std::is_permutation
Defined in header <algorithm>
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template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(1) | (since C++11) (constexpr since C++20) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > |
(2) | (since C++11) (constexpr since C++20) |
template< class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first1, ForwardIt1 last1, |
(3) | (since C++14) (constexpr since C++20) |
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > |
(4) | (since C++14) (constexpr since C++20) |
Checks whether [
first1,
last1)
is a permutation of a range starting from first2:
- For overloads (1,2), the second range has std::distance(first1, last1) elements.
- For overloads (3,4), the second range is
[
first2,
last2)
.
If ForwardIt1
and ForwardIt2
have different value types, the program is ill-formed.
If the comparison function is not an equivalence relation, the behavior is undefined.
Contents |
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 Type1 &a, const Type2 &b); While the signature does not need to have const &, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) |
Type requirements | ||
-ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator.
|
Return value
true if the range [
first1,
last1)
is a permutation of the range [
first2,
last2)
, false otherwise.
Complexity
Given N as std::distance(first1, last1):
ForwardIt1
and ForwardIt2
are both LegacyRandomAccessIterator, and last1 - first1 != last2 - first2 is true, no comparison will be 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 = 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; } |
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.
Example
#include <algorithm> #include <iostream> template<typename Os, typename V> Os& operator<<(Os& os, const V& v) { os << "{ "; for (const auto& 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::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 permutation of a range of elements (function template) | |
generates the next smaller lexicographic permutation of a range of elements (function template) | |
(C++20) |
specifies that a relation imposes an equivalence relation (concept) |
(C++20) |
determines if a sequence is a permutation of another sequence (niebloid) |