Last Updated : 08 Mar, 2024
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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 111Value of Counter Subset
000 -> Empty set
001 -> a
010 -> b
011 -> ab
100 -> c
101 -> ac
110 -> bc
111 -> abc
Recommended Practice
Power Set
Try It!
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
ababcacbcabc
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
abcabacbcabc
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
cbacbcacbaba
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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