Print all permutations of selected length characters with repetition in CPP
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generate all possible combinations of a set of characters
This code just prints numbers! What should I do to print string?!
for example, I choose 5 for length with ABC and the output should be:
aaaaa aaaab aaaac aaaba aaabb aaabc aaaca aaacb aaacc aabaa aabab aabac aabba aabbb aabbc aabca aabcb aabcc aacaa aacab aacac aacba aacbb aacbc aacca aaccb aaccc abaaa abaab abaac ababa ababb ababc abaca abacb abacc abbaa abbab abbac abbba abbbb abbbc abbca abbcb abbcc abcaa abcab abcac abcba abcbb abcbc abcca abccb abccc acaaa acaab acaac acaba acabb acabc acaca acacb acacc acbaa acbab acbac acbba acbbb acbbc acbca acbcb acbcc accaa accab accac accba accbb accbc accca acccb acccc baaaa baaab baaac baaba baabb baabc baaca baacb baacc babaa babab babac babba babbb babbc babca babcb babcc bacaa bacab bacac bacba bacbb bacbc bacca baccb baccc bbaaa bbaab bbaac bbaba bbabb bbabc bbaca bbacb bbacc bbbaa bbbab bbbac bbbba bbbbb bbbbc bbbca bbbcb bbbcc bbcaa bbcab bbcac bbcba bbcbb bbcbc bbcca bbccb bbccc bcaaa bcaab bcaac bcaba bcabb bcabc bcaca bcacb bcacc bcbaa bcbab bcbac bcbba bcbbb bcbbc bcbca bcbcb bcbcc bccaa bccab bccac bccba bccbb bccbc bccca bcccb bcccc caaaa caaab caaac caaba caabb caabc caaca caacb caacc cabaa cabab cabac cabba cabbb cabbc cabca cabcb cabcc cacaa cacab cacac cacba cacbb cacbc cacca caccb caccc cbaaa cbaab cbaac cbaba cbabb cbabc cbaca cbacb cbacc cbbaa cbbab cbbac cbbba cbbbb cbbbc cbbca cbbcb cbbcc cbcaa cbcab cbcac cbcba cbcbb cbcbc cbcca cbccb cbccc ccaaa ccaab ccaac ccaba ccabb ccabc ccaca ccacb ccacc ccbaa ccbab ccbac ccbba ccbbb ccbbc ccbca ccbcb ccbcc cccaa cccab cccac cccba cccbb cccbc cccca ccccb ccccc
Code:
#include <stdio.h> #include <list> #include <string> #include <iostream> using namespace std; int outputDigit; string input[1000]; int num; string final; int z=0; struct Generator { public: Generator(int s): cSlots(s) { a = new int[s]; for (int i = 0; i < cSlots - 1; i++) { a[i] = 1; } a[cSlots - 1] = 0; nextInd = cSlots; } ~Generator() { delete a; } bool doNext() { for (;;) { if (a[nextInd - 1] == cValues) { nextInd--; if (nextInd == 0) return false; } else { a[nextInd - 1]++; while (nextInd < cSlots) { nextInd++; a[nextInd - 1] = 1; } return true; } } } void doPrint() { for (int i = 0; i < cSlots; i++) { num=a[i]; cout<<num; } printf("\n"); } public: int *a; int cSlots; int cValues; int nextInd; }; int main() { cout<<"enter outputDigits"; cin>>outputDigit; Generator g(outputDigit); while (g.doNext()) { g.doPrint(); } return 0; }
but the output is
enter outputDigits5 11121 11122 11123 11124 11125 11126 11127 11128 11129 111210 111211 111212 111213 111214 111215 111216 111217 111218 111219 111220 111221 111222 111223 111224 111225 111226 111227 111228 111229 111230 111231 111232 111233 111234 111235......
You can do a simple exhaustive search.
#include <iostream> #include <string> using namespace std; void enumerate(const string& s, int n, string t = "") { if (n == 0) cout << t << endl; else { for (char c : s) enumerate(s, n - 1, t + c); } } int main() { enumerate("abc", 5); }
Complexity: O(s.size()^n)
Print all permutations with repetition of characters in C++, Print all permutations with repetition of characters in C++ Repeating of characters of the string is allowed. The printing of permutation should be done in alphabetical order (lexicographically sorted order). To solve this problem, we need to use fix and recur logic. Note that there are n! permutations and it requires O(n) time to print a a permutation. Note : The above solution prints duplicate permutations if there are repeating characters in input string. Please see below link for a solution that prints only distinct permutations even if there are duplicates in input.
