The following graph G is called a Petersen graph and its vertices have been numbered from 0 to 9. Some letters have also been assigned to vertices of G, as can be seen from the following picture:
Let’s consider a walk W in graph G, which consists of L vertices W1, W2, …, WL. A string S of L letters ‘A’ – ‘E’ is realized by walking W if the sequence of letters written along W is equal to S. Vertices can be visited multiple times while walking along W.
For example, S = ‘ABBECCD’ is realized by W = (0, 1, 6, 9, 7, 2, 3). Determine whether there is a walk W that realizes a given string S in graph G and if so then find the lexicographically least such walk. The only line of input contains one string S. If there is no walk W which realizes S, then output -1 otherwise, you should output the least lexicographical walk W which realizes S.
Example of a Petersen Graph
Input : s = 'ABB'
Explanation: As we can see in the graph
the path from ABB is 016.
Input : s = 'AABE'
Explanation: As there is no path that
exists, hence output is -1.
begin res := starting vertex for each character c in S except the first one, do if there is an edge between v and c in outer graph, then v := c else if there is an edge between v and c+5 in inner graph, then v := c + 5 else return false end if put v into res done return true end
Below is the implementation of the above algorithm:
The time complexity of the above program is O(N), where N is the length of the given string S. We are applying Breadth First Search here, which runs in linear time.
Space complexity: O(N)
The space complexity of the above program is O(N), where N is the length of the given string S. We are using two auxiliary arrays – result and S to store the path and the given string, respectively. Both of them require linear space.
This article is contributed by Sunidhi Chaudhary. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Login to comment...