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.

The overloads (1) and (3) use operator== for equality, whereas the overloads (2) and (4) use the binary predicate p.

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

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)