Power Set

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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 2ⁿ elements

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

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

Method 1:
For a given set[] S, the power set can be found by generating all binary numbers between 0 and 2ⁿ-1, where n is the size of the set.
For example, for the set S {x, y, z}, generate all binary numbers from 0 to 2³-1 and for each generated number, the corresponding set can be found by considering set bits in the number.

Below is the implementation of the above approach.

C++

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print all the power set 
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() 
{ 
    char set[] = { 'a', 'b', 'c' }; 
    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. 

Javascript

<script> 
// javascript program for power setpublic  
  
    function printPowerSet(set, set_size)  
    { 
  
        /* 
         * set_size of power set of a set with set_size n is (2**n -1) 
         */
        var pow_set_size = parseInt(Math.pow(2, set_size)); 
        var 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) 
                    document.write(set[j]); 
            } 
            document.write("<br/>"); 
        } 
    } 
  
    // Driver program to test printPowerSet 
        let set = [ 'a', 'b', 'c' ]; 
        printPowerSet(set, 3); 
  
// This code is contributed by shikhasingrajput  
</script> 

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(n2ⁿ)
Auxiliary Space: O(1)

Method 2: (sorted by cardinality)

In auxiliary array of bool set all elements to 0. That represent an empty set. Set first element of auxiliary array to 1 and generate all permutations to produce all subsets with one element. Then set the second element to 1 which will produce all subsets with two elements, repeat until all elements are included.

Below is the implementation of the above approach.

C++

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print all the power set 
void printPowerSet(char set[], int n) 
{ 
    bool *contain = new bool[n]{0}; 
      
    // Empty subset 
    cout << "" << endl; 
    for(int i = 0; i < n; i++) 
    { 
        contain[i] = 1; 
        // All permutation 
        do
        { 
            for(int j = 0; j < n; j++) 
                if(contain[j]) 
                    cout << set[j]; 
            cout << endl; 
        } while(prev_permutation(contain, contain + n)); 
    } 
} 
  
/*Driver code*/
int main() 
{ 
    char set[] = {'a','b','c'}; 
    printPowerSet(set, 3); 
    return 0; 
} 
  
// This code is contributed by zlatkodamijanic

Java

// Java program for the above approach 
  
import java.util.*; 
  
class GFG 
{ 
  
  // A function to reverse only the indices in the 
  // range [l, r] 
  static int[] reverse(int[] arr, int l, int r) 
  { 
    int d = (r - l + 1) / 2; 
    for (int i = 0; i < d; i++) { 
      int t = arr[l + i]; 
      arr[l + i] = arr[r - i]; 
      arr[r - i] = t; 
    } 
    return arr; 
  } 
  // A function which gives previous 
  // permutation of the array 
  // and returns true if a permutation 
  // exists. 
  static boolean prev_permutation(int[] str) 
  { 
    // Find index of the last 
    // element of the string 
    int n = str.length - 1; 
  
    // Find largest index i such 
    // that str[i - 1] > str[i] 
    int i = n; 
    while (i > 0 && str[i - 1] <= str[i]) { 
      i--; 
    } 
  
    // If string is sorted in 
    // ascending order we're 
    // at the last permutation 
    if (i <= 0) { 
      return false; 
    } 
  
    // Note - str[i..n] is sorted 
    // in ascending order Find 
    // rightmost element's index 
    // that is less than str[i - 1] 
    int j = i - 1; 
    while (j + 1 <= n && str[j + 1] < str[i - 1]) { 
      j++; 
    } 
  
    // Swap character at i-1 with j 
    int temper = str[i - 1]; 
    str[i - 1] = str[j]; 
    str[j] = temper; 
  
    // Reverse the substring [i..n] 
    str = reverse(str, i, str.length - 1); 
  
    return true; 
  } 
  
  // Function to print all the power set 
  static void printPowerSet(char[] set, int n) 
  { 
  
    int[] contain = new int[n]; 
    for (int i = 0; i < n; i++) 
      contain[i] = 0; 
  
