std::is_permutation - cppreference.com

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# std::is_permutation

From cppreference.com
< cpp‎ | algorithm

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`.

### [edit] 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.

### [edit] Return value

true if the range `[first1, last1)` is a permutation of the range `[first2, last2)`.

### [edit] 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.

### [edit] 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;
    }

### [edit] 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

### [edit] See also

next_permutation | generates the next greater lexicographic permutation of a range of elements (function template) [edit]
prev_permutation | generates the next smaller lexicographic permutation of a range of elements (function template) [edit]

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