Generating Lyndon words of length n
Last Updated : 09 Sep, 2022
Given an integer n and an array of characters S, the task is to generate Lyndon words of length n having characters from S.
A Lyndon word is a string which is strictly less than all of its rotations in lexicographic order. For example, the string "012" is a Lyndon word as it is less than its rotations "120" and "201", but "102" is not a Lyndon word as it is greater than its rotation "021".
Note: "000" is not considered to be a Lyndon word as it is equal to the string obtained by rotating it.
Examples:
Input: n = 2, S = {0, 1, 2}
Output: 01 02 12
Other possible strings of length 2 are "00", "11", "20", "21", and "22". All of these are either greater than or equal to one of their rotations.
Input: n = 1, S = {0, 1, 2}
Output: 0 1 2
Approach:
There exists an efficient approach to generate Lyndon words which was given by Jean-Pierre Duval, which can be used to generate all the Lyndon words upto length n in time proportional to the number of such words. (Please refer to the paper "Average cost of Duval’s algorithm for generating Lyndon words" by Berstel et al. for the proof) The algorithm generates the Lyndon words in lexicographic order.
If w is a Lyndon word, the next word is obtained by the following steps:
- Repeat w to form a string v of length n, such that v[i] = w[i mod |w|].
- While the last character of v is the last one in the sorted ordering of S, remove it.
- Replace the last character of v by its successor in the sorted ordering of S.
For example, if n = 5, S = {a, b, c, d}, and w = "add" then we get v = "addad". Since 'd' is the last character in the sorted ordering of S, we remove it to get "adda" and then replace the last 'a' by its successor 'b' to get the Lyndon word "addb".
Below is the implementation of the above approach:
C++ // C++ implementation of // the above approach #include<bits/stdc++.h> using namespace std; int main() { int n = 2; char S[] = {'0', '1', '2' }; int k = 3; sort(S, S + 3); // To store the indices // of the characters vector<int> w; w.push_back(-1); // Loop till w is not empty while(w.size() > 0) { // Incrementing the last character w[w.size()-1]++; int m = w.size(); if(m == n) { string str; for(int i = 0; i < w.size(); i++) { str += S[w[i]]; } cout << str << endl; } // Repeating w to get a // n-length string while(w.size() < n) { w.push_back(w[w.size() - m]); } // Removing the last character // as long it is equal to // the largest character in S while(w.size() > 0 && w[w.size() - 1] == k - 1) { w.pop_back(); } } return 0; } // This code is contributed by AdeshSingh1 // Java program to check if a string is two time // rotation of another string. import java.util.*; public class Main { // Driver method public static void main(String[] args) { int n=2; char S[]={'0', '1', '2' }; int k = 3; Arrays.sort(S); // To store the indices // of the characters ArrayList<Integer> w = new ArrayList<Integer>(); w.add(-1); // Loop till w is not empty while(w.size() > 0) { // Incrementing the last character Integer value = w.get(w.size()-1); // get value value = value + 1; // increment value w.set(w.size()-1, value); // replace value int m = w.size(); if(m == n) { String str=""; for(int i = 0; i < w.size(); i++) { value = w.get(i); str += S[value]; } System.out.println(str); } // Repeating w to get a // n-length string while(w.size() < n) { value = w.size() - m; value = w.get(value); w.add(value); } // Removing the last character // as long it is equal to // the largest character in S while(w.size() > 0 && w.get(w.size() - 1) == k - 1) { w.remove(w.size()-1); } } } } // This code is contributed by Aarti_Rathi # Python implementation of # the above approach n = 2 S = ['0', '1', '2'] k = len(S) S.sort() # To store the indices # of the characters w = [-1] # Loop till w is not empty while w: # Incrementing the last character w[-1] += 1 m = len(w) if m == n: print(''.join(S[i] for i in w)) # Repeating w to get a # n-length string while len(w) < n: w.append(w[-m]) # Removing the last character # as long it is equal to # the largest character in S while w and w[-1] == k - 1: w.pop() using System; using System.Collections.Generic; class GFG { // Driver code public static void Main () { int n=2; char [] S={'0', '1', '2' }; int k = 3; Array.Sort(S); // To store the indices // of the characters List<int> w = new List<int>(); w.Add(-1); // Loop till w is not empty while(w.Count > 0) { // Incrementing the last character w[w.Count-1]++; int m = w.Count; if(m == n) { string str=""; for(int i = 0; i < w.Count; i++) { str += S[w[i]]; } Console.WriteLine(str); } // Repeating w to get a // n-length string while(w.Count < n) { w.Add(w[w.Count - m]); } // Removing the last character // as long it is equal to // the largest character in S while(w.Count > 0 && w[w.Count - 1] == k - 1) { w.RemoveAt(w.Count-1); } } } } // This code is contributed by Aarti_Rathi <script> // JavaScript implementation of // the above approach // driver code let n = 2; let S = ['0', '1', '2']; let k = 3; S.sort(); // To store the indices // of the characters let w = []; w.push(-1); // Loop till w is not empty while(w.length > 0) { // Incrementing the last character w[w.length-1]++; let m = w.length; if(m == n) { let str = ""; for(let i = 0; i < w.length; i++) { str += S[w[i]]; } document.write(str,"</br>"); } // Repeating w to get a // n-length string while(w.length < n) { w.push(w[w.length - m]); } // Removing the last character // as long it is equal to // the largest character in S while(w.length > 0 && w[w.length - 1] == k - 1) { w.pop(); } } // This code is contributed by shinjanpatra </script> Similar Reads
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