###
Print power set of any given set OR Print all subsets of a set OR

Find all subsets of a set OR Print all subset of an array.

Given a set of numbers, print all the possible subsets of it including empty set.

###
**What is Power Set?**

In mathematics, the power set (or powerset) of any set S, written P(S), is the set of all subsets of S, including the empty set and S itself.Let's understand it with help of example.

**Case 1:**

Input: Set(S) = [1,2]

Output: PowerSet P(S) = [[ ], [2], [1], [2, 1]]

**Case 2:**

Input: Set(S) = [a,b,c]

Output: PowerSet P(S) = [[ ], [a], [b], [c], [a,b], [a, c], [b, c], [a, b, c]]

Solution

Solution

There can be many approach to solve this problem, here we will look at 2 simple approaches.

1.

**Using Tree representation approach.**

2.

**Using Bit manipulation approach.**

###
**Tree Representation Approach**

In this approach, we will draw a Tree like shown below,

There are 2 possibilities for each item of a set, It will be either included in a subset or not included.

Based on this fact,

**Lets draw a Tree for given Set [a, b, c].**

There are 2 possibilities of "a". It will be either included in a subset or not included.

Similarly, For each possibilities of "a", again there are 2 possibilities for "b".

"b" will be included for each possibilities of "a" or not included.

Similarly, For each possibilities of "b", again there are 2 possibilities for "c".

Again, For "c", there are 2 possibilities either it will be considered or will be ignored in a subset.

**Stop as there are no more items in a set remaining. Our Tree is complete and will look as shown in image above.**

Lets understand our example and draw a Tree for it.

**Set [a, b, c].**

**"a" is included.**

**1.**"b" will be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set.

**2.**"c" will not be included in a subset. END- No more element in Set.

**1.**"b" will not be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set.

**2.**"c" will not be included in a subset. END- No more element in Set.

**"a" is not included.**

**1.**"b" will be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set.

**2.**"c" will not be included in a subset. END- No more element in Set.

**1.**"b" will not be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set.

**2.**"c" will not be included in a subset. END- No more element in Set.

**Now, read the Tree from Bottom to Root. Wherever there is Yes, include the item in a subset, ignore otherwise.**

**"a" is included.**

**1.**"b" will be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set. =

**[ c, b, a ]**

**2.**"c" will not be included in a subset. END- No more element in Set. =

**[ b, a ]**

**1.**"b" will not be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set. =

**[ c, a ]**

**2.**"c" will not be included in a subset. END- No more element in Set. =

**[ a ]**

**"a" is not included.**

**1.**"b" will be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set. =

**[ c, b ]**

**2.**"c" will not be included in a subset. END- No more element in Set. =

**[ b ]**

**1.**"b" will not be included in a subset.

**2.**"c" will be included in a subset. END- No more element in Set. =

**[ c ]**

**2.**"c" will not be included in a subset. END- No more element in Set. =

**[ ]**

**Subset or PowerSet of [a, b, c] is [ [ c, b, a ], [ b, a ], [ c, a ], [ a ], [ c, b ], [ b ], [ c ], [ ] ]**

Understanding Program

Understanding Program

Algorithm we will use for printing subset of a set is,

**Step 1.**Collect empty subset ([ ]) first because empty subset will be subset of any set including null set.

**Step 2.**Now, For each item of a Set, Iterate subsets already found and

**1. Include a item in each subset and**

**2.**

**Not include a item in each subset.**

### Java Program for printing Subsets of set using Tree representation approach.

package javabypatel; import java.util.ArrayList; import java.util.Iterator; import java.util.List; public class FindAllSubsetOfSet { public static void main(String[] args) { System.out.println("\n"+createSubsetUsingTree("A")); } private static List<String> createSubsetUsingTree(String str){ List<String> result = new ArrayList<String>(); // take set if you want unique results. result.add("[]"); //If str is not null, then process further otherwise return empty set. if(str!=null && str.length()>0){ //Iterate each element of a set for (int i = 0; i < str.length(); i++) { //Working on str.charAt(i); //Store the result of subset of str.charAt(i) in tempList. List<String> tempList = new ArrayList<String>(); //Add str.charAt(i) in each item of result. Iterator<String> iter = result.iterator(); while(iter.hasNext()){ String val = iter.next(); // If val is [], it means str.charAt(i) is not included, So include it in result. if(val.equals("[]")){ tempList.add("[" + str.charAt(i) + "]"); }else{ //For each item, there will be 2 subset, one including it and one without including it. //If val is not [], it means it already contain some subset without str.charAt(i), So just include it. tempList.add("[" + val.substring(1,val.length()-1) + ", " + str.charAt(i) + "]"); } } //Add all subsets present in tempList to final result. result.addAll(tempList); } } return result; } }

