NAME¶

std::is_permutation - std::is_permutation

Synopsis¶

Defined in header <algorithm>
template< class ForwardIt1, class ForwardIt2 >

bool is_permutation( ForwardIt1 first1, ForwardIt1 (1) (since C++11)
last1, (constexpr since C++20)

ForwardIt2 first2 );
template< class ForwardIt1, class ForwardIt2,

class BinaryPredicate >
bool is_permutation( ForwardIt1 first1, ForwardIt1 (2) (since C++11)
last1, (constexpr since C++20)

ForwardIt2 first2, BinaryPredicate
p );
template< class ForwardIt1, class ForwardIt2 >

bool is_permutation( ForwardIt1 first1, ForwardIt1 (since C++14)
last1, (3) (constexpr since C++20)

ForwardIt2 first2, ForwardIt2 last2
);
template< class ForwardIt1, class ForwardIt2,

class BinaryPredicate >
bool is_permutation( ForwardIt1 first1, ForwardIt1 (since C++14)
last1, (4) (constexpr since C++20)
ForwardIt2 first2, ForwardIt2
last2,

BinaryPredicate p );

Checks whether [first1, last1) is a permutation of a range starting from first2:

* For overloads (1,2), the second range has std::distance(first1, last1) elements.
* For overloads (3,4), the second range is [first2, last2).
1,3) Elements are compared using operator==.
2,4) Elements are compared using the given binary predicate p.

If ForwardIt1 and ForwardIt2 have different value types, the program is ill-formed.

If the comparison function is not an equivalence relation, the behavior is
undefined.

Parameters¶

first1, last1 - the range of elements to compare
first2, last2 - the second range to compare
binary predicate which returns true if the elements should be
treated as equal.

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

bool pred(const Type1 &a, const Type2 &b);

p - 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 objects of types
InputIt1 and InputIt2 can be dereferenced and then implicitly
converted to Type1 and Type2 respectively.

Type requirements¶

-
ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator.

Return value¶

true if the range [first1, last1) is a permutation of the range [first2, last2),
false otherwise.

Complexity¶

Given \(\scriptsize N\)N as std::distance(first1, last1):

1) Exactly \(\scriptsize N\)N comparisons using operator== if the two ranges are
equal, otherwise \(\scriptsize O(N^2)\)O(N2
) comparisons in the worst case.
2) Exactly \(\scriptsize N\)N applications of the predicate p if the two ranges are
equal, otherwise \(\scriptsize O(N^2)\)O(N2
) applications in the worst case.
3,4) If ForwardIt1 and ForwardIt2 are both LegacyRandomAccessIterator, and last1 -
first1 != last2 - first2 is true, no comparison will be made.
Otherwise:
3) Exactly \(\scriptsize N\)N comparisons using operator== if the two ranges are
equal, otherwise \(\scriptsize O(N^2)\)O(N2
) comparisons in the worst case.
4) Exactly \(\scriptsize N\)N applications of the predicate p if the two ranges are
equal, otherwise \(\scriptsize O(N^2)\)O(N2
) applications in the worst case.

Possible implementation¶

template<class ForwardIt1, class ForwardIt2>
bool is_permutation(ForwardIt1 first, ForwardIt1 last,
ForwardIt2 d_first)
{
// skip common prefix
std::tie(first, d_first) = std::mismatch(first, last, d_first);

// iterate over the rest, counting how many times each element
// from [first, last) appears in [d_first, d_last)
if (first != last)
{
ForwardIt2 d_last = std::next(d_first, std::distance(first, last));
for (ForwardIt1 i = first; i != last; ++i)
{
if (i != std::find(first, i, *i))
continue; // this *i has been checked

auto m = std::count(d_first, d_last, *i);
if (m == 0 || std::count(i, last, *i) != m)
return false;
}
}
return true;
}

Note¶

The std::is_permutation can be used in testing, namely to check the correctness of
rearranging algorithms (e.g. sorting, shuffling, partitioning). If x is an original
range and y is a permuted range then std::is_permutation(x, y) == true means that y
consist of "the same" elements, maybe staying at other positions.

Example¶

// Run this code

#include <algorithm>
#include <iostream>

template<typename Os, typename V>
Os& operator<<(Os& os, const V& v)
{
os << "{ ";
for (const auto& e : v)
os << e << ' ';
return os << '}';
}

int main()
{
static constexpr auto v1 = {1, 2, 3, 4, 5};
static constexpr auto v2 = {3, 5, 4, 1, 2};
static constexpr auto v3 = {3, 5, 4, 1, 1};

std::cout << v2 << " is a permutation of " << v1 << ": " << std::boolalpha
<< std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n'
<< v3 << " is a permutation of " << v1 << ": "
<< std::is_permutation(v1.begin(), v1.end(), v3.begin()) << '\n';
}

Output:¶

{ 3 5 4 1 2 } is a permutation of { 1 2 3 4 5 }: true
{ 3 5 4 1 1 } is a permutation of { 1 2 3 4 5 }: false

See also¶

generates the next greater lexicographic permutation of a
next_permutation range of elements
(function template)
generates the next smaller lexicographic permutation of a
prev_permutation range of elements
(function template)
equivalence_relation specifies that a relation imposes an equivalence relation
(C++20) (concept)
ranges::is_permutation determines if a sequence is a permutation of another sequence
(C++20) (niebloid)