Find the smallest number with n set and m unset bits

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:

Calculate num = (1 << (n + m)) – 1. This will produce a number num having (n + m) number of bits and all are set.
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# 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:

Time Complexity : O(1)

Space Complexity : O(1)
For greater values of n and m, you can use long int and long long int datatypes to generate the required number.

Find the smallest number with n set and m unset bits

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