Print bitwise AND set of a number N

Last Updated : 02 Aug, 2023

Given a number N, print all the numbers which are a bitwise AND set of the binary representation of N. Bitwise AND set of a number N is all possible numbers x smaller than or equal N such that N & i is equal to x for some number i.

Examples :

Input : N = 5
Output : 0, 1, 4, 5 
Explanation: 0 & 5 = 0 
             1 & 5 = 1
             2 & 5 = 0
             3 & 5 = 1
             4 & 5 = 4
             5 & 5 = 5  
So we get 0, 1, 4 and 5 in the 
bitwise subsets of N.
Input : N = 9
Output : 0, 1, 8, 9

Simple Approach: A naive approach is to iterate from all numbers from 0 to N and check if (N&i == i). Print the numbers which satisfy the specified condition.

Below is the implementation of above idea:

C++

// CPP program to print all bitwise
// subsets of N (Naive approach)
#include <bits/stdc++.h>
using namespace std;

// function to find bitwise subsets
// Naive approach
void printSubsets(int n) {
  for (int i = 0; i <= n; i++)
    if ((n & i) == i)
      cout << i << " ";
}

// Driver Code
int main() {
  
  int n = 9;
  printSubsets(n);
  return 0;
}

Java

// JAVA program to print all bitwise
// subsets of N (Naive approach)
class GFG {
    
    // function to find bitwise subsets
    // Naive approach
    static void printSubsets(int n)
    {
        
        for (int i = 0; i <= n; i++)
            if ((n & i) == i)
                System.out.print(i + " ");
    }
    
    // Driver function
    public static void main(String[] args)
    {
        int n = 9;
        
        printSubsets(n);
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python program to print all bitwise 
# subsets of N (Naive approach)
def printSubsets(n):
    
    for i in range(n + 1):
        
        if ((n & i) == i):
            print(i ," ", end = "")

# Driver code
n = 9
printSubsets(n)

# This code is contributed by Anant Agarwal.

// C# program to print all bitwise
// subsets of N (Naive approach)
using System;

class GFG {
    
    // function to find bitwise subsets
    // Naive approach
    static void printSubsets(int n)
    {
        
        for (int i = 0; i <= n; i++)
            if ((n & i) == i)
                Console.Write(i + " ");
    }
    
    // Driver function
    public static void Main()
    {
        int n = 9;
        
        printSubsets(n);
    }
}

// This code is contributed by vt_m.

JavaScript

<script>

// JavaScript program to print all bitwise
// subsets of N (Efficient approach)

// function to find bitwise 
    // subsets Efficient approach
    function printSubsets(n)
    {
    for (let i = n; i > 0; i = (i - 1) & n)
  
        document.write(i + " ");
        document.write(" 0 ");
      
    }
// Driver code

    let n = 9;
    printSubsets(n);

// This code is contributed by souravghosh0416.
</script>

PHP

<?php
// PHP program to print all bitwise
// subsets of N (Naive approach)

// function to find bitwise subsets
// Naive approach
function printSubsets($n) 
{
for ($i = 0; $i <= $n; $i++)
    if (($n & $i) == $i)
    echo $i." ";
}

// Driver Code
$n = 9;
printSubsets($n);

// This code is contributed by mits 
?>

Output

0 1 8 9

Time Complexity : O(N)

Efficient Solution: An efficient solution is to use bitwise operators to find the subsets. Instead of iterating for every i, we can simply iterate for the bitwise subsets only. Iterating backward for i=(i-1)&n gives us every bitwise subset, where i starts from n and ends at 1.

Below is the implementation of above idea:

C++

// CPP program to print all bitwise
// subsets of N (Efficient approach)

#include <bits/stdc++.h>
using namespace std;

// function to find bitwise subsets
// Efficient approach
void printSubsets(int n) {
  
  for (int i = n; i > 0; i = (i - 1) & n)
    cout << i << " ";
  cout << 0;
}

// Driver Code
int main() {  
  int n = 9;  
  printSubsets(n);  
  return 0;
}

Java

// Java program to print all bitwise
// subsets of N (Efficient approach)

class GFG 
{

    // function to find bitwise 
    // subsets Efficient approach
    static void printSubsets(int n)
    {
    for (int i = n; i > 0; i = (i - 1) & n)

        System.out.print(i + " ");
        System.out.print(" 0 ");
    
    }

// Driver Code
public static void main(String[] args)
{
    int n = 9;
    printSubsets(n);
}
}

// This code is contributed by ajit.

