Generate all binary strings of length n with k bits set

This method will generate all integers with exactly N '1' bits.

From https://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation

Compute the lexicographically next bit permutation

Suppose we have a pattern of N bits set to 1 in an integer and we want the next permutation of N 1 bits in a lexicographical sense. For example, if N is 3 and the bit pattern is 00010011, the next patterns would be 00010101, 00010110, 00011001, 00011010, 00011100, 00100011, and so forth. The following is a fast way to compute the next permutation.

unsigned int v; // current permutation of bits
unsigned int w; // next permutation of bits

unsigned int t = v | (v - 1); // t gets v's least significant 0 bits set to 1
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));

The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros. If you are using Microsoft compilers for x86, the intrinsic is _BitScanForward. These both emit a bsf instruction, but equivalents may be available for other architectures. If not, then consider using one of the methods for counting the consecutive zero bits mentioned earlier. Here is another version that tends to be slower because of its division operator, but it does not require counting the trailing zeros.

unsigned int t = (v | (v - 1)) + 1;
w = t | ((((t & -t) / (v & -v)) >> 1) - 1);

Thanks to Dario Sneidermanis of Argentina, who provided this on November 28, 2009.

One problem with many of the standard solutions to this problem is that the entire set of strings is generated and then those are iterated through, which may exhaust the stack. It quickly becomes unwieldy for any but the smallest sets. In addition, in many instances, only a partial sampling is needed, but the standard (recursive) solutions generally chop the problem into pieces that are heavily biased to one direction (eg. consider all the solutions with a zero starting bit, and then all the solutions with a one starting bit).

In many cases, it would be more desireable to be able to pass a bit string (specifying element selection) to a function and have it return the next bit string in such a way as to have a minimal change (this is known as a Gray Code) and to have a representation of all the elements.

Donald Knuth covers a whole host of algorithms for this in his Fascicle 3A, section 7.2.1.3: Generating all Combinations.

There is an approach for tackling the iterative Gray Code algorithm for all ways of choosing k elements from n at http://answers.yahoo.com/question/index?qid=20081208224633AA0gdMl with a link to final PHP code listed in the comment (click to expand it) at the bottom of the page.

Well for this question (where you need to iterate over all the submasks in increasing order of their number of set bits), which has been marked as a duplicate of this.

We can simply iterate over all the submasks add them to a vector and sort it according to the number of set bits.

vector<int> v;
for(ll i=mask;i>0;i=(i-1)&mask)
    v.push_back(i);
auto cmp = [](const auto &a, const auto &b){
    return __builtin_popcountll(a) < __builtin_popcountll(b);
}
v.sort(v.begin(), v.end(), cmp);

Another way would be to iterate over all the submasks N times and add a number to the vector if the number of set bits is equal to i in the ith iteration.

Both ways have complexity of O(n*2^n)