std::is_permutation

Defined in header `<algorithm>`
	(1)
`template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 );`		(since C++11) (until C++20)
`template< class ForwardIt1, class ForwardIt2 > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 );`		(since C++20)
	(2)
`template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p );`		(since C++11) (until C++20)
`template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p );`		(since C++20)
	(3)
`template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 );`		(since C++14) (until C++20)
`template< class ForwardIt1, class ForwardIt2 > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 );`		(since C++20)
	(4)
`template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, BinaryPredicate p );`		(since C++14) (until C++20)
`template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > constexpr bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2, BinaryPredicate p );`		(since C++20)

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==. The behavior is undefined if it is not an equivalence relation.

2,4) Elements are compared using the given binary predicate p. The behavior is undefined if it is not an equivalence relation.

Parameters

first1, last1	-	the range of elements to compare
first2, last2	-	the second range to compare
p	-	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 Type &a, const Type &b);` `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` must meet the requirements of 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

At most \(\scriptsize \mathcal{O}(N^2)\)O(N²) applications of the predicate, or exactly \(\scriptsize N\)N if the sequences are already equal, where \(\scriptsize N\)N is std::distance(first1, last1).

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.

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.

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;
}

Example

#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

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

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