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In this article, we have explored what is the meaning of Power Set of String and the algorithm to generate Power Set of String in Lexicographic order.

## Table of contents

- Power Set
- Lexicographic order
- What is Power set of string in Lexicographic order?
- Algorithm
- Complextiy

## Power Set

We are all familiar with term "SET" at somewhere, probably in set theory (or in mathematics).

What is set? The answer is "It is a collection of well defined unordered distinct objects".

Example:

- The collection of all tall boys in the class.

It is not a set, because we do not know that which height we can treat as tall. - The collection of all tall boys whose height is greater or equal to 165cm in the class.

It is a set.

## Question

#### Which of the following is set?

What is Power Set ? "It is a collection of all possible subsets of a given set is called Power Set". The power set of 'A' is denoted by P(A).

Example:

- Given set S: {a, b}

Then Power set P(S): {" ", a, b, ab}

Note: Taking " " as NULLSET. - Set S: {a, b, c}

Power set P(S): {" ", a, b, c, ab, ac, bc, abc}

Now, Formula to find the number of elements in Power Set,

Lets take Set A: {a, b, c}, Number of elements in Set 'A' are 3, and the power set P(A): {" ", a, b, c, ab, ac, bc, abc} and Number of elements in P(A) are C(3,0) + C(3,1) + C(3,2) + C(3,3) = 8.

Therefore, If Number of elements in Set A are 'N', then Number of elements in Power Set P(A) are ( C(N,0) + C(N,1) + C(N,2) + . . . + C(N,N) ) = 2^N.

## Question

#### Which of the option is the power set of set A= { {1}, {2,3} }?

## Lexicographic order

Lexicographic order is a way of ordering words in alphabatical order based on their components alphabets. It is also known as Dictionary Order.

Example: algorithm, worst, average, best

Lexicographic order: algorithm, average, best, worst

## Power Set of String in Lexicographic order

In this, we are given a string S, we have to find the power set of this string in lexicographical order and print them.

Example:

Say S: {a, b, c}

Power set P(S): {" ", a, b, c, ab, ac, bc, abc}

Output should look like: [ " ", a, ab, abc, ac, b, bc, c ]

### Algorithm

###### Using Recursion

Arguments passed in Power set function are:

- S: Given input String
- Size: Length of string S
- pos: Index in current permutation
- sset: Stores current Permutation Subset

Begin power-set

- power-set function is having four arguments given input string 'S', size of S, current position "pos", and subset string "sset" which will stores subset.
- Base condition: if current pos is equal to the size of given string S then return.
- print the subset string.
- Run a for-loop for i: current pos +1 to size of give string.
- concatenate the element of S[i] into subset string(sset).
- Now call the power-set function by giving S, size, i+1, sset
- remove one element from the last of sset.

- return

End power-set

```
Using Recursion method
power-set(S, size, pos, sset){
if(pos is equals to the size)
then
return
end if
print string sset
for i: pos + 1 to size
concatenate the element of S[i] into sset
power-set(S, size, i+1, sset)
remove one(1) elements from string sset from end
end for
return
}
```

###### Code

```
#include <bits/stdc++.h>
using namespace std;
//-----------Power-Set Function---------------//
void power_set(string S, int n, int pos = -1, string sset = "") {
//Base case condition
if (pos == n)
return;
//Print string in Subset
cout<<sset<<"\n";
for (int i = pos + 1; i < n; i++) {
sset+= S[i];
power_set(S, n, i, sset);
//BackTracking
sset = sset.erase(sset.size() - 1);
}
return;
}
//----------Remove Duplicate Elements--------//
string duplicate(string s){
if(s.size()==0){
return "";
}
string x="";
int n=s.size(),i=0;
x+=s[0];
for(int j=1;j<n;j++){
if(x[i]!=s[j]){
i++;
x+=s[j];
}
}
return x;
}
//-----------Driver Function---------------//
int main() {
string S;
cout<<"Enter string: \n";
cin>>S;
cout<<"Power Set of the string '"<<S<<"' is :\n";
//sort the elements
sort(S.begin(),S.end());
//call duplicate() to remove duplicate elements
S=duplicate(S);
//calling power-set function
power_set(S, S.size());
return 0;
}
```

###### Output

```
Enter string:
merge
Power Set of the string 'merge' is :
e
eg
egm
egmr
egr
em
emr
er
g
gm
gmr
gr
m
mr
r
```

###### Using Iteration

Keep some logical things in mind, given below:

- '&'(Bitwise And) works as it takes two operands and does And on every bit of two numbers.

Example: Take operand-1 as 2: 010(in binary) and operand-2 as 6: 110(in binary). Now, take these two operands and do bitwise and operation i.e. (110)&(010)= 10 (in decimal it is '2'). - '<<'(Bitwise left shift) works as it shifts the bit of first operand the specified number of bits.

Example: if x=2(10 in binary), then y = x<<2 = 8(1000 in binary).

Begin power-set(s)

- Where s: given string
- take variable to store elements of power set, lets say sub
- for i: 0 to 2^(length of s)
- take a string variable to store current permutation subset, lets say sset
- for j:0 to (length of s)
- condition: if(jth bit of i is set"1"), then insert the element on s[j] into sset

- insert the sset into sub

- sort the elements of sub
- Now print all the elements of sub

End power-set

```
power-set(s, length)
take a variable which will store all the string of power set, say sub
take a variable, lets say power, which will run up 2^length
for i:0 to power
take a string variable say sset to store current permutation subset
for j:0 to length
condition: if(i & (1<<j))
then insert s[j] into sset
end for
insert the sset into sub
end for
sort the sub
print the elements of sub using loop
return
```

Where s: given input string

length: size of String s

###### Code

```
#include <bits/stdc++.h>
using namespace std;
//-----------Power-Set Function---------------//
void power_set(string s,int n) {
vector<string> sub;
int power;
power=pow(2,n);
for(int i=0;i<power;i++){
string sset="";
for(int j=0;j<n;j++){
if(i & (1<<j)){
sset+=s[j];
}
}
//push the subsequence into sub
sub.push_back(sset);
}
//sort the vector string
sort(sub.begin(),sub.end());
//print the Power-set
int len=sub.size();
for(int i=0;i<len;i++){
cout<<sub[i]<<endl;
}
return;
}
//----------Remove Duplicate Elements--------//
string duplicate(string s){
if(s.size()==0){
return "";
}
string x="";
int n=s.size(),i=0;
x+=s[0];
for(int j=1;j<n;j++){
if(x[i]!=s[j]){
i++;
x+=s[j];
}
}
return x;
}
//-----------Driver Function---------------//
int main() {
string S;
cout<<"Enter string: \n";
cin>>S;
cout<<"Power Set of the string '"<<S<<"' is :\n";
//sort the elements
sort(S.begin(),S.end());
//call duplicate() to remove duplicate elements
S=duplicate(S);
//calling power-set function
power_set(S, S.size());
return 0;
}
```

###### OUTPUT

```
Enter string:
algo
Power Set of the string 'algo' is :
a
ag
agl
aglo
ago
al
alo
ao
g
gl
glo
go
l
lo
o
```

### Complexity

###### Using Recursion

Time Complexity: O(N*(2^N)), N is the size of the given input string and 2^N elements in power-set.

Space Complexity: O(N), N is the size of string.

In the worst case scenario, the depth of the recursion tree would be â€˜Nâ€™. Hence, the overall space complexity due to the recursion call stack is O(N).

###### Using Iterartion

Time Complexity: O(N*(2^N)), N is the size of the given input string and 2^N elements in power-set.

Space Complexity: O(1), constant space complexity.