# Find the smallest number with n set and m unset bits

Given two non-negative numbers n and m. The problem is to find the smallest number having n number of set bits and m number of unset bits in its binary representation.

Constraints: 1 <= n, 0 <= m, (m+n) <= 31

Note : 0 bits before leading 1 (or leftmost 1) in binary representation are counted

Examples:

```Input : n = 2, m = 2
Output : 9
(9)10 = (1001)2
We can see that in the binary representation of 9
there are 2 set and 2 unsets bits and it is the
smallest number.

Input : n = 4, m = 1
Output : 23
```

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

Approach: 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 bits in the range from n to (n+m-1) in num, i.e, to toggle bits from the rightmost nth bit to the rightmost (n+m-1)th bit and then return the toggled number. Refer this post.

## C/C++

```// C++ implementation to find the smallest number
// with n set and m unset bits
#include <bits/stdc++.h>

using namespace std;

// function to toggle bits in the given range
unsigned int toggleBitsFromLToR(unsigned int n,
unsigned int l,
unsigned int r)
{
// for invalid range
if (r < l)
return n;

// calculating a number 'num' having 'r'
// number of bits and bits in the range l
// to r are the only set bits
int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);

// toggle bits in the range l to r in 'n'
// and return the number
return (n ^ num);
}

// function to find the smallest number
// with n set and m unset bits
unsigned int smallNumWithNSetAndMUnsetBits(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 smallest number
}

// Driver program to test above
int main()
{
unsigned int n = 2, m = 2;
cout << smallNumWithNSetAndMUnsetBits(n, m);
return 0;
}
```

## Java

```// Java implementation to find the smallest number
// with n set and m unset bits

class GFG
{
// Function to toggle bits in the given range
static int toggleBitsFromLToR(int n, int l, int r)
{
// for invalid range
if (r < l)
return n;

// calculating a number 'num' having 'r'
// number of bits and bits in the range l
// to r are the only set bits
int num = ((1 << r) - 1) ^ ((1 << (l - 1)) - 1);

// toggle bits in the range l to r in 'n'
// and return the number
return (n ^ num);
}

// Function to find the smallest number
// with n set and m unset bits
static int smallNumWithNSetAndMUnsetBits(int n, int m)
{
// calculating a number 'num' having '(n+m)' bits
// and all are set
int num = (1 << (n + m)) - 1;

// required smallest number
}

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

// Contributed by Pramod Kumar
```

Output:

```9
```

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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