Given a natural number n, print all the subsets of the set 
without using any array or loop (only the use of recursion is allowed).
Examples:
Input : n = 4
Output : { 1 2 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 2 }
{ 1 3 4 }
{ 1 3 }
{ 1 4 }
{ 1 }
{ 2 3 4 }
{ 2 3 }
{ 2 4 }
{ 2 }
{ 3 4 }
{ 3 }
{ 4 }
{ }
Input : n = 2
Output : { 1 2 }
{ 1 }
{ 2 }
{ }
Approach:
- Start from
upto 0. - Consider the binary representation of num with n bits.
- Start from the leftmost bit which represents 1, the second bit represents 2, and so on until nth bit which represents n.
- Print the number corresponding to the bit if it is set.
- Perform the above steps for all values of num until it is equal to 0.
Let’s understand the above approach through an example:
Considering input n = 4, start from .
and so on … until num = 0.
Below is the implementation of the above approach:
C++
// C++ code to print all subsets// of {1, 2, 3, n} without using// array or loop, just recursion.#include <bits/stdc++.h>using namespace std;void subset(int, int, int);// This recursive function calls subset// function to print the subsets one by one.// numBits --> number of bits needed to// represent the number (simply input value n).// num --> Initially equal to 2 ^ n - 1 and// decreases by 1 every recursion until 0.void printSubsets(int numOfBits, int num) { if (num >= 0) { cout << "{ "; // Print the subset corresponding to // binary representation of num. subset(numOfBits - 1, num, numOfBits); cout << "}" << endl; // Call the function recursively to // print the next subset. printSubsets(numOfBits, num - 1); } else return;}// This function recursively prints the// subset corresponding to the binary// representation of num.// nthBit --> nth bit from right side// starting from n and decreases until 0void subset(int nthBit, int num, int numOfBits){ if (nthBit >= 0) { // Print number in given subset only // if the bit corresponding to it // is set in num. if (num & (1 << nthBit)) { cout << numOfBits - nthBit << " "; } // Check for the next bit subset(nthBit - 1, num, numOfBits); } else return;}// Driver Codeint main(){ int n = 4; printSubsets(n, pow(2, n) - 1);}// This code is contributed by// sanjeev2552 |
Java
// Java code to print all subsets // of {1, 2, 3, n} without using// array or loop, just recursion.class GfG { // This recursive function calls subset // function to print the subsets one by one. // numBits --> number of bits needed to // represent the number (simply input value n). // num --> Initially equal to 2 ^ n - 1 and // decreases by 1 every recursion until 0. static void printSubSets(int numOfBits, int num) { if (num >= 0) { System.out.print("{ "); // Print the subset corresponding to // binary representation of num. subset(numOfBits - 1, num, numOfBits); System.out.println("}"); // Call the function recursively to // print the next subset. printSubSets(numOfBits, num - 1); } else return; } // This function recursively prints the // subset corresponding to the binary // representation of num. // nthBit --> nth bit from right side // starting from n and decreases until 0. static void subset(int nthBit, int num, int numOfBits) { if (nthBit >= 0) { // Print number in given subset only // if the bit corresponding to it // is set in num. if ((num & (1 << nthBit)) != 0) { System.out.print(numOfBits - nthBit + " "); } // Check for the next bit subset(nthBit - 1, num, numOfBits); } else return; } // Driver code public static void main(String[] args) { int n = 4; printSubSets(n, (int) (Math.pow(2, n)) -1); }}// This code is contributed by laststringx |
Python3
# Python3 code to print all subsets # of {1, 2, 3, …n} without using# array or loop, just recursion.# This recursive function calls subset# function to print the subsets one by one. # numBits --> number of bits needed to # represent the number (simply input value n).# num --> Initially equal to 2 ^ n - 1 and # decreases by 1 every recursion until 0.def printSubsets(numOfBits, num): if num >= 0: print("{", end = " ") # Print the subset corresponding to # binary representation of num. subset(numOfBits-1, num, numOfBits) print("}") # Call the function recursively to # print the next subset. printSubsets(numOfBits, num-1) else: return# This function recursively prints the # subset corresponding to the binary # representation of num.# nthBit --> nth bit from right side # starting from n and decreases until 0.def subset(nthBit, num, numOfBits): if nthBit >= 0: # Print number in given subset only # if the bit corresponding to it # is set in num. if num & (1 << nthBit) != 0: print(numOfBits - nthBit, end = " ") # Check for the next bit subset(nthBit-1, num, numOfBits) else: return# Driver Code n = 4printSubsets(n, 2**n - 1) |
C#
// C# code to print all subsets // of {1, 2, 3, n} without using // array or loop, just recursion.using System;class GfG { // This recursive function calls subset // function to print the subsets one by one. // numBits --> number of bits needed to // represent the number (simply input value n). // num --> Initially equal to 2 ^ n - 1 and // decreases by 1 every recursion until 0. static void printSubSets(int numOfBits, int num) { if (num >= 0) { Console.Write("{ "); // Print the subset corresponding to // binary representation of num. subset(numOfBits - 1, num, numOfBits); Console.WriteLine("}"); // Call the function recursively to // print the next subset. printSubSets(numOfBits, num - 1); } else return; } // This function recursively prints the // subset corresponding to the binary // representation of num. // nthBit --> nth bit from right side // starting from n and decreases until 0. static void subset(int nthBit, int num, int numOfBits) { if (nthBit >= 0) { // Print number in given subset only // if the bit corresponding to it // is set in num. if ((num & (1 << nthBit)) != 0) { Console.Write(numOfBits - nthBit + " "); } // Check for the next bit subset(nthBit - 1, num, numOfBits); } else return; } // Driver codeM public static void Main(String[] args) { int n = 4; printSubSets(n, (int) (Math.Pow(2, n)) -1); } }// This code is contributed by Srathore |
Javascript
<script>// Javascript code to print all subsets// of {1, 2, 3, n} without using// array or loop, just recursion.// This recursive function calls subset// function to print the subsets one by one.// numBits --> number of bits needed to// represent the number (simply input value n).// num --> Initially equal to 2 ^ n - 1 and// decreases by 1 every recursion until 0.function printSubsets(numOfBits, num) { if (num >= 0) { document.write( "{ "); // Print the subset corresponding to // binary representation of num. subset(numOfBits - 1, num, numOfBits); document.write( "}<br>" ); // Call the function recursively to // print the next subset. printSubsets(numOfBits, num - 1); } else return;}// This function recursively prints the// subset corresponding to the binary// representation of num.// nthBit --> nth bit from right side// starting from n and decreases until 0function subset(nthBit, num, numOfBits){ if (nthBit >= 0) { // Print number in given subset only // if the bit corresponding to it // is set in num. if (num & (1 << nthBit)) { document.write( numOfBits - nthBit + " "); } // Check for the next bit subset(nthBit - 1, num, numOfBits); } else return;}// Driver Codevar n = 4;printSubsets(n, Math.pow(2, n) - 1);</script> |
Output:
{ 1 2 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 2 }
{ 1 3 4 }
{ 1 3 }
{ 1 4 }
{ 1 }
{ 2 3 4 }
{ 2 3 }
{ 2 4 }
{ 2 }
{ 3 4 }
{ 3 }
{ 4 }
{ }
Time Complexity:
Auxiliary Space: O(n) for call stack
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