Skip to content
Related Articles

Related Articles

Improve Article

Find the smallest number with n set and m unset bits

  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2021

Given two non-negative numbers n and m. The problem is to find the smallest number having n number of set bits and m number of unset bits in its binary representation.
Constraints: 1 <= n, 0 <= m, (m+n) <= 31
Note : 0 bits before leading 1 (or leftmost 1) in binary representation are counted

Examples: 

Input : n = 2, m = 2
Output : 9
(9)10 = (1001)2
We can see that in the binary representation of 9 
there are 2 set and 2 unsets bits and it is the
smallest number. 

Input : n = 4, m = 1
Output : 23

Approach: Following are the steps: 

  1. Calculate num = (1 << (n + m)) – 1. This will produce a number num having (n + m) number of bits and all are set.
  2. Now, toggle bits in the range from n to (n+m-1) in num, i.e, to toggle bits from the rightmost nth bit to the rightmost (n+m-1)th bit and then return the toggled number. Refer this post.

C++




// C++ implementation to find the smallest number
// with n set and m unset bits
#include <bits/stdc++.h>
 
using namespace std;
 
// function to toggle bits in the given range
unsigned int toggleBitsFromLToR(unsigned int n,
                                unsigned int l,
                                unsigned int r)
{
    // for invalid range
    if (r < l)
        return n;
 
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
    // toggle bits in the range l to r in 'n'
    // and return the number
    return (n ^ num);
}
 
// function to find the smallest number
// with n set and m unset bits
unsigned int smallNumWithNSetAndMUnsetBits(unsigned int n,
                                           unsigned int m)
{
    // calculating a number 'num' having '(n+m)' bits
    // and all are set
    unsigned int num = (1 << (n + m)) - 1;
 
    // required smallest number
    return toggleBitsFromLToR(num, n, n + m - 1);
}
 
// Driver program to test above
int main()
{
    unsigned int n = 2, m = 2;
    cout << smallNumWithNSetAndMUnsetBits(n, m);
    return 0;
}

Java




// Java implementation to find the smallest number
// with n set and m unset bits
 
class GFG
{
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // for invalid range
        if (r < l)
            return n;
  
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
  
        // toggle bits in the range l to r in 'n'
        // and return the number
        return (n ^ num);
    }
     
    // Function to find the smallest number
    // with n set and m unset bits
    static int smallNumWithNSetAndMUnsetBits(int n, int m)
    {
        // calculating a number 'num' having '(n+m)' bits
        // and all are set
        int num = (1 << (n + m)) - 1;
  
        // required smallest number
        return toggleBitsFromLToR(num, n, n + m - 1);
    }
     
    // driver program
    public static void main (String[] args)
    {
        int n = 2, m = 2;
        System.out.println(smallNumWithNSetAndMUnsetBits(n, m));
    }
}
 
// Contributed by Pramod Kumar

Python3




# Python3 implementation to find
# the smallest number with n set
# and m unset bits
 
# function to toggle bits in the
# given range
def toggleBitsFromLToR(n, l, r):
 
    # for invalid range
    if (r < l):
        return n
  
    # calculating a number 'num'
    # having 'r' number of bits
    # and bits in the range l
    # to r are the only set bits
    num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1)
  
    # toggle bits in the range
    # l to r in 'n' and return the number
    return (n ^ num)
 
# function to find the smallest number
# with n set and m unset bits
def smallNumWithNSetAndMUnsetBits(n, m):
 
    # calculating a number 'num' having
    # '(n+m)' bits and all are set
    num = (1 << (n + m)) - 1
  
    # required smallest number
    return toggleBitsFromLToR(num, n, n + m - 1);
 
  
# Driver program to test above
n = 2
m = 2
 
ans = smallNumWithNSetAndMUnsetBits(n, m)
print (ans)
 
# This code is contributed by Saloni Gupta

C#




// C# implementation to find the smallest number
// with n set and m unset bits
using System;
 
class GFG
{
    // Function to toggle bits in the given range
    static int toggleBitsFromLToR(int n, int l, int r)
    {
        // for invalid range
        if (r < l)
            return n;
 
        // calculating a number 'num' having 'r'
        // number of bits and bits in the range l
        // to r are the only set bits
        int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);
 
        // toggle bits in the range l to r in 'n'
        // and return the number
        return (n ^ num);
    }
     
    // Function to find the smallest number
    // with n set and m unset bits
    static int smallNumWithNSetAndMUnsetBits(int n, int m)
    {
        // calculating a number 'num' having '(n+m)' bits
        // and all are set
        int num = (1 << (n + m)) - 1;
 
        // required smallest number
        return toggleBitsFromLToR(num, n, n + m - 1);
    }
     
    // Driver program
    public static void Main ()
    {
        int n = 2, m = 2;
        Console.Write(smallNumWithNSetAndMUnsetBits(n, m));
    }
}
 
// This code is contributed by Sam007

PHP




<?php
// PHP  implementation to find the smallest number
// with n set and m unset bits
 
// function to toggle bits in the given range
 
function toggleBitsFromLToR($n,$l,$r)
{
    // for invalid range
    if ($r < $l)
        return $n;
 
    // calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    $num = ((1 << $r) - 1) ^ ((1 << ($l - 1)) - 1);
 
    // toggle bits in the range l to r in 'n'
    // and return the number
    return ($n ^ $num);
}
 
// function to find the smallest number
// with n set and m unset bits
function smallNumWithNSetAndMUnsetBits($n, $m)
{
    // calculating a number 'num' having '(n+m)' bits
    // and all are set
    $num = (1 << ($n + $m)) - 1;
 
    // required smallest number
    return toggleBitsFromLToR($num, $n, $n + $m - 1);
}
 
// Driver program to test above
     $n = 2; $m = 2;
     echo  smallNumWithNSetAndMUnsetBits($n, $m);
 
 
// This Code is Contributed by ajit
?>

Javascript




<script>
 
// Javascript implementation to find
// the smallest number with n set and
// m unset bits
 
// Function to toggle bits in the given range
function toggleBitsFromLToR(n, l, r)
{
     
    // For invalid range
    if (r < l)
        return n;
 
    // Calculating a number 'num' having 'r'
    // number of bits and bits in the range l
    // to r are the only set bits
    let num = ((1 << r) - 1) ^
              ((1 << (l - 1)) - 1);
 
    // Toggle bits in the range l to r in 'n'
    // and return the number
    return (n ^ num);
}
   
// Function to find the smallest number
// with n set and m unset bits
function smallNumWithNSetAndMUnsetBits(n, m)
{
     
    // Calculating a number 'num' having
    // '(n+m)' bits and all are set
    let num = (1 << (n + m)) - 1;
 
    // Required smallest number
    return toggleBitsFromLToR(num, n, n + m - 1);
}
 
// Driver code
let n = 2, m = 2;
 
document.write(smallNumWithNSetAndMUnsetBits(n, m));
 
// This code is contributed by suresh07
 
</script>

Output: 

9

For greater values of n and m, you can use long int and long long int datatypes to generate the required number.
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :