# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

```// 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
}

// 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
}

// driver program
public static void main (String[] args)
{
int n = 2, m = 2;
System.out.println(largeNumWithNSetAndMUnsetBits(n, m));
}
}

// Contributed by Pramod Kumar
```

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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