Python3

# Python 3 program to
# print all bitwise
# subsets of N
# (Efficient approach)

# function to find
# bitwise subsets
# Efficient approach
def printSubsets(n):
    i=n
    while(i != 0):
        print(i,end=" ")
        i=(i - 1) & n
    print("0")

# Driver Code
n = 9 
printSubsets(n) 

# This code is contributed by
# Smith Dinesh Semwal

// C# program to print all bitwise
// subsets of N (Efficient approach)
using System;

public class GFG {

    // function to find bitwise subsets
    // Efficient approach
    static void printSubsets(int n) {
    
    for (int i = n; i > 0; i = (i - 1) & n)
        Console.Write(i +" ");
        Console.WriteLine("0");
    }
    
    // Driver Code
    static public void Main () {
        
        int n = 9;
        
        printSubsets(n);
    }
}

// This code is contributed by vt_m.

JavaScript

<script>

// Javascript program to print all bitwise
// subsets of N (Efficient approach)

// Function to find bitwise subsets
// Efficient approach
function printSubsets(n)
{
    for(let i = n; i > 0; i = (i - 1) & n)
        document.write(i +" ");
        document.write("0" + "</br>");
}

// Driver code
let n = 9;
     
printSubsets(n);

// This code is contributed by divyesh072019

</script>

PHP

<?php
// PHP program to print all bitwise
// subsets of N (Efficient approach)

// function to find bitwise subsets
// Efficient approach
function printSubsets($n)
{

    for ($i = $n; $i > 0; 
         $i = ($i - 1) & $n)
         
        echo $i." ";
    echo "0";
}

// Driver Code
$n = 9; 
printSubsets($n); 

// This code is contributed by mits 
?>

Output

9 8 1 0

Output :

9 8 1 0

Time Complexity: O(K), where K is the number of bitwise subsets of N.

Approach: Optimized Bit Manipulation

Steps:

Initialize an empty list res with 0 as the first element.
Initialize a variable x as 1.
While x is less than or equal to N, do the following:
a. If the x-th bit of N is 1, then for each element r in res, append r | x to res.
b. Multiply x by 2.
Return the final list res.

C++

#include <iostream>
#include <vector>
using namespace std;

vector<int> bitwise_and_set(int N) {
    vector<int> res = {0};
    int x = 1;
    while (x <= N) {
        if (N & x) {
            vector<int> temp;
            for (int r : res) {
                temp.push_back(r | x);
            }
            res.insert(res.end(), temp.begin(), temp.end());
        }
        x <<= 1;
    }
    return res;
}

int main() {
    int N = 5;
    vector<int> result = bitwise_and_set(N);
    for (int r : result) {
        cout << r << " ";
    }
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class Main {
    public static List<Integer> bitwiseAndSet(int N) {
        List<Integer> res = new ArrayList<>();
        res.add(0);
        int x = 1;
        while (x <= N) {
            if ((N & x) != 0) {
                List<Integer> temp = new ArrayList<>();
                for (int r : res) {
                    temp.add(r | x);
                }
                res.addAll(temp);
            }
            x <<= 1;
        }
        return res;
    }

    public static void main(String[] args) {
        int N = 5;
        List<Integer> result = bitwiseAndSet(N);
        for (int r : result) {
            System.out.print(r + " ");
        }
    }
}

Python3

def bitwise_and_set(N):
    res = [0]
    x = 1
    while x <= N:
        if N & x:
            res += [r | x for r in res]
        x <<= 1
    return res


# Example Usage
N = 5
print(bitwise_and_set(N))  # Output: [0, 1, 4, 5]

using System;
using System.Collections.Generic;

public class Program {
    public static List<int> BitwiseAndSet(int N) {
        List<int> res = new List<int> {0};
        int x = 1;
        while (x <= N) {
            if ((N & x) != 0) {
                List<int> temp = new List<int>();
                foreach (int r in res) {
                    temp.Add(r | x);
                }
                res.AddRange(temp);
            }
            x <<= 1;
        }
        return res;
    }

    public static void Main() {
        int N = 5;
        List<int> result = BitwiseAndSet(N);
        foreach (int r in result) {
            Console.Write(r + " ");
        }
    }
}

JavaScript

function bitwiseAndSet(N) {
  var res = [0]; // Initialize the result array with 0
  var x = 1; // Initialize x as 1

  while (x <= N) { // Loop while x is less than or equal to N
    if (N & x) { // If the bitwise AND of N and x is non-zero
      var temp = []; // Create a temporary array

      for (var r of res) { // Iterate over the elements of res array
        temp.push(r | x); // Perform bitwise OR operation and add the result to temp array
      }

      res = res.concat(temp); // Concatenate temp array to res array
    }

    x <<= 1; // Left shift x by 1
  }

  return res; // Return the resulting array
}

var N = 5;
var result = bitwiseAndSet(N);

for (var r of result) { // Iterate over the elements of result array
  console.log(r + " "); // Print each element followed by a space
}

Output

0 1 4 5

Time Complexity: O(log N)
Auxiliary Space: O(2^log N)

Print bitwise AND set of a number N

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Print bitwise AND set of a number N

Approach: Optimized Bit Manipulation

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