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Power Set in Lexicographic order

Last Updated : 23 May, 2024
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This article is about generating Power set in lexicographical 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 <bits/stdc++.h>
using namespace std;

// str : Stores input string
// n : Length of str.
void func(string s, vector<string>& 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<string> 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 */
Python 
# 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, curr
def 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 order
def powerSet(string):
    string = ''.join(sorted(string))
    permuteRec(string, len(string))

# Driver Code
if __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 
// javascript 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, curr
    function permuteRec( str , n , index,  curr) {
        // base case
        if (index == n) {
            return;
        }
        console.log(curr+" ");
        for (var 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.
    function powerSet(str) {
        var arr = str.split("");
        arr.sort();
        permuteRec(arr, str.length, -1, "");
    }

    // Driver code
    
        var str = "cab";
        powerSet(str);

// This code contributed by umadevi9616
PHP 
<?php
// PHP 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, curr
function permuteRec($str, $n, $index = -1, 
                              $curr = "")
{
    // base case
    if ($index == $n) 
        return;

    echo $curr."\n";
    for ($i = $index + 1; $i < $n; $i++) 
    {

        $curr=$curr.$str[$i];
        permuteRec($str, $n, $i, $curr);

        // backtracking
        $curr =""; 
    }
    return;
}

// Generates power set in lexicographic
// order.
function powerSet($str)
{

    $str = str_split($str);
    sort($str);
    permuteRec($str, sizeof($str));
}

// Driver code
$str = "cab";
powerSet($str);

// This code is contributed by Mithun Kumar
?>

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.
C++ 
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;

void generate_power_set(string s)
{
    // Sort the set in lexicographical order
    sort(s.begin(), s.end());

    int n = s.length();
    vector<string> subsets;

    // Generate all possible binary strings of length n
    for (int i = 0; i < (1 << n); i++) {
        string binary = "";
        int num = i;

        // Convert the integer i to a binary string of
        // length n
        for (int j = 0; j < n; j++) {
            binary = char('0' + num % 2) + binary;
            num /= 2;
        }

        // Generate the subset based on the binary string
        string subset = "";
        for (int j = 0; j < n; j++) {
            if (binary[j] == '1') {
                subset += s[j];
            }
        }
        subsets.push_back(subset);
    }

    // Sort the subsets in lexicographically order
    sort(subsets.begin(), subsets.end());

    // Print the subsets in sorted order
    for (auto& subset : subsets) {
        cout << subset << " ";
    }
    cout << endl;
}

int main()
{
    string s = "abc";
    generate_power_set(s);
    return 0;
}

// This code is contributed by abhinav_m22
Java 
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Program {
    // Function to generate the power set of a string
    static void generatePowerSet(String s) {
        // Convert the string to a character array
        char[] charArray = s.toCharArray();
        // Sort the character array
        Arrays.sort(charArray);
        String sortedString = new String(charArray);

        int n = sortedString.length();
        List<String> subsets = new ArrayList<>();

        // Generate all possible binary strings of length n
        for (int i = 0; i < (1 << n); i++) {
            StringBuilder binary = new StringBuilder();
            int num = i;

            // Convert the integer i to a binary string of length n
            for (int j = 0; j < n; j++) {
                binary.insert(0, num % 2);
                num /= 2;
            }

            // Generate the subset based on the binary string
            StringBuilder subset = new StringBuilder();
            for (int j = 0; j < n; j++) {
                if (binary.charAt(j) == '1') {
                    subset.append(sortedString.charAt(j));
                }
            }
            subsets.add(subset.toString());
        }

        // Sort the subsets in lexicographically order
        subsets.sort(null);

        // Print the subsets in sorted order
        for (String subset : subsets) {
            System.out.print(subset + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        String s = "abc";
        generatePowerSet(s);
    }
}
Python 
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 usage
s = 'abc'
generate_power_set(s)
C# 
using System;
using System.Collections.Generic;
using System.Linq;

class Program {
    // Function to generate the power set of a string
    static void GeneratePowerSet(string s)
    {
        // Sort the set in lexicographical order
        char[] charArray = s.ToCharArray();
        Array.Sort(charArray);
        string sortedString = new string(charArray);

        int n = sortedString.Length;
        List<string> subsets = new List<string>();

        // Generate all possible binary strings of length n
        for (int i = 0; i < (1 << n); i++) {
            string binary = "";
            int num = i;

            // Convert the integer i to a binary string of
            // length n
            for (int j = 0; j < n; j++) {
                binary = (num % 2) + binary;
                num /= 2;
            }

            // Generate the subset based on the binary
            // string
            string subset = "";
            for (int j = 0; j < n; j++) {
                if (binary[j] == '1') {
                    subset += sortedString[j];
                }
            }
            subsets.Add(subset);
        }

        // Sort the subsets in lexicographically order
        subsets.Sort();

        // Print the subsets in sorted order
        foreach(var subset in subsets)
        {
            Console.Write(subset + " ");
        }
        Console.WriteLine();
    }

    static void Main()
    {
        string s = "abc";
        GeneratePowerSet(s);
    }
}
JavaScript 
function generatePowerSet(s) {
    // Convert the string to an array of characters
    let charArray = Array.from(s);
    // Sort the character array
    charArray.sort();
    let sortedString = charArray.join('');

    let n = sortedString.length;
    let subsets = [];

    // Generate all possible binary strings of length n
    for (let i = 0; i < (1 << n); i++) {
        let binary = '';
        let num = i;

        // Convert the integer i to a binary string of length n
        for (let j = 0; j < n; j++) {
            binary = (num % 2) + binary;
            num = Math.floor(num / 2);
        }

        // Generate the subset based on the binary string
        let subset = '';
        for (let j = 0; j < n; j++) {
            if (binary.charAt(j) === '1') {
                subset += sortedString.charAt(j);
            }
        }
        subsets.push(subset);
    }

    // Sort the subsets in lexicographically order
    subsets.sort();

    // Print the subsets in sorted order
    console.log(subsets.join(' '));
}

let s = 'abc';
generatePowerSet(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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