    // Empty subset 
    System.out.println(); 
    for (int i = 0; i < n; i++) { 
      contain[i] = 1; 
  
      // To avoid changing original 'contain' 
      // array creating a copy of it i.e. 
      // "Contain" 
      int[] Contain = new int[n]; 
      for (int indx = 0; indx < n; indx++) { 
        Contain[indx] = contain[indx]; 
      } 
  
      // All permutation 
      do { 
        for (int j = 0; j < n; j++) { 
          if (Contain[j] != 0) { 
            System.out.print(set[j]); 
          } 
        } 
        System.out.print("\n"); 
  
      } while (prev_permutation(Contain)); 
    } 
  } 
  
  /*Driver code*/
  public static void main(String[] args) 
  { 
    char[] set = { 'a', 'b', 'c' }; 
    printPowerSet(set, 3); 
  } 
} 
  
// This code is contributed by phasing17

Python3

# Python3 program for the above approach 
  
# A function which gives previous 
# permutation of the array 
# and returns true if a permutation 
# exists. 
def prev_permutation(str): 
  
    # Find index of the last 
    # element of the string 
    n = len(str) - 1
  
    # Find largest index i such 
    # that str[i - 1] > str[i] 
    i = n 
    while (i > 0 and str[i - 1] <= str[i]): 
        i -= 1
  
    # If string is sorted in 
    # ascending order we're 
    # at the last permutation 
    if (i <= 0): 
        return False
  
    # Note - str[i..n] is sorted 
    # in ascending order Find 
    # rightmost element's index 
    # that is less than str[i - 1] 
    j = i - 1
    while (j + 1 <= n and str[j + 1] < str[i - 1]): 
        j += 1
  
    # Swap character at i-1 with j 
    temper = str[i - 1] 
    str[i - 1] = str[j] 
    str[j] = temper 
  
    # Reverse the substring [i..n] 
    size = n-i+1
    for idx in range(int(size / 2)): 
        temp = str[idx + i] 
        str[idx + i] = str[n - idx] 
        str[n - idx] = temp 
  
    return True
  
# Function to print all the power set 
def printPowerSet(set, n): 
  
    contain = [0 for _ in range(n)] 
  
    # Empty subset 
    print() 
  
    for i in range(n): 
        contain[i] = 1
  
        # To avoid changing original 'contain' 
        # array creating a copy of it i.e. 
        # "Contain" 
        Contain = contain.copy() 
  
        # All permutation 
        while True: 
            for j in range(n): 
                if (Contain[j]): 
                    print(set[j], end="") 
            print() 
            if not prev_permutation(Contain): 
                break
  
# Driver code 
set = ['a', 'b', 'c'] 
printPowerSet(set, 3) 
  
# This code is contributed by phasing17 

C#

// C# program for the above approach 
  
using System; 
using System.Linq; 
using System.Collections.Generic; 
  
class GFG 
{ 
  
  // A function which gives previous 
  // permutation of the array 
  // and returns true if a permutation 
  // exists. 
  static bool prev_permutation(int[] str) 
  { 
    // Find index of the last 
    // element of the string 
    int n = str.Length - 1; 
  
    // Find largest index i such 
    // that str[i - 1] > str[i] 
    int i = n; 
    while (i > 0 && str[i - 1] <= str[i]) { 
      i--; 
    } 
  
    // If string is sorted in 
    // ascending order we're 
    // at the last permutation 
    if (i <= 0) { 
      return false; 
    } 
  
    // Note - str[i..n] is sorted 
    // in ascending order Find 
    // rightmost element's index 
    // that is less than str[i - 1] 
    int j = i - 1; 
    while (j + 1 <= n && str[j + 1] < str[i - 1]) { 
      j++; 
    } 
  
