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# 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 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

Recommended Practice

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 ``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 ``#include ` `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<

## 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

 ``

## PHP

 ``

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 ``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(``"
"``);``    ``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(``"
"``);``            ` `        ``} ``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 ``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 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 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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