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std::is_permutation - cppreference.com

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.

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 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. ​

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

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.

If

and

are both

, and

is

, 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.

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;
}

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.

#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

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