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#
// 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
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.
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