# 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 ` ` `  `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 ` `    ``return` `toggleBitsFromLToR(num, n, n + m - 1); ` `} ` ` `  `// 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 ` `        ``return` `toggleBitsFromLToR(num, n, n + m - ``1``); ` `    ``} ` `     `  `    ``// driver program ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``2``, m = ``2``; ` `        ``System.out.println(smallNumWithNSetAndMUnsetBits(n, m)); ` `    ``} ` `} ` ` `  `// Contributed by Pramod Kumar `

## Python3

 `# Python3 implementation to find ` `# the smallest number with n set ` `# and m unset bits ` ` `  `# function to toggle bits in the ` `# given range ` `def` `toggleBitsFromLToR(n, l, 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 ` `    ``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 ` `def` `smallNumWithNSetAndMUnsetBits(n, m): ` ` `  `    ``# calculating a number 'num' having ` `    ``# '(n+m)' bits and all are set ` `    ``num ``=` `(``1` `<< (n ``+` `m)) ``-` `1` `  `  `    ``# required smallest number ` `    ``return` `toggleBitsFromLToR(num, n, n ``+` `m ``-` `1``); ` ` `  `  `  `# Driver program to test above ` `n ``=` `2` `m ``=` `2` ` `  `ans ``=` `smallNumWithNSetAndMUnsetBits(n, m) ` `print` `(ans) ` ` `  `# This code is contributed by Saloni Gupta `

## C#

 `// C# implementation to find the smallest number ` `// with n set and m unset bits ` `using` `System; ` ` `  `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 ` `        ``return` `toggleBitsFromLToR(num, n, n + m - 1); ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 2, m = 2; ` `        ``Console.Write(smallNumWithNSetAndMUnsetBits(n, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 ` `

Output:

```9
```

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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