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# De Bruijn sequence | Set 1

• Difficulty Level : Expert
• Last Updated : 10 Jul, 2021

Given an integer n and a set of characters A of size k, find a string S such that every possible string on A of length n appears exactly once as a substring in S. Such a string is called de Bruijn sequence.

Examples:

Input: n = 3, k = 2, A = {0, 1)
Output: 0011101000
All possible strings of length three (000, 001, 010, 011, 100, 101, 110 and 111) appear exactly once as sub-strings in A.

Input: n = 2, k = 2, A = {0, 1)
Output: 01100

Approach:
We can solve this problem by constructing a directed graph with kn-1 nodes with each node having k outgoing edges. Each node corresponds to a string of size n-1. Every edge corresponds to one of the k characters in A and adds that character to the starting string.

For example, if n=3 and k=2, then we construct the following graph:

• The node ’01’ is connected to node ’11’ through edge ‘1’, as adding ‘1’ to ’01’ (and removing the first character) gives us ’11’.
• We can observe that every node in this graph has equal in-degree and out-degree, which means that a Eulerian circuit exists in this graph.
• The Eulerian circuit will correspond to a de Bruijn sequence as every combination of a node and an outgoing edge represents a unique string of length n.
• The de Bruijn sequence will contain the characters of the starting node and the characters of all the edges in the order they are traversed in.
• Therefore the length of the string will be kn+n-1. We will use Hierholzer’s Algorithm to find the Eulerian circuit. The time complexity of this approach is O(kn).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `unordered_set seen;``vector<``int``> edges;` `// Modified DFS in which no edge``// is traversed twice``void` `dfs(string node, ``int``& k, string& A)``{``    ``for` `(``int` `i = 0; i < k; ++i) {``        ``string str = node + A[i];``        ``if` `(seen.find(str) == seen.end()) {``            ``seen.insert(str);``            ``dfs(str.substr(1), k, A);``            ``edges.push_back(i);``        ``}``    ``}``}` `// Function to find a de Bruijn sequence``// of order n on k characters``string deBruijn(``int` `n, ``int` `k, string A)``{` `    ``// Clearing global variables``    ``seen.clear();``    ``edges.clear();` `    ``string startingNode = string(n - 1, A[0]);``    ``dfs(startingNode, k, A);` `    ``string S;` `    ``// Number of edges``    ``int` `l = ``pow``(k, n);``    ``for` `(``int` `i = 0; i < l; ++i)``        ``S += A[edges[i]];``    ``S += startingNode;` `    ``return` `S;``}` `// Driver code``int` `main()``{``    ``int` `n = 3, k = 2;``    ``string A = ``"01"``;` `    ``cout << deBruijn(n, k, A);` `    ``return` `0;``}`

## Java

 `// Java implementation of``// the above approach``import` `java.util.*;` `class` `GFG``{` `    ``static` `Set seen = ``new` `HashSet();``    ``static` `Vector edges = ``new` `Vector();` `    ``// Modified DFS in which no edge``    ``// is traversed twice``    ``static` `void` `dfs(String node, ``int` `k, String A)``    ``{``        ``for` `(``int` `i = ``0``; i < k; ++i)``        ``{``            ``String str = node + A.charAt(i);``            ``if` `(!seen.contains(str))``            ``{``                ``seen.add(str);``                ``dfs(str.substring(``1``), k, A);``                ``edges.add(i);``            ``}``        ``}``    ``}` `    ``// Function to find a de Bruijn sequence``    ``// of order n on k characters``    ``static` `String deBruijn(``int` `n, ``int` `k, String A)``    ``{` `        ``// Clearing global variables``        ``seen.clear();``        ``edges.clear();` `        ``String startingNode = string(n - ``1``, A.charAt(``0``));``        ``dfs(startingNode, k, A);` `        ``String S = ``""``;` `        ``// Number of edges``        ``int` `l = (``int``) Math.pow(k, n);``        ``for` `(``int` `i = ``0``; i < l; ++i)``            ``S += A.charAt(edges.get(i));``        ``S += startingNode;` `        ``return` `S;``    ``}` `    ``private` `static` `String string(``int` `n, ``char` `charAt)``    ``{``        ``String str = ``""``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``str += charAt;``        ``return` `str;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``3``, k = ``2``;``        ``String A = ``"01"``;` `        ``System.out.print(deBruijn(n, k, A));``    ``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of``# the above approach``import` `math` `seen ``=` `set``()``edges ``=` `[]` `# Modified DFS in which no edge``# is traversed twice``def` `dfs( node, k, A):``    ` `    ``for` `i ``in` `range``(k):``        ``str` `=` `node ``+` `A[i]``        ``if` `(``str` `not` `in` `seen):``            ``seen.add(``str``)``            ``dfs(``str``[``1``:], k, A)``            ``edges.append(i)` `# Function to find a de Bruijn sequence``# of order n on k characters``def` `deBruijn(n, k, A):``    ` `    ``# Clearing global variables``    ``seen.clear()``    ``edges.clear()``    ` `    ``startingNode ``=` `A[``0``] ``*` `(n ``-` `1``)``    ``dfs(startingNode, k, A)``    ` `    ``S ``=` `""``    ` `    ``# Number of edges``    ``l ``=` `int``(math.``pow``(k, n))``    ``for` `i ``in` `range``(l):``        ``S ``+``=` `A[edges[i]]``        ` `    ``S ``+``=` `startingNode``    ``return` `S` `# Driver code``n ``=` `3``k ``=` `2``A ``=` `"01"` `print``(deBruijn(n, k, A))` `# This code is contributed by shubhamsingh10`

## C#

 `// C# implementation of``// the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``static` `HashSet seen = ``new` `HashSet();``    ``static` `List<``int``> edges = ``new` `List<``int``>();` `    ``// Modified DFS in which no edge``    ``// is traversed twice``    ``static` `void` `dfs(String node, ``int` `k, String A)``    ``{``        ``for` `(``int` `i = 0; i < k; ++i)``        ``{``            ``String str = node + A[i];``            ``if` `(!seen.Contains(str))``            ``{``                ``seen.Add(str);``                ``dfs(str.Substring(1), k, A);``                ``edges.Add(i);``            ``}``        ``}``    ``}` `    ``// Function to find a de Bruijn sequence``    ``// of order n on k characters``    ``static` `String deBruijn(``int` `n, ``int` `k, String A)``    ``{` `        ``// Clearing global variables``        ``seen.Clear();``        ``edges.Clear();` `        ``String startingNode = strings(n - 1, A[0]);``        ``dfs(startingNode, k, A);` `        ``String S = ``""``;` `        ``// Number of edges``        ``int` `l = (``int``) Math.Pow(k, n);``        ``for` `(``int` `i = 0; i < l; ++i)``            ``S += A[edges[i]];``        ``S += startingNode;` `        ``return` `S;``    ``}` `    ``private` `static` `String strings(``int` `n, ``char` `charAt)``    ``{``        ``String str = ``""``;``        ``for` `(``int` `i = 0; i < n; i++)``            ``str += charAt;``        ``return` `str;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `n = 3, k = 2;``        ``String A = ``"01"``;` `        ``Console.Write(deBruijn(n, k, A));``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`0011101000`

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