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890. Find and Replace Pattern

Description

Given a list of strings words and a string pattern, return a list of words[i] that match pattern. You may return the answer in any order.

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.

 

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.

Example 2:

Input: words = ["a","b","c"], pattern = "a"
Output: ["a","b","c"]

 

Constraints:

• 1 <= pattern.length <= 20
• 1 <= words.length <= 50
• words[i].length == pattern.length
• pattern and words[i] are lowercase English letters.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public List<String> findAndReplacePattern(String[] words, String pattern) {
          List<String> ans = new ArrayList<>();
          for (String word : words) {
              if (match(word, pattern)) {
                  ans.add(word);
              }
          }
          return ans;
      }
  
      private boolean match(String s, String t) {
          int[] m1 = new int[128];
          int[] m2 = new int[128];
          for (int i = 0; i < s.length(); ++i) {
              char c1 = s.charAt(i);
              char c2 = t.charAt(i);
              if (m1[c1] != m2[c2]) {
                  return false;
              }
              m1[c1] = i + 1;
              m2[c2] = i + 1;
          }
          return true;
      }
  }
  

• class Solution {
  public:
      vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
          vector<string> ans;
          auto match = [](string& s, string& t) {
              int m1[128] = {0};
              int m2[128] = {0};
              for (int i = 0; i < s.size(); ++i) {
                  if (m1[s[i]] != m2[t[i]]) return 0;
                  m1[s[i]] = i + 1;
                  m2[t[i]] = i + 1;
              }
              return 1;
          };
          for (auto& word : words)
              if (match(word, pattern)) ans.emplace_back(word);
          return ans;
      }
  };
  

• class Solution:
      def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
          def match(s, t):
              m1, m2 = [0] * 128, [0] * 128
              for i, (a, b) in enumerate(zip(s, t), 1):
                  if m1[ord(a)] != m2[ord(b)]:
                      return False
                  m1[ord(a)] = m2[ord(b)] = i
              return True
  
          return [word for word in words if match(word, pattern)]
  
  

• func findAndReplacePattern(words []string, pattern string) []string {
  	match := func(s, t string) bool {
  		m1, m2 := make([]int, 128), make([]int, 128)
  		for i := 0; i < len(s); i++ {
  			if m1[s[i]] != m2[t[i]] {
  				return false
  			}
  			m1[s[i]] = i + 1
  			m2[t[i]] = i + 1
  		}
  		return true
  	}
  	var ans []string
  	for _, word := range words {
  		if match(word, pattern) {
  			ans = append(ans, word)
  		}
  	}
  	return ans
  }
  

• function findAndReplacePattern(words: string[], pattern: string): string[] {
      return words.filter(word => {
          const map1 = new Map<string, number>();
          const map2 = new Map<string, number>();
          for (let i = 0; i < word.length; i++) {
              if (map1.get(word[i]) !== map2.get(pattern[i])) {
                  return false;
              }
              map1.set(word[i], i);
              map2.set(pattern[i], i);
          }
          return true;
      });
  }
  
  

• use std::collections::HashMap;
  impl Solution {
      pub fn find_and_replace_pattern(words: Vec<String>, pattern: String) -> Vec<String> {
          let pattern = pattern.as_bytes();
          let n = pattern.len();
          words
              .into_iter()
              .filter(|word| {
                  let word = word.as_bytes();
                  let mut map1 = HashMap::new();
                  let mut map2 = HashMap::new();
                  for i in 0..n {
                      if map1.get(&word[i]).unwrap_or(&n) != map2.get(&pattern[i]).unwrap_or(&n) {
                          return false;
                      }
                      map1.insert(word[i], i);
                      map2.insert(pattern[i], i);
                  }
                  true
              })
              .collect()
      }
  }
  
  

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