    // Swap character at i-1 with j 
    var temper = str[i - 1]; 
    str[i - 1] = str[j]; 
    str[j] = temper; 
  
    // Reverse the substring [i..n] 
    int size = n - i + 1; 
    Array.Reverse(str, i, size); 
  
    return true; 
  } 
  
  // Function to print all the power set 
  static void printPowerSet(char[] set, int n) 
  { 
  
    int[] contain = new int[n]; 
    for (int i = 0; i < n; i++) 
      contain[i] = 0; 
  
    // Empty subset 
    Console.WriteLine(); 
    for (int i = 0; i < n; i++) { 
      contain[i] = 1; 
  
      // To avoid changing original 'contain' 
      // array creating a copy of it i.e. 
      // "Contain" 
      int[] Contain = new int[n]; 
      for (int indx = 0; indx < n; indx++) { 
        Contain[indx] = contain[indx]; 
      } 
  
      // All permutation 
      do { 
        for (int j = 0; j < n; j++) { 
          if (Contain[j] != 0) { 
            Console.Write(set[j]); 
          } 
        } 
        Console.Write("\n"); 
  
      } while (prev_permutation(Contain)); 
    } 
  } 
  
  /*Driver code*/
  public static void Main(string[] args) 
  { 
    char[] set = { 'a', 'b', 'c' }; 
    printPowerSet(set, 3); 
  } 
} 
  
// This code is contributed by phasing17

Javascript

// JavaScript program for the above approach 
  
// A function which gives previous  
// permutation of the array 
// and returns true if a permutation  
// exists.  
function prev_permutation(str){     
    // Find index of the last 
    // element of the string 
    let n = str.length - 1; 
   
    // Find largest index i such 
    // that str[i - 1] > str[i] 
    let i = n; 
    while (i > 0 && str[i - 1] <= str[i]){ 
        i--; 
    }    
        
    // If string is sorted in 
    // ascending order we're 
    // at the last permutation 
    if (i <= 0){ 
        return false; 
    } 
   
    // Note - str[i..n] is sorted 
    // in ascending order Find 
    // rightmost element's index 
    // that is less than str[i - 1] 
    let j = i - 1; 
    while (j + 1 <= n && str[j + 1] < str[i - 1]){ 
        j++; 
    } 
      
    // Swap character at i-1 with j 
    const temper = str[i - 1]; 
    str[i - 1] = str[j]; 
    str[j] = temper; 
      
    // Reverse the substring [i..n] 
    let size = n-i+1; 
    for (let idx = 0; idx < Math.floor(size / 2); idx++) { 
        let temp = str[idx + i]; 
        str[idx + i] = str[n - idx]; 
        str[n - idx] = temp; 
    } 
      
    return true; 
} 
  
// Function to print all the power set 
function printPowerSet(set, n){    
  
    let contain = new Array(n).fill(0); 
  
    // Empty subset 
    document.write("<br>"); 
    for(let i = 0; i < n; i++){ 
        contain[i] = 1;  
          
        // To avoid changing original 'contain'  
        // array creating a copy of it i.e. 
        // "Contain" 
        let Contain = new Array(n); 
        for(let indx = 0; indx < n; indx++){ 
            Contain[indx] = contain[indx]; 
        } 
  
        // All permutation 
        do{ 
            for(let j = 0; j < n; j++){                 
                if(Contain[j]){ 
                    document.write(set[j]); 
                }   
            } 
            document.write("<br>"); 
              
        } while(prev_permutation(Contain)); 
    } 
} 
  
/*Driver code*/
const set = ['a','b','c']; 
printPowerSet(set, 3); 
  
// This code is contributed by Gautam goel (gautamgoel962)

Output

a
b
c
ab
ac
bc
abc

Time Complexity: O(n2ⁿ)
Auxiliary Space: O(n)

Method 3:

We can use backtrack here, we have two choices first consider that element then don’t consider that element.

Below is the implementation of the above approach.

