Given two non-negative numbers n and m. The problem is to find the largest number having n number of set bits and m number of unset bits in its binary representation.
Note : 0 bits before leading 1 (or leftmost 1) in binary representation are counted
Constraints: 1 <= n, 0 <= m, (m+n) <= 31
Examples :
Input : n = 2, m = 2 Output : 12 (12)10 = (1100)2 We can see that in the binary representation of 12 there are 2 set and 2 unsets bits and it is the largest number. Input : n = 4, m = 1 Output : 30
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 the last m bits of num and then return the toggled number. Refer this post.
C++
// C++ implementation to find the largest number // with n set and m unset bits #include <bits/stdc++.h> using namespace std;
// function to toggle the last m bits unsigned int toggleLastMBits(unsigned int n,
unsigned int m)
{ // if no bits are required to be toggled
if (m == 0)
return n;
// calculating a number 'num' having 'm' bits
// and all are set
unsigned int num = (1 << m) - 1;
// toggle the last m bits and return the number
return (n ^ num);
} // function to find the largest number // with n set and m unset bits unsigned int largeNumWithNSetAndMUnsetBits(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 largest number
return toggleLastMBits(num, m);
} // Driver program to test above int main()
{ unsigned int n = 2, m = 2;
cout << largeNumWithNSetAndMUnsetBits(n, m);
return 0;
} |
Java
// Java implementation to find the largest number // with n set and m unset bits import java.io.*;
class GFG
{ // Function to toggle the last m bits
static int toggleLastMBits( int n, int m)
{
// if no bits are required to be toggled
if (m == 0 )
return n;
// calculating a number 'num' having 'm' bits
// and all are set
int num = ( 1 << m) - 1 ;
// toggle the last m bits and return the number
return (n ^ num);
}
// Function to find the largest number
// with n set and m unset bits
static int largeNumWithNSetAndMUnsetBits( int n, int m)
{
// calculating a number 'num' having '(n+m)' bits
// and all are set
int num = ( 1 << (n + m)) - 1 ;
// required largest number
return toggleLastMBits(num, m);
}
// driver program
public static void main (String[] args)
{
int n = 2 , m = 2 ;
System.out.println(largeNumWithNSetAndMUnsetBits(n, m));
}
} // Contributed by Pramod Kumar |
Python3
# Python implementation to # find the largest number # with n set and m unset bits # function to toggle # the last m bits def toggleLastMBits(n,m):
# if no bits are required
# to be toggled
if (m = = 0 ):
return n
# calculating a number
# 'num' having 'm' bits
# and all are set
num = ( 1 << m) - 1
# toggle the last m bits
# and return the number
return (n ^ num)
# function to find # the largest number # with n set and m unset bits def largeNumWithNSetAndMUnsetBits(n,m):
# calculating a number
# 'num' having '(n+m)' bits
# and all are set
num = ( 1 << (n + m)) - 1
# required largest number
return toggleLastMBits(num, m)
# Driver code n = 2
m = 2
print (largeNumWithNSetAndMUnsetBits(n, m))
# This code is contributed # by Anant Agarwal. |
C#
// C# implementation to find the largest number // with n set and m unset bits using System;
class GFG
{ // Function to toggle the last m bits
static int toggleLastMBits( int n, int m)
{
// if no bits are required to be toggled
if (m == 0)
return n;
// calculating a number 'num' having 'm' bits
// and all are set
int num = (1 << m) - 1;
// toggle the last m bits and return the number
return (n ^ num);
}
// Function to find the largest number
// with n set and m unset bits
static int largeNumWithNSetAndMUnsetBits( int n, int m)
{
// calculating a number 'num' having '(n+m)' bits
// and all are set
int num = (1 << (n + m)) - 1;
// required largest number
return toggleLastMBits(num, m);
}
// Driver program
public static void Main ()
{
int n = 2, m = 2;
Console.Write(largeNumWithNSetAndMUnsetBits(n, m));
}
} // This code is contributed by Sam007 |
PHP
<?php // PHP implementation to find // the largest number with n // set and m unset bits // function to toggle // the last m bits function toggleLastMBits( $n , $m )
{ // if no bits are required
// to be toggled
if ( $m == 0)
return $n ;
// calculating a number 'num'
// having 'm' bits and all are set
$num = (1 << $m ) - 1;
// toggle the last m bits
// and return the number
return ( $n ^ $num );
} // function to find the largest number // with n set and m unset bits function largeNumWithNSetAndMUnsetBits( $n ,
$m )
{ // calculating a number 'num'
// having '(n+m)' bits and all are set
$num = (1 << ( $n + $m )) - 1;
// required largest number
return toggleLastMBits( $num , $m );
} // Driver Code $n = 2; $m = 2;
echo largeNumWithNSetAndMUnsetBits( $n , $m );
// This code is contributed by vt_m. ?> |
Javascript
<script> // Javascript implementation to find the largest number // with n set and m unset bits // function to toggle the last m bits function toggleLastMBits(n, m)
{ // if no bits are required to be toggled
if (m == 0)
return n;
// calculating a number 'num' having 'm' bits
// and all are set
var num = (1 << m) - 1;
// toggle the last m bits and return the number
return (n ^ num);
} // function to find the largest number // with n set and m unset bits function largeNumWithNSetAndMUnsetBits(n, m)
{ // calculating a number 'num' having '(n+m)' bits
// and all are set
num = (1 << (n + m)) - 1;
// required largest number
return toggleLastMBits(num, m);
} // Driver program to test above var n = 2, m = 2;
document.write( largeNumWithNSetAndMUnsetBits(n, m)); </script> |
Output :
12
Time Complexity : O(1)
Auxiliary Space: 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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