Std::is_permutation (3) Linux Manual Page

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

std::is_permutation (3) Linux Manual Page