⭐ Must Read: web//https://www.geeksforgeeks.org/power set

Power Set Power set P(S) of a set S is the set of all subsets of S. For example S = {a, b, c} then P(s) = {{}, {a}, {b}, {c}, {a,b}, {a, c}, {b, c}, {a, b, c}}.

If S has n elements in it then P(s) will have 2^n elements

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Algorithm:

Input: Set[], set_size
1. Get the size of power set
    powet_set_size = pow(2, set_size)
2  Loop for counter from 0 to pow_set_size
     (a) Loop for i = 0 to set_size
          (i) If ith bit in counter is set
               Print ith element from set for this subset
     (b) Print separator for subsets i.e., newline

Example:

Set  = [a,b,c]
power_set_size = pow(2, 3) = 8
Run for binary counter = 000 to 111

Value of Counter            Subset
    000                    -> Empty set
    001                    -> a
    010                    -> b
    011                    -> ab
    100                    -> c
    101                    -> ac
    110                    -> bc
    111                    -> abc

Method 1:

C++

// C++ Program of above approach 
#include <iostream> 
#include <math.h> 
using namespace std; 
  
class gfg 
{ 
      
public: 
void printPowerSet(char *set, int set_size) 
{ 
    /*set_size of power set of a set with set_size 
    n is (2**n -1)*/
    unsigned int pow_set_size = pow(2, set_size); 
    int counter, j; 
  
    /*Run from counter 000..0 to 111..1*/
    for(counter = 0; counter < pow_set_size; counter++) 
    { 
    for(j = 0; j < set_size; j++) 
    { 
        /* Check if jth bit in the counter is set 
            If set then print jth element from set */
        if(counter & (1 << j)) 
            cout << set[j]; 
    } 
    cout << endl; 
    } 
} 
}; 
  
/*Driver code*/
int main() 
{ 
    gfg g; 
    char set[] = {'a','b','c'}; 
    g.printPowerSet(set, 3); 
    return 0; 
} 
  
// This code is contributed by SoM15242 

C

#include <stdio.h> 
#include <math.h> 
  
void printPowerSet(char *set, int set_size) 
{ 
    /*set_size of power set of a set with set_size 
      n is (2**n -1)*/
    unsigned int pow_set_size = pow(2, set_size); 
    int counter, j; 
  
    /*Run from counter 000..0 to 111..1*/
    for(counter = 0; counter < pow_set_size; counter++) 
    { 
      for(j = 0; j < set_size; j++) 
       { 
          /* Check if jth bit in the counter is set 
             If set then print jth element from set */
          if(counter & (1<<j)) 
            printf("%c", set[j]); 
       } 
       printf("\n"); 
    } 
} 
  
/*Driver program to test printPowerSet*/
int main() 
{ 
    char set[] = {'a','b','c'}; 
    printPowerSet(set, 3); 
    return 0; 
} 

Java

// Java program for power set 
import java .io.*; 
  
public class GFG { 
      
    static void printPowerSet(char []set, 
                            int set_size) 
    { 
          
        /*set_size of power set of a set 
        with set_size n is (2**n -1)*/
        long pow_set_size =  
            (long)Math.pow(2, set_size); 
        int counter, j; 
      
        /*Run from counter 000..0 to 
        111..1*/
        for(counter = 0; counter <  
                pow_set_size; counter++) 
        { 
            for(j = 0; j < set_size; j++) 
            { 
                /* Check if jth bit in the  
                counter is set If set then  
                print jth element from set */
                if((counter & (1 << j)) > 0) 
                    System.out.print(set[j]); 
            } 
              
            System.out.println(); 
        } 
    } 
      
    // Driver program to test printPowerSet 
    public static void main (String[] args) 
    { 
        char []set = {'a', 'b', 'c'}; 
        printPowerSet(set, 3); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# python3 program for power set 
  
import math; 
  
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  
            # print jth element from set  
            if((counter & (1 << j)) > 0): 
                print(set[j], end = ""); 
        print(""); 
  
# Driver program to test printPowerSet 
set = ['a', 'b', 'c']; 
printPowerSet(set, 3); 
  
# This code is contributed by mits. 

C#

// C# program for power set 
using System; 
  
class GFG { 
      
    static void printPowerSet(char []set, 
                            int set_size) 
    { 
        /*set_size of power set of a set 
        with set_size n is (2**n -1)*/
        uint pow_set_size =  
              (uint)Math.Pow(2, set_size); 
        int counter, j; 
      
        /*Run from counter 000..0 to 
        111..1*/
        for(counter = 0; counter <  
                   pow_set_size; counter++) 
        { 
            for(j = 0; j < set_size; j++) 
            { 
                /* Check if jth bit in the  
                counter is set If set then  
                print jth element from set */
                if((counter & (1 << j)) > 0) 
                    Console.Write(set[j]); 
            } 
              
            Console.WriteLine(); 
        } 
    } 
      
    // Driver program to test printPowerSet 
    public static void Main () 
    { 
        char []set = {'a', 'b', 'c'}; 
        printPowerSet(set, 3); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program for power set 
  
function printPowerSet($set, $set_size) 
{ 
  
    // set_size of power set of 
    // a set with set_size 
    // n is (2**n -1) 
    $pow_set_size = pow(2, $set_size); 
    $counter; $j; 
  
    // Run from counter 000..0 to 
    // 111..1 
    for($counter = 0; $counter < $pow_set_size; 
                                    $counter++) 
    { 
        for($j = 0; $j < $set_size; $j++) 
        { 
              
            /* Check if jth bit in  
               the counter is set 
               If set then print  
               jth element from set */
            if($counter & (1 << $j)) 
                echo $set[$j]; 
        } 
          
    echo "\n"; 
    } 
} 
  
    // Driver Code 
    $set = array('a','b','c'); 
    printPowerSet($set, 3); 
  
// This code is contributed by Vishal Tripathi 
?> 

Output:

a
b
ab
c
ac
bc
abc

Time Complexity: O(n2^n)

Method 2:
This method is specific to the python programming language. We can iterate a loop over 0 to the length of the set to obtain and generate all possible combinations of that string with the iterable length. The program below will give the implementation of the above idea.

#Python program to find powerset 
from itertools import combinations 
def print_powerset(string): 
    for i in range(0,len(string)+1): 
        for element in combinations(string,i): 
            print(''.join(element)) 
string=['a','b','c'] 
print_powerset(string)  

Output:


a
b
c
ab
ac
bc
abc

Recursive program to generate power set

Refer Power Set in Java for implementation in Java and more methods to print power set.

References:
http://en.wikipedia.org/wiki/Power_set

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Sep	OCT	Nov
	27
2019	2020	2021

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

Recommended Posts: