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++
// 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 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
# 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# 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 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.
?>
<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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