std::is_permutation (3) - Linux Manuals

std::is_permutation: std::is_permutation

NAME

std::is_permutation - std::is_permutation

Synopsis

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

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
                binary predicate which returns true if the elements should be treated as equal.
p - 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 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 LegacyRandomAccessIterator 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

// Run this code

  #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