Description

You have a list of words and a pattern, and you want to know which words in words matches the pattern.

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 desired word.

(Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.)

Return a list of the words in words that match the given pattern.

You may return the answer in any order.

Example 1:

Input: words = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb"
Output: ["mee","aqq"]
Explanation: "mee" matches the pattern because there is a permutation {a -> m, b -> e, ...}. 
"ccc" does not match the pattern because {a -> c, b -> c, ...} is not a permutation,
since a and b map to the same letter.

Note:

  • 1 <= words.length <= 50
  • 1 <= pattern.length = words[i].length <= 20

Solutions

  找到所有与 pattern 形式相同的字符串保存下来。

1. Greedy + Dict

  理解题意就可以了,不是很难,但是写出来的代码就不够优雅,一堆 if。用字典存映射关系,然后挨个字符去找对应!

  找对应的时候注意,如果字符串的字符不在字典中,还要判断是不是在已经存进字典的 values 中。

# Time: O(n^2)
# Space: O(n)
class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        res = []
        n = len(pattern)
        for item in words:
            dic = {}
            Flag = True
            for i in range(n):
                if pattern[i] in dic:
                    if item[i] != dic[pattern[i]]:
                        Flag = False
                        break
                else:
                    if item[i] in dic.values():
                        Flag = False
                        break
                    dic[pattern[i]] = item[i]
            if Flag:
                res.append(item)
        return res
# 46/46 cases passed (32 ms)
# Your runtime beats 79 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.7 MB)

References

  1. 890. Find and Replace Pattern