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std::is_permutation

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

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 `ForwardIterator`.
-`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 O(N²) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first1, last1).

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

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 = d_first;
       std::advance(d_last, std::distance(first, last));
       for (ForwardIt1 i = first; i != last; ++i) {
            if (i != std::find(first, i, *i)) continue; // already counted this *i
 
            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 <vector>
#include <iostream>
int main()
{
    std::vector<int> v1{1,2,3,4,5};
    std::vector<int> v2{3,5,4,1,2};
    std::cout << "3,5,4,1,2 is a permutation of 1,2,3,4,5? "
              << std::boolalpha
              << std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n';
 
    std::vector<int> v3{3,5,4,1,1};
    std::cout << "3,5,4,1,1 is a permutation of 1,2,3,4,5? "
              << std::boolalpha
              << 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