Find the largest number with n set and m unset bits

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:

1. Calculate num = (1 << (n + m)) – 1. This will produce a number num having (n + m) number of bits and all are set.
2. 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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