Syntax error Program to find number of steps required to change one word to another in Python

Program to find number of steps required to change one word to another in Python

Suppose we have a list of words called dictionary and we have another two strings start and end. We want to reach from start to end by changing one character at a time and each resulting word should also be in the dictionary. Words are case-sensitive. So we have to find the minimum number of steps it would take to reach at the end. If it is not possible then return -1.

So, if the input is like dictionary = ["may", "ray", "rat"] start = "rat" end = "may", then the output will be 3, as we can select this path: ["rat", "ray", "may"].

To solve this, we will follow these steps −

dictionary := a new set with all unique elements present in
q = a double ended queue with a pair (start, 1)
while q is not empty, do
   (word, distance) := left element of q, and delete the left element
   if word is same as end, then
      return distance
      for i in range 0 to size of word - 1, do
         for each character c in "abcdefghijklmnopqrstuvwxyz", do
            next_word := word[from index 0 to i - 1] concatenate c concatenate word[from index (i + 1) to end]
            if next_word is in dictionary, then
               delete next_word from dictionary
               insert (next_word, distance + 1) at the end of q
return -1

Example (Python)

Let us see the following implementation to get better understanding −

 Live Demo

from collections import deque
class Solution:
   def solve(self, dictionary, start, end):
      dictionary = set(dictionary)
      q = deque([(start, 1)])
      while q:
         word, distance = q.popleft()
         if word == end:
            return distance
         for i in range(len(word)):
            for c in "abcdefghijklmnopqrstuvwxyz":
               next_word = word[:i] + c + word[i + 1 :]
               if next_word in dictionary:
                  dictionary.remove(next_word)
                  q.append((next_word, distance + 1))
      return -1
ob = Solution()
dictionary = ["may", "ray", "rat"]
start = "rat"
end = "may"
print(ob.solve(dictionary, start, end))

Input

["may", "ray", "rat"], "rat", "may"

Output

3
Updated on: 2020-12-12T09:08:44+05:30

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