NAME

std::next_permutation - std::next_permutation

Synopsis

Defined in header <algorithm>
template< class BidirIt >
bool next_permutation( BidirIt first, BidirIt (until C++20)
last );
template< class BidirIt >
constexpr bool next_permutation( BidirIt first, (since C++20)
BidirIt last ); (1)
template< class BidirIt, class Compare >
bool next_permutation( BidirIt first, BidirIt (until C++20)
last, Compare comp ); (2)
template< class BidirIt, class Compare >
constexpr bool next_permutation( BidirIt first, (since C++20)
BidirIt last, Compare comp );

Permutes the range [first, last) into the next permutation, where the set of all
permutations is ordered lexicographically with respect to operator< or comp. Returns
true if such a "next permutation" exists; otherwise transforms the range into the
lexicographically first permutation (as if by std::sort(first, last, comp)) and
returns false.

Parameters

first, last - the range of elements to permute
comparison function object (i.e. an object that satisfies the
requirements of Compare) which returns true if the first argument
is less than the second.

The signature of the comparison function should be equivalent to the
following:

bool cmp(const Type1 &a, const Type2 &b);
comp -
While the signature does not need to have const &, the function must
not modify the objects passed to it and must be able to accept all
values of type (possibly const) Type1 and Type2 regardless of value
category (thus, Type1 & is not allowed
, nor is Type1 unless for Type1 a move is equivalent to a copy
(since C++11)).
The types Type1 and Type2 must be such that an object of type BidirIt
can be dereferenced and then implicitly converted to both of them.

Type requirements

-
BidirIt must meet the requirements of ValueSwappable and
LegacyBidirectionalIterator.

Return value

true if the new permutation is lexicographically greater than the old. false if the
last permutation was reached and the range was reset to the first permutation.

Exceptions

Any exceptions thrown from iterator operations or the element swap.

Complexity

At most N/2 swaps, where N = std::distance(first, last). Averaged over the entire
sequence of permutations, typical implementations use about 3 comparisons and 1.5
swaps per call.

Notes

Implementations (e.g. MSVC STL) may enable vectorization when the iterator type
satisfies LegacyContiguousIterator and swapping its value type calls neither
non-trivial special member function nor ADL-found swap.

Possible implementation

template<class BidirIt>
bool next_permutation(BidirIt first, BidirIt last)
{
auto r_first = std::make_reverse_iterator(last);
auto r_last = std::make_reverse_iterator(first);
auto left = std::is_sorted_until(r_first, r_last);
if(left != r_last){
auto right = std::upper_bound(r_first, left, *left);
std::iter_swap(left, right);
}
std::reverse(left.base(), last);
return left != r_last;
}

Example

The following code prints all three permutations of the string "aba"

// Run this code

#include <algorithm>
#include <string>
#include <iostream>

int main()
{
std::string s = "aba";
std::sort(s.begin(), s.end());
do {
std::cout << s << '\n';
} while(std::next_permutation(s.begin(), s.end()));
}

Output:

aab
aba
baa