Finding all subsets of a given set in Java
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.

// 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 obe
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);
}
}
Output:
{ } { a } { b } { a b } { c } { a c } { b c } { a b c }
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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