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

Contraints: 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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.

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C#

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// 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

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PHP

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<?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.
?>

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Output :

12

For greater values of n and m, you can use long int and long long int datatypes to generate the required number.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m



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