The word 'permutation' also refers to the act or process of changing the linear order of an ordered set. Interestingly, there's an example of implementation on cppreference. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. Now if you want to reinvent the C++ wheel, the best thing would be to re-implement std::next_permutation: an algorithm that does its work incrementally, in place, and with iterators (meaning that you can compute the permutations of strings, arrays, double-linked lists and everything that exposes bidirectional iterators). The function returns true if the next higher permutation exists otherwise, it returns false to indicate that the object is already at the highest possible. I understand the need to reinvent the wheel but in this case you reinvented an other kind of wheel: functional-style combinatorics aren't very well suited to C++ and the high-performance / low-memory usage it is well known for. The STL provides std::nextpermutation, which returns the next permutation in lexicographic order by in-place rearranging the specified object as a lexicographically greater permutation. So instead of writing for(.) if(.) v.push_back(x) just write for(.) You are more likely to make mistakes when you put multiple statements on the same line. You don't need std::to_string, just print b. Std::copy(b.begin(), b.end(), std::back_inserter(res)) Std::vector > merge(const std::vector >& a, const std::vector >& b) Std::cout remove_item(const std::vector& a, int id)Īssert(id >= 0 & id res(a.begin(), a.begin() + id) Your problem is solved by std::next_permutation, so you can proceed along the lines of: #include Vector > permutations(const vector& b, const vector& a=)įor(unsigned int i=0 i<< "Sol Num = " << res.size() << endl įor(const auto& b : a) cout << to_string(b) << " " Ī cleaner solution is to trust the standard library and try to re-use the generic components already available there. Vector > merge(const vector >& a, const vector >& b) Vector remove_item (const vector& a, const unsigned int id)įor(unsigned int i=0 i add_item(const vector& a, const int b) The recursion ends when there are no more elements to place hence b.empty() is trueĮach recursion set ends with a valid sequence hence they are all merged together in a final list of sequences Inside the recursion, a loop places the N(i) (with i recursion index and) remaining elements producing the same amount of new root sequences and a new recursion is started so that N(i+1)=N(i)-1 hence meaning the overall complexity is O(N!) as expected The remaining elements set b which is the set of elements still to be placed The root sequence a which is the set of elements already placed The strategy is based on recursion + iterations Step-5 reverse the array from the pivot+1.The task is to compute all the permutations for a given vector of integers (but of course the specific integer type is not relevant for the solution) Step-4 swap the pivot number with its exact largest one.Īfter the swap, the array looks like this num Step-3 if the first number pivot is found, find the exact largest number than the first number pivot from the right side.įrom the num array the exact largest number of pivot=1 is 2 Step-2 if the pivot is not found, all numbers are in ascending order from the right side, which means the given permutation is the last permutation. Step-1 find the first number pivot which not increasing in ascending order, from the right side.įrom the num array, 1 is the number that is not increasing in ascending order Lets understand the algo above one in a clear way with an example. A permutation of a set, in mathematics, is an ordering of its elements into a series or linear sequence or a reordering of its elements if such a set is. Find the largest index j greater than i such that Lj > L. Find the largest index k such that nums k such that nums The following algorithm is presented by a man named Narayan Pandita in the 14th century. we have to rearrange the array in the form of the next greater permutation in lexicographically or dictionary order. In this problem, we have given an int array. Today we will understand the 3rd problem from the SDE-Sheet which is the Next Permutation.
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