Print all submasks of a given mask

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Given an integer N, the task is to print all the subsets of the set formed by the set bits present in the binary representation of N.

Examples:

Input: N = 5
Output: 5 4 1
Explanation:
Binary representation of N is “101”, Therefore all the required subsets are {“101”, “100”, “001”, “000”}.

Input: N = 25
Output: 25 24 17 16 9 8 1
Explanation:
Binary representation of N is “11001”. Therefore, all the required subsets are {“11001”, “11000”, “10001”, “10000”, “01001”, “01000”, “0001”, “0000”}.

Naive Approach: The simplest approach is to traverse every mask in the range [0, 1 << (count of set bit in N)] and check if no other bits are set in it except for the bits in N. Then, print it.
Time Complexity: O(2^{(count of set bit in N)})
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by only traversing the submasks which are the subset of mask N.

Suppose S is the current submask which is the subset of mask N. Then, it can be observed that by assigning S = (S – 1) & N, the next submask of N can be obtained which is less than S.

In S – 1, it flips all the bits present on the right of the rightmost set bit including rightmost set bit of S.

Therefore, after performing Bitwise & with N, a submask of N is obtained.

Therefore, S = (S – 1) & N gives the next submask of N which is less than S.

Follow the steps below to solve the problem:

Initialize a variable, say S = N.
Iterate while S > 0 and in each iteration, print the value of S.
Assign S = (S – 1) & N.

Below is the implementation of the above approach:

C++

// C++ Program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the submasks of N
void SubMasks(int N)
{
    for (int S = N; S; S = (S - 1) & N) {
        cout << S << " ";
    }
}
 
// Driver Code
int main()
{
    int N = 25;
    SubMasks(N);
 
    return 0;
}

Java

// Java Program for above approach
import java.util.*;
class GFG
{
   
// Function to print the submasks of N
static void SubMasks(int N)
{
    for (int S = N; S > 0; S = (S - 1) & N) 
    {
        System.out.print(S + " ");
    }
}
 
// Driver Code
public static void main(String args[])
{
    int N = 25;
    SubMasks(N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Python3

# Python3 program for the above approach
 
# Function to print the submasks of N
def SubMasks(N) :
    S = N 
    while S > 0:
        print(S,end=' ')
        S = (S - 1) & N
 
# Driven Code
if __name__ == '__main__':
    N = 25
    SubMasks(N)
     
    # This code is contributed by bgangwar59.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to print the submasks of N
static void SubMasks(int N)
{
    for (int S = N; S > 0; S = (S - 1) & N) 
    {
        Console.Write(S + " ");
    }
}
 
// Driver Code
static public void Main()
{
    int N = 25;
    SubMasks(N);
}
}
 
// This code is contributed by Code_hunt.

Javascript

<script>
 
// JavaScript program of the above approach
 
// Function to print the submasks of N
function SubMasks(N)
{
    for (let S = N; S > 0; S = (S - 1) & N)
    {
        document.write(S + " ");
    }
}
 
    // Driver Code
     
    let N = 25;
    SubMasks(N);
 
</script>

Output:

25 24 17 16 9 8 1

Time Complexity: O(2^{(count of set bit in N)})
Auxiliary Space: O(1)

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