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

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.

Output: 

9

Time Complexity : O(1)

Space Complexity : O(1)
For greater values of n and m, you can use long int and long long int datatypes to generate the required number.