Find the largest number with n set and m unset bits
Last Updated :
31 May, 2022
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++
#include <bits/stdc++.h>
using namespace std;
unsigned int toggleLastMBits(unsigned int n,
unsigned int m)
{
if (m == 0)
return n;
unsigned int num = (1 << m) - 1;
return (n ^ num);
}
unsigned int largeNumWithNSetAndMUnsetBits(unsigned int n,
unsigned int m)
{
unsigned int num = (1 << (n + m)) - 1;
return toggleLastMBits(num, m);
}
int main()
{
unsigned int n = 2, m = 2;
cout << largeNumWithNSetAndMUnsetBits(n, m);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int toggleLastMBits( int n, int m)
{
if (m == 0 )
return n;
int num = ( 1 << m) - 1 ;
return (n ^ num);
}
static int largeNumWithNSetAndMUnsetBits( int n, int m)
{
int num = ( 1 << (n + m)) - 1 ;
return toggleLastMBits(num, m);
}
public static void main (String[] args)
{
int n = 2 , m = 2 ;
System.out.println(largeNumWithNSetAndMUnsetBits(n, m));
}
}
|
Python3
def toggleLastMBits(n,m):
if (m = = 0 ):
return n
num = ( 1 << m) - 1
return (n ^ num)
def largeNumWithNSetAndMUnsetBits(n,m):
num = ( 1 << (n + m)) - 1
return toggleLastMBits(num, m)
n = 2
m = 2
print (largeNumWithNSetAndMUnsetBits(n, m))
|
C#
using System;
class GFG
{
static int toggleLastMBits( int n, int m)
{
if (m == 0)
return n;
int num = (1 << m) - 1;
return (n ^ num);
}
static int largeNumWithNSetAndMUnsetBits( int n, int m)
{
int num = (1 << (n + m)) - 1;
return toggleLastMBits(num, m);
}
public static void Main ()
{
int n = 2, m = 2;
Console.Write(largeNumWithNSetAndMUnsetBits(n, m));
}
}
|
PHP
<?php
function toggleLastMBits( $n , $m )
{
if ( $m == 0)
return $n ;
$num = (1 << $m ) - 1;
return ( $n ^ $num );
}
function largeNumWithNSetAndMUnsetBits( $n ,
$m )
{
$num = (1 << ( $n + $m )) - 1;
return toggleLastMBits( $num , $m );
}
$n = 2; $m = 2;
echo largeNumWithNSetAndMUnsetBits( $n , $m );
?>
|
Javascript
<script>
function toggleLastMBits(n, m)
{
if (m == 0)
return n;
var num = (1 << m) - 1;
return (n ^ num);
}
function largeNumWithNSetAndMUnsetBits(n, m)
{
num = (1 << (n + m)) - 1;
return toggleLastMBits(num, m);
}
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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