std::is_permutation() algorithm

od C++20

// (1)
template< class ForwardIt1, class ForwardIt2 >
constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 );

// (2)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >
constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p );

// (3)
template< class ForwardIt1, class ForwardIt2 >
constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 );

// (4)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >
constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1,
                               ForwardIt2 first2, ForwardIt2 last2, BinaryPredicate p );

od C++14

// (1)
template< class ForwardIt1, class ForwardIt2 >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 );

// (2)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p );

// (3)
template< class ForwardIt1, class ForwardIt2 >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 );

// (4)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, BinaryPredicate p );

od C++11

// (1)
template< class ForwardIt1, class ForwardIt2 >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 );

// (2)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >
bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p );

Returns true if there exists a permutation of the elements in the range [first1; last1) that makes that range equal to the range [first2; last2), where last2 denotes first2 + (last1 - first1) if it was not given.

• (1, 3) Elements are compared using operator==.
• (2, 4) Elements are compared using the given binary predicate p.

Undefined Behaviour

The behavior is undefined

if it is not an equivalence relation.

Parameters

first1 last1	The range of elements to compare.
first1 last1	The second range to compare.
p	Unary predicate which returns true if the element value should be replaced. Type should be the value type of both ForwardIt1 and ForwardIt2. The signature does not need to have const&, but the function must not modify the objects passed to it.

Type requirements

ForwardIt1 ForwardIt2	LegacyForwardIterator
ForwardIt1 ForwardIt2	Must have the same value type.

Return value

true if the range [first1; last1) is a permutation of the range [first2; last2).

Complexity

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

At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N is .

However if ForwardIt1 and ForwardIt2 meet the requirements of LegacyRandomAccessIterator and std::distance(first1, last1) != std::distance(first2, last2) no applications of the predicate are made.

Exceptions

The overloads with a template parameter named ExecutionPolicy report errors as follows:

• If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the standard policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
• If the algorithm fails to allocate memory, std::bad_alloc is thrown.

Possible implementation

is_permutation (1)

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;
}

Notes

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.

Examples

Main.cpp

#include <algorithm>
#include <iostream>

template<typename Os, typename V>
Os& operator<<(Os& os, V const& v)
{
    os << "{ ";
    for (auto const& 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