Problem: Find all the subsets of a given set.
Input: S = {a, b, c, d} Output: {}, {a} , {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, {a,b,c,d}
The total number of subsets of any given set is equal to 2^ (no. of elements in the set). If we carefully notice it is nothing but binary numbers from 0 to 15 which can be shown as below:
0000 |
{} |
0001 |
{a} |
0010 |
{b} |
0011 |
{a, b} |
0100 |
{c} |
0101 |
{a, c} |
0110 |
{b, c} |
0111 |
{a, b, c} |
1000 |
{d} |
1001 |
{a, d} |
1010 |
{b, d} |
1011 |
{a, b, d} |
1100 |
{c, d} |
1101 |
{a, c, d} |
1110 |
{b, c, d} |
1111 |
{a, b, c, d} |
Starting from right, 1 at ith position shows that the ith element of the set is present as 0 shows that the element is absent. Therefore, what we have to do is just generate the binary numbers from 0 to 2^n – 1, where n is the length of the set or the numbers of elements in the set.
Implementation:
Java
// A Java program to print all subsets of a set import java.io.IOException; class Main { // Print all subsets of given set[] static void printSubsets( char set[]) { int n = set.length; // Run a loop for printing all 2^n // subsets one by one for ( int i = 0 ; i < ( 1 <<n); i++) { System.out.print( "{ " ); // Print current subset for ( int j = 0 ; j < n; j++) // (1<<j) is a number with jth bit 1 // so when we 'and' them with the // subset number we get which numbers // are present in the subset and which // are not if ((i & ( 1 << j)) > 0 ) System.out.print(set[j] + " " ); System.out.println( "}" ); } } // Driver code public static void main(String[] args) { char set[] = { 'a' , 'b' , 'c' }; printSubsets(set); } } |
{ } { a } { b } { a b } { c } { a c } { b c } { a b c }
Time complexity: O(n * (2^n)) as the outer loop runs for O(2^n) and the inner loop runs for O(n).
Auxiliary Space :O(1), since no extra space is used.
Related Post:
Finding all subsets of a Set in C/C++
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