for each integer
Each binary representation is a set of N bits (for lesser numbers, add leading zeros). Each bit corresponds, if the certain set member is included in current subset.
Example, 3 numbers: a, b, c
number binary subset
0 000 {}
1 001 {c}
2 010 {b}
3 011 {b,c}
4 100 {a}
5 101 {a,c}
6 110 {a,b}
7 111 {a,b,c}

Answer 4

Well, you need to generate all subsets. For a set of size n, there are
2n subsets.
One way would be to iterate over the numbers from 0 to 2n - 1
and convert each to a list of binary digits, where 0 means exclude
that element and 1 means include it.
Another way would be with recursion, divide and conquer.

Answer 5

Generating all combination of a set (By including or not an item).
explain by example:
3 items in a set (or list). The possible subset will be:
000
100
010
001
110
101
011
111
The result is 2^(number of elements in the set).
As such we can generate all combinations of N items (with python) as follows:
def powerSet(items):
N = len(items)
for i in range(2**N):
comb=[]
for j in range(N):
if (i >> j) % 2 == 1:
comb.append(items[j])
yield comb
for x in powerSet([1,2,3]):
print (x)

Answer 6

You Get Something Like This by Implementing the top rated Answer.
def printPowerSet(set,set_size):
# set_size of power set of a set
# with set_size n is (2**n -1)
pow_set_size = (int) (math.pow(2, set_size));
counter = 0;
j = 0;
# Run from counter 000..0 to 111..1
for counter in range(0, pow_set_size):
for j in range(0, set_size):
# Check if jth bit in the
# counter is set If set then
# pront jth element from set
if((counter & (1 << j)) > 0):
print(set[j], end = "");
print("");

Answer 7

C# Solution
Time Complexity and Space Complexity: O(n*2^n)
public class Powerset
{
/*
P[1,2,3] = [[],[1],[2],[3],[1,2],[1,3],[2,3],[1,2,3]]
*/
public List<List<int>> PowersetSoln(List<int> array)
{
/*
We will start with an empty subset
loop through the number in the array
loop through subset generated till and add the number to each subsets
*/
var subsets = new List<List<int>>();
subsets.Add(new List<int>());
for (int i = 0; i < array.Count; i++)
{
int subsetLen = subsets.Count;
for (int innerSubset = 0; innerSubset < subsetLen; innerSubset++)
{
var newSubset = new List<int>(subsets[innerSubset]);
newSubset.Add(array[i]);
subsets.Add(newSubset);
}
}
return subsets;
}
}

Answer 8

Sample Java Code:
void printPowerSetHelper(String s, String r) {
if (s.length() > 0) {
printPowerSetHelper(s.substring(1), r + s.charAt(0));
printPowerSetHelper(s.substring(1), r);
}
if (r.length() > 0) System.out.println(r);
}
void printPowerSet(String s) {
printPowerSetHelper(s,"");
}