A quick note on all permutations

Marius Bancila [Bancila 2018] (Chapter 6, problem 52) poses the next permutation problem as follows:

Write a function that, prints on the console all the possible permutations of a given string.

and provides a recursive version of the solution as follows:

void next_permutation(std::string str, std::string perm)
{
if (str.empty()) std::cout << perm << std::endl;
else
{
for (size_t i = 0; i< str.size(); ++i)
{
next_permutation(str.substr(1), perm + str);
std::rotate(std::begin(str),
std::begin(str) + 1,
std::end(str));
}
}
}

void print_permutations_recursive(std::string str)
{
next_permutation(str, "");
}

A better solution would be as follows

void next_permutation(std::string& str, size_t n)
{
if (n==1) std::cout << str << std::endl;
else
{
for (size_t i = 0; i<n; ++i)
{
next_permutation(str,n-1);

std::rotate( std::end(str)-n ,
std::end(str)-n+1,
std::end(str));
}
}
}

void print_permutations_recursive(std::string str)
{
if(!str.empty())
next_permutation(str, str.length());
}

This avoids string copy over recursive calls. More importantly the second parameter
has a meaning in that it is the length of the suffix that needs to be permuted.
The other thing to notice is that rotate is O(n) which can be reduced to O(1) by swapping as in

for (size_t i = 0; i < n; ++i)
{
std::swap(*(str.end()-n), *(str.end()-n+i));
next_permutation(str,n-1);
std::swap(*(str.end()-n), *(str.end()-n+i));
}

Reference
[Bancila 2018] Marius Bancila "The Modern C++ Challenge" Packt Publishing Birmingham-Mumbai 2018. 