#ANOTHER WORD FOR SORT THROUGH CODE#

Change each letter at START to the letter to STOP one by one and check if such word exists. Finish to change the word when START equals STOP. List is our 'way' from words START to STOP, so, we add the original word to it first. Please, check this pseudocode to see my idea: Two words are given: START and STOP. My idea is to load the dictionary as a HashSet and use contains() method to find out if the word exists in the dictionary or not. I don't think we need graph or some other complicated data structure. Public static List FindPath(List input, string start, string end)Ĭhar allcharacters = įor (int i = 0 i ExtractPath(string start,string end)

Static Dictionary parents = new Dictionary() dictionary used to find the parent in every node in the graph and to avoid traversing an already traversed node Static HashSet Dictionary = new HashSet()

#ANOTHER WORD FOR SORT THROUGH CODE#

This is C# code to solve the problem using BFS: //use a hash set for a fast check if a word is already in the dictionary To make the path tracing easier, we can add extra information of parent string in each node. If this set contains the target string, stop further processing of nodes/strings.Add all these strings to visitedSetOfStrings.Add non-visited valid strings to next level which are at edit distance +1 from the strings of current level.Using BFS, we can simplify this process by a great deal. Then we have to keep track of the levels at which it is found and the reference to that node, and finally find the minimum level possible and then trace it from root.įor example, consider this conversion from -> zoom: fromĭram drom food << To traverse upto this level is enough If we go with DFS, there are chances that we meet target string (or to_string) far deeper in the graph. Transform(english_words, 'damp', 'lame', solution is correct but it misses on the opportunity to simplify and optimize the solution. If is_diff_one(w, start) and w not in potential_ans:Įnglish_words = set() # ĭef transform(english_words, start, end, count): # transform(english_words, 'damp', 'like') # def transform(english_words, start, end): The new word you get in each step must be in the # Given two words of equal length that are in a dictionary, write a method to transform one word into another word by changing only You could simply use recursive back-tracking but this is far from the most optimal solution.

Another word for sort through