###
**Recursive Approach**

package javabypatel; static void printSubsequences(int arr[], int index, String str) { if (index >= arr.length) { System.out.println("[" + str + "]"); return; } printSubsequences1(arr, index + 1, str + arr[index]); printSubsequences1(arr, index + 1, str); }

###
**Another Recursive Approach**

package javabypatel; static void printSubsequences(int arr[], int index, String str) { if (index >= arr.length) { System.out.println(str); return; } System.out.println(str); for (int i = index; i < arr.length; i++) { String str2 = str + arr[i]; printSubsequences(arr, i + 1, str2); } }

###
**Bit Manipulation Approach**

In this approach we will use Bits Manipulation for printing Subset of a set.

**Before going ahead, first lets understand how to check any bit is set in a byte.**

Binary representation of a number 1 = 0001, 2 = 0010, ...., 5 = 0101, ....

**How to identify whether LSB, that is bit at position 0 is set in 5(0101)?**

**Solution:**

Do a Logical AND of 1 and 5 that is (0001 & 0101).

If the result is greater than 0, then LSB, that is bit at position 0 is set in value 5.

**How to identify whether bit at position 1 is set in 5(0101)?**

**Solution:**

Do a Logical AND of 2 and 5 that is (0010 & 0101).

If the result is greater than 0, then bit at position 1 is set

So, to generalize it, if we want to check nth bit is set in a number,

So, to generalize it, if we want to check nth bit is set in a number,

**We will first take binary representation of 1 (0001),**

Step 1.

Step 1.

**Step 2.**Move the LSB in binary representation of 1 to nth position.

**Eg:**If we want to check whether bit at position 2 is set, then move 1 two times on right side,

= 1 << 2

= 0001 << 2

= 0100

**Step 3.**Once the LSB of binary 1 is moved to nth position, which we want to check,

just do logical AND of binary representation of 1 with the number,

If the result is greater than 0, then the bit is set, not otherwise.

**If we want to find subset of 3 characters [a, b, c], then how many total subset can be made out of it?**

If you observe, the number of subsets equals to 2 to the power of the number of elements in the set.

Number of subsets that can be made from set of 3 characters [a, b, c] is 2^3 = 8 subsets.

Number of subsets that can be made from set of 4 characters [a, b, c, d] is 2^4 = 16 subsets.

**So, If there are n characters [a, b, c.... n] in a set, then there will be 2^n total subsets.**

**How to find total number of subsets possible using bit manipulation?**

By shifting the bit representation of the number 1 by n, we will get 2^n.

Thus, if we shift the bit string of 1 by the number of elements in the given set, we'll get the number of subsets for that set.

For example, if we have S = [a, b, c], length of set here is 3,

So, for finding total subsets, we have to shift binary 1, 3 times,

1 << 3 = (0001 << 3) = (1000) = 2^3 = 8 subsets of set S.

For example, if we have S = [a, b, c], length of set here is 3,

So, for finding total subsets, we have to shift binary 1, 3 times,

1 << 3 = (0001 << 3) = (1000) = 2^3 = 8 subsets of set S.

If we compare subset of [a, b, c] to binaries representation of numbers from 0 to 7, we can find a relation with the bit sets in each number to subsets of [a, b, c].

**Understanding Program**

### Java Program for printing Subsets of set using Bit Manipulation approach.

package string; import java.util.ArrayList; import java.util.HashSet; import java.util.List; import java.util.Set; public class FindAllSubsetOfSet { public static void main(String[] args) { List<Object> list = new ArrayList<Object>(); list.add("a"); list.add("b"); list.add("c"); System.out.println(getSubsetUsingBitMap(list)); } private static Set<Set<Object>> getSubsetUsingBitMap(List<Object> list){ Set<Set<Object>> result = new HashSet<Set<Object>>(); int numOfSubsets = 1 << list.size(); //OR Math.pow(2, list.size()) // For i from 0 to 7 in case of [a, b, c], // we will pick 0(0,0,0) and check each bits to see any bit is set, // If set then element at corresponding position in a given Set need to be included in a subset. for(int i = 0; i < numOfSubsets; i++){ Set<Object> subset = new HashSet<Object>(); int mask = 1; // we will use this mask to check any bit is set in binary representation of value i. for(int k = 0; k < list.size(); k++){ if((mask & i) != 0){ // If result is !=0 (or >0) then bit is set. subset.add(list.get(k)); // include the corresponding element from a given set in a subset. } // check next bit in i. mask = mask << 1; } // add all subsets in final result. result.add(subset); } return result; } }

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**Enjoy !!!!**

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