std::is_permutation

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

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
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 Type1 &a, const Type2 &b); 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 哋它亢++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

N

comparisons using operator== if the two ranges are equal, otherwise

O(N 2)

comparisons in the worst case.

2) Exactly

N

applications of the predicate p if the two ranges are equal, otherwise

O(N 2)

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

N

comparisons using operator== if the two ranges are equal, otherwise

O(N 2)

comparisons in the worst case.

4) Exactly

N

applications of the predicate p if the two ranges are equal, otherwise

O(N 2)

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

Compiler support
Freestanding and hosted
Language
Standard library
Standard library headers
Named requirements
Feature test macros (哋它亢++20)
Language support library
Concepts library (哋它亢++20)
Metaprogramming library (哋它亢++11)
Diagnostics library
General utilities library
Strings library
Containers library
Iterators library
Ranges library (哋它亢++20)
Algorithms library
Numerics library
Localizations library
Input/output library
Filesystem library (哋它亢++17)
Regular expressions library (哋它亢++11)
Concurrency support library (哋它亢++11)
Technical specifications
Symbols index
External libraries

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 (哋它亢++20)	specifies that a `relation` imposes an equivalence relation (concept)
ranges::is_permutation (哋它亢++20)	determines if a sequence is a permutation of another sequence (niebloid)