Find and Replace Pattern in Python

Suppose we have a list of words and a pattern, and we have to find which words in words matches the pattern. Here a word matches the pattern if there exists a permutation of letters p so that after replacing every letter x in the pattern with p(x), we get the target word. We have to find a list of the words in words that match the given pattern.

So for example, if the input is like ["abc","deq","mee","aqq","dkd","ccc"] and pattern is “abb”, then the output will be [“mee”, “aqq”], here mee and aqq are matching the style of the pattern “abb”. but “ccc” is not a pattern, because this is not a permutation.

To solve this, we will follow these steps −

• define one convert() method. This will take word as input, this will act like −
• counter := 1, s := empty string
• s := s + string equivalent of counter
• for i in range 1 to length of word – 1
  • j := i – 1
  • while j>=0
    • if word[j] is word[i], then break
    • decrease j by 1
  • if j > -1, then s := s + s[j], otherwise increase counter by 1 and s := s + counter value as string
• return s
• the actual method will be like
• make one array word_num, and this is empty, make another empty array res
• for each element i in words −
  • insert convert(i) into word_num
• pattern := convert(pattern)
• for i in range 0 to length of words – 1
  • if words_num[i] = pattern, then insert words[i] into res
• return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution(object):    def findAndReplacePattern(self, words, pattern):       words_num = []       result = []       for i in words:          words_num.append(self.convert(i))       pattern = self.convert(pattern)       for i in range(len(words)):          if words_num[i] == pattern:             result.append(words[i])       return result    def convert(self,word):       counter = 1       s = ""       s+=str(counter)       for i in range(1,len(word)):          j= i -1          while j>=0:             if word[j] == word[i]:                break             j-=1          if j >-1:             s+=s[j]          else:             counter+=1             s+=str(counter)       return s ob = Solution() print(ob.findAndReplacePattern(["abc","deq","mee","aqq","dkd","ccc"],"abb"))

Input

["abc","deq","mee","aqq","dkd","ccc"] "abb"

Output

['mee', 'aqq']