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# Power Set in Lexicographic order

Examples :

Input : abc
Output : a ab abc ac b bc c

The idea is to sort array first. After sorting, one by one fix characters and recursively generates all subsets starting from them. After every recursive call, we remove last character so that next permutation can be generated.

Implementation:

## C++

 // CPP program to generate power set in// lexicographic order.#include using namespace std; // str : Stores input string// n : Length of str.void func(string s, vector& str, int n, int pow_set){    int i, j;    for (i = 0; i < pow_set; i++) {        string x;        for (j = 0; j < n; j++) {            if (i & 1 << j) {                x = x + s[j];            }        }        if (i != 0)            str.push_back(x);    }}int main(){    int n;    string s;    vector str;    s = "cab";    n = s.length();    int pow_set = pow(2, n);    func(s, str, n, pow_set);    sort(str.begin(), str.end());    for (int i = 0; i < str.size(); i++)        cout << str[i] << " ";    cout << endl;     return 0;}

## Java

 // Java program to generate power set in// lexicographic order.import java.util.*; class GFG {     // str : Stores input string    // n : Length of str.    // curr : Stores current permutation    // index : Index in current permutation, curr    static void permuteRec(String str, int n,                           int index, String curr)    {        // base case        if (index == n) {            return;        }        System.out.println(curr);        for (int i = index + 1; i < n; i++) {             curr += str.charAt(i);            permuteRec(str, n, i, curr);             // backtracking            curr = curr.substring(0, curr.length() - 1);        }        return;    }     // Generates power set in lexicographic    // order.    static void powerSet(String str)    {        char[] arr = str.toCharArray();        Arrays.sort(arr);        permuteRec(new String(arr), str.length(), -1, "");    }     // Driver code    public static void main(String[] args)    {        String str = "cab";        powerSet(str);    }} /* This code contributed by PrinciRaj1992 */

## Python3

 # Python3 program to generate power# set in lexicographic order. # str : Stores input string# n : Length of str.# curr : Stores current permutation# index : Index in current permutation, currdef permuteRec(string, n, index = -1, curr = ""):     # base case    if index == n:        return     if len(curr) > 0:        print(curr)     for i in range(index + 1, n):        curr += string[i]        permuteRec(string, n, i, curr)         # backtracking        curr = curr[:len(curr) - 1] # Generates power set in lexicographic orderdef powerSet(string):    string = ''.join(sorted(string))    permuteRec(string, len(string)) # Driver Codeif __name__ == "__main__":    string = "cab"    powerSet(string) # This code is contributed by vibhu4agarwal

## C#

 // C# program to generate power set in// lexicographic order.using System; class GFG {     // str : Stores input string    // n : Length of str.    // curr : Stores current permutation    // index : Index in current permutation, curr    static void permuteRec(String str, int n,                           int index, String curr)    {        // base case        if (index == n) {            return;        }        Console.WriteLine(curr);        for (int i = index + 1; i < n; i++) {             curr += str[i];            permuteRec(str, n, i, curr);             // backtracking            curr = curr.Substring(0, curr.Length - 1);        }        return;    }     // Generates power set in lexicographic    // order.    static void powerSet(String str)    {        char[] arr = str.ToCharArray();        Array.Sort(arr);        permuteRec(new String(arr), str.Length, -1, "");    }     // Driver code    public static void Main(String[] args)    {        String str = "cab";        powerSet(str);    }} // This code contributed by Rajput-Ji



## Javascript



Output

a ab b c ca cab cb

Time Complexity: O(n*2n
Auxiliary Space: O(1)

Method (binary numbers)

The idea is to use binary numbers  to generate the power set of a given set of elements in lexicographical order

• Sort the given set in lexicographical order.
• Define a variable “n” to represent the size of the set.
• Use a loop to generate all possible binary numbers of length “n”.
• For each binary number, convert it to a string of 0s and 1s,
• Add the current subset to the output list.
• Sort the output list in lexicographical order.
• Print the sorted list of subsets.

## Python3

 def generate_power_set(s):    # Sort the set in lexicographical order    s = ''.join(sorted(s))     n = len(s)    subsets = []    # Generate all possible binary strings of length n    for i in range(2**n):        # Convert the integer i to a binary string of length n        binary = bin(i)[2:].zfill(n)        # Generate the subset based on the binary string        subset = ''.join([s[j] for j in range(n) if binary[j] == '1'])        subsets.append(subset)    # Sort the subsets in lexicographically order    subsets.sort()    # Print the subsets in sorted order    for subset in subsets:        print(subset) # Example usages = 'abc'generate_power_set(s)

Output

a
ab
abc
ac
b
bc
c

Time complexity :O(2^n * n), where n is the length of the input set.
Space complexity  :O(2^n * n), since the output list of subsets can potentially contain 2^n elements

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