C++

#include <bits/stdc++.h> 
using namespace std; 
  
void findPowerSet(char* s, vector<char> &res, int n){ 
        if (n == 0) { 
            for (auto i: res)  
              cout << i; 
            cout << "\n"; 
            return; 
              
        } 
         res.push_back(s[n - 1]); 
         findPowerSet(s, res, n - 1); 
         res.pop_back();                     
         findPowerSet(s, res, n - 1); 
    } 
      
void printPowerSet(char* s, int n){ 
    vector<char> ans; 
    findPowerSet(s, ans, n); 
} 
  
  
int main() 
{ 
    char set[] = { 'a', 'b', 'c' }; 
    printPowerSet(set, 3); 
    return 0; 
}

Java

import java.util.*; 
   
class Main 
{ 
    public static void findPowerSet(char []s, Deque<Character> res,int n){ 
        if (n == 0){ 
          for (Character element : res) 
             System.out.print(element); 
          System.out.println(); 
            return; 
        } 
        res.addLast(s[n - 1]); 
        findPowerSet(s, res, n - 1); 
        res.removeLast();                     
        findPowerSet(s, res, n - 1); 
    } 
   
    public static void main(String[] args) 
    { 
        char []set = {'a', 'b', 'c'}; 
        Deque<Character> res = new ArrayDeque<>(); 
        findPowerSet(set, res, 3); 
    } 
} 

Python3

# Python3 program to implement the approach 
  
# Function to build the power sets 
def findPowerSet(s, res, n): 
    if (n == 0): 
        for i in res: 
            print(i, end="") 
        print() 
        return
  
    # append the subset to result 
    res.append(s[n - 1]) 
    findPowerSet(s, res, n - 1) 
    res.pop() 
    findPowerSet(s, res, n - 1) 
  
# Function to print the power set 
def printPowerSet(s, n): 
    ans = [] 
    findPowerSet(s, ans, n) 
  
# Driver code 
set = ['a', 'b', 'c'] 
printPowerSet(set, 3) 
  
# This code is contributed by phasing17 

C#

// C# code to implement the approach 
  
using System; 
using System.Collections.Generic; 
  
class GFG  
{ 
    // function to build the power set 
    public static void findPowerSet(char[] s, 
                                    List<char> res, int n) 
    { 
  
        // if the end is reached 
        // display all elements of res 
        if (n == 0) { 
            foreach(var element in res) 
                Console.Write(element); 
            Console.WriteLine(); 
            return; 
        } 
  
        // append the subset to res 
        res.Add(s[n - 1]); 
        findPowerSet(s, res, n - 1); 
        res.RemoveAt(res.Count - 1); 
        findPowerSet(s, res, n - 1); 
    } 
  
    // Driver code 
    public static void Main(string[] args) 
    { 
        char[] set = { 'a', 'b', 'c' }; 
        List<char> res = new List<char>(); 
  
        // Function call 
        findPowerSet(set, res, 3); 
    } 
} 
  
// This code is contributed by phasing17

Javascript

// JavaScript program to implement the approach 
  
// Function to build the power sets 
function findPowerSet(s, res, n) 
{ 
    if (n == 0) 
    { 
        for (var i of res) 
            process.stdout.write(i + ""); 
        process.stdout.write("\n"); 
        return; 
    } 
  
    // append the subset to result 
    res.push(s[n - 1]); 
    findPowerSet(s, res, n - 1); 
    res.pop(); 
    findPowerSet(s, res, n - 1); 
} 
  
// Function to print the power set 
function printPowerSet(s, n) 
{ 
    let ans = []; 
    findPowerSet(s, ans, n); 
} 
  
  
// Driver code 
let set = ['a', 'b', 'c']; 
printPowerSet(set, 3); 
  
  
// This code is contributed by phasing17 

Output

cba
cb
ca
c
ba
b
a

Time Complexity: O(2^n)
Auxiliary Space: O(n)

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

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