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

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']