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
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
If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
