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std::is_permutation(3) C++ Standard Libary std::is_permutation(3) 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, (until ForwardIt1 last1, C++20) ForwardIt2 first2 ); template< class ForwardIt1, class ForwardIt2 > (since constexpr bool is_permutation( ForwardIt1 C++20) first1, ForwardIt1 last1, ForwardIt2 first2 ); template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > (since C++11) bool is_permutation( ForwardIt1 first1, (until ForwardIt1 last1, C++20) ForwardIt2 first2, BinaryPredicate p ); template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > (since constexpr bool is_permutation( ForwardIt1 C++20) first1, ForwardIt1 last1, ForwardIt2 first2, BinaryPredicate p ); template< class ForwardIt1, class ForwardIt2 (1) > (since C++14) bool is_permutation( ForwardIt1 first1, (until ForwardIt1 last1, C++20) ForwardIt2 first2, ForwardIt2 last2 ); template< class ForwardIt1, class ForwardIt2 (2) > (since constexpr bool is_permutation( ForwardIt1 C++20) first1, ForwardIt1 last1, ForwardIt2 first2, ForwardIt2 last2 ); template< class ForwardIt1, class (3) ForwardIt2, class BinaryPredicate > (since bool is_permutation( ForwardIt1 first1, C++14) ForwardIt1 last1, (until ForwardIt2 first2, ForwardIt2 last2, C++20) BinaryPredicate p ); (4) template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > constexpr bool is_permutation( ForwardIt1 (since first1, ForwardIt1 last1, 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 unde- fined 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. The signature of the predicate function should be equivalent to the following: p - 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 LegacyFor- wardIterator. - 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 \(\scriptsize \mathcal{O}(N^2)\)O(N^2) applications of the predicate, or exactly \(\scriptsize N\)N if the sequences are already equal, where \(\scriptsize N\)N is 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. Note The std::is_permutation can be used in testing, namely to check the correctness of rearranging algorithms (e.g. sorting, shuffling, partitioning). If x is an original range and y is a permuted range then std::is_permutation(x, y) == true means that y consist of "the same" elements, maybe staying at other positions. 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 = std::next(d_first, std::distance(first, last)); for (ForwardIt1 i = first; i != last; ++i) { if (i != std::find(first, i, *i)) continue; // this *i has been checked 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 <iostream> #include <algorithm> template<typename Os, typename V> Os& operator<< (Os& os, V const& v) { os << "{ "; for (auto const& e : v) os << e << ' '; return os << "}"; } int main() { static constexpr auto v1 = {1,2,3,4,5}; static constexpr auto v2 = {3,5,4,1,2}; static constexpr auto v3 = {3,5,4,1,1}; std::cout << v2 << " is a permutation of " << v1 << ": " << std::boolalpha << std::is_permutation(v1.begin(), v1.end(), v2.begin()) << '\n' << v3 << " is a permutation of " << v1 << ": " << 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 generates the next greater lexicographic per- mutation of a next_permutation range of elements (function template) generates the next smaller lexicographic per- mutation of a prev_permutation range of elements (function template) equivalence_relation specifies that a relation imposes an equiva- lence relation (C++20) (concept) ranges::is_permutation determines if a sequence is a permutation of another sequence (C++20) (niebloid) http://cppreference.com 2022.07.31 std::is_permutation(3)
NAME | Synopsis | Parameters | Type requirements | Return value | Complexity | Note | Possible implementation | Example | Output: | See also
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