NAME¶

std::next_permutation - std::next_permutation

Synopsis¶

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

Permutes the range [first, last) into the next permutation. Returns true if such a
“next permutation” exists; otherwise transforms the range into the lexicographically
first permutation (as if by std::sort) and returns false.

1) The set of all permutations is ordered lexicographically with respect to
operator<
(until C++20)
std::less{}
(since C++20).
2) The set of all permutations is ordered lexicographically with respect to comp.

If
the type of *first is not Swappable
(until C++11)
BidirIt is not ValueSwappable
(since C++11), the behavior is undefined.

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 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.

Complexity¶

Given \(\scriptsize N\)N as std::distance(first, last):

1,2) At most \(\scriptsize \frac{N}{2}\)

N
2

swaps.

Exceptions¶

Any exceptions thrown from iterator operations or the element 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;
}

Notes¶

Averaged over the entire sequence of permutations, typical implementations use about
3 comparisons and 1.5 swaps per call.

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.

Example¶

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

// Run this code

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

int main()
{
std::string s = "aba";

do
{
std::cout << s << '\n';
}
while (std::next_permutation(s.begin(), s.end()));

std::cout << s << '\n';
}

Output:¶

aba
baa
aab

See also¶

is_permutation determines if a sequence is a permutation of another
(C++11) sequence
(function template)
generates the next smaller lexicographic permutation of a
prev_permutation range of elements
(function template)
ranges::next_permutation generates the next greater lexicographic permutation of a
(C++20) range of elements
(niebloid)