NAME¶

std::prev_permutation - std::prev_permutation

Synopsis¶

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

Transforms the range [first, last) into the previous permutation. Returns true if
such permutation exists, otherwise transforms the range into the last permutation
(as if by std::sort followed by std::reverse) 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 ValueSwappable and
LegacyBidirectionalIterator.

Return value¶

true if the new permutation precedes the old in lexicographical order. false if the
first permutation was reached and the range was reset to the last permutation.

Exceptions¶

Any exceptions thrown from iterator operations or the element swap.

Complexity¶

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

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

N
2

swaps.

Possible implementation¶

template<class BidirIt>
bool prev_permutation(BidirIt first, BidirIt last)
{
if (first == last)
return false;
BidirIt i = last;
if (first == --i)
return false;

while (1)
{
BidirIt i1, i2;

i1 = i;
if (*i1 < *--i)
{
i2 = last;
while (!(*--i2 < *i))
;
std::iter_swap(i, i2);
std::reverse(i1, last);
return true;
}

if (i == first)
{
std::reverse(first, last);
return false;
}
}
}

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 six permutations of the string "cab" in reverse order.

// Run this code

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

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

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

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

Output:¶

cab bca bac acb abc cba

See also¶

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