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 k^{n-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 k
^{n}+n-1. We will use Hierholzer’s Algorithm to find the Eulerian circuit. The time complexity of this approach is O(k^{n}).

Below is the implementation of the above approach:

`// C++ implementation of ` `// the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `unordered_set<string> 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

0011101000

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