What's cValues
?
I know that this answer will not give you the right answer directly. However, I'll advise you to make sure that you:
• initialize all attributes in the constructor.
• avoid using global members.
• use more descriptive names.
• try to reproduce your bugs with an input of smaller size first (e.g. 2 or 3).
Regards,
All permutations of length k from n characters with repetition in CPP , The idea is to start from an empty output string (we call it prefix in following code). One by one add all characters to prefix. For every character added, print all possible strings with current prefix by recursively calling for "length" equals to "length"-1. Update: this update is writen answring the following spec. Generate all binary permutations such that there are more or equal 1's than 0's before every point in all permutations Print consecutive characters together in a line Print all valid words that are possible using Characters of Array
A hard code solution would be:
const char choices[] = {'a', 'b', 'c'}; for (int a1 : choices) { for (int a2 : choices) { for (int a3 : choices) { for (int a4 : choices) { for (int a5 : choices) { do_job(a1, a2, a3, a4, a5); } } } } }
You may use the following for a generic way (putting all a
s into vector):
template <typename T> bool increase(const std::vector<T>& v, std::vector<std::size_t>& it) { for (std::size_t i = 0, size = it.size(); i != size; ++i) { const std::size_t index = size - 1 - i; ++it[index]; if (it[index] >= v.size()) { it[index] = 0; } else { return true; } } return false; } template <typename T> void do_job(const std::vector<T>& v, const std::vector<std::size_t>& it) { for (const auto e : it) { std::cout << v[e] << " "; } std::cout << std::endl; } template <typename T> void iterate(const std::vector<T>& v, std::size_t size) { std::vector<std::size_t> it(size, 0); do { do_job(v, it); } while (increase(v, it)); }
Permutations with repetitions, In case you ever want to just display the string (if you have something like "AAA"): // if a different character from the first is not found // std::string::npos� In this problem, we are given a string of n characters and we have to print all permutations of characters of the string. Repeating of characters of the string is allowed. The printing of permutation should be done in alphabetical order (lexicographically sorted order). Let’s take an example to understand the topic better : Input: XY
Print all permutations with repetition of characters, Once all permutations starting with the first character are printed, fix the second Dec 08, 2007 � We know that the number of permutations with repetition is n^r where n is When the repetition of items is allowed, at every step of selection from the set of 'n' Print all permutations with repetition of characters in C++ Print all� Print all permutations of a string in Java; Check if a binary string contains all permutations of length k; Generate all permutations of a string that follow given constraints; Write a program to print all permutations of a given string; Check if given string can be formed by two other strings or their permutations; Print all lexicographical
All permutations with repetition, Print all of the possible ways to write a string of length N from the characters in To refer to combinations in which repetition is allowed, the terms k-selection, How to generate all Permutations - posted in C/C++ Tutorials: Problem: Given a� Print all the permutations of a string without repetition using Collections in Java Last Updated: 03-09-2019 Given a string str , the task is to print all the permutations of str .
Generate all possible combinations of a set of characters javascript, This is a C++ program to generate all possible combinations of given list of This algorithm print all the possible combination of each length from the given array. First is, we select it, means put true in check[] and increment currLen and start. And we have to make all the permutations of the digits 2, 3 and 4. So, we will make the permutations of 2, 3 and 4 by keeping 2 fixed. Thus the numbers obtained are: 1234 1243. Now, we will fix 3 out of 2, 3 and 4. 1324 1342. Again, keeping 4 fixed out of 2, 3 and 4. 1423 1432. Now, we have all the numbers which can be made by keeping 1 at the
Comments
- Please edit to remove unneeded empty lines from you code and choose a smaller example, maybe with 4 or 3 instead of 5. Both to make the post more readable and get potential helpers into a better and more helpful mood.
- @Some I usually like your contributions, but OP could not have done that edit without being told by system to give more helpful prose with all that code... And I am sure that would have been a good thing, see my above comment.
- actually, I searched a lot but it wasn't what I want or it wasn't written by CPP anyway thank you it works
- do you need
v
inincrease
, and not justmax_index
? - @Caleth:
v.size()
is indeed enough. I adapted one of my answer to similar but different question which use the vector.