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std::is_permutation

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<td>
Defined in header <algorithm>
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<td >
template< class ForwardIt1, class ForwardIt2 >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first );
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<td > (1) </td> <td > (since C++11) </td> </tr> <tr class="t-dcl ">

<td >
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first, BinaryPredicate p );
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<td > (2) </td> <td > (since C++11) </td> </tr> Template:ddcl list end

Returns true if there exists a permutation of the elements in the range [first1, last1) that makes that range equal to the range beginning at d_first. The first version uses operator== for equality, the second version uses the binary predicate p

Contents

Parameters

first, last - the range of elements to compare
d_first - the beginning of 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 Type1 &a, const Type2 &b);

The signature does not need to have const &, but the function must not modify the objects passed to it.
The types Type1 and Type2 must be such that objects of types ForwardIt1 and ForwardIt2 can be dereferenced and then implicitly converted to Type1 and Type2 respectively.

Type requirements
-
ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator.

Return value

true if the range [first, last) is a permutation of the range beginning at d_first.

Complexity

At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first, last).

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

See also

Template:cpp/algorithm/dcl list next permutationTemplate:cpp/algorithm/dcl list prev permutation