[20180411] medium problems

Author: iamminji

README.md

@@ -134,6 +134,7 @@
 | [0841](https://leetcode.com/problems/keys-and-rooms/)                 | Keys and Rooms                 | medium       | [python](🤔medium/0841.Keys-and-Rooms/keys_and_rooms.py)         |   dfs, graph        |
 | [0872](https://leetcode.com/problems/leaf-similar-trees/)                        | Leaf-Similar Trees                        | easy       | python         |           |
 | [0876](https://leetcode.com/problems/middle-of-the-linked-list/)                        | Middle of the Linked List                        | easy       | [python](🙂easy/0876.Middle-of-the-Linked-List/middle_of_the_linked_list.py)         |           |
+| [0890](https://leetcode.com/problems/find-and-replace-pattern/)                        | Find and Replace Pattern                        | medium       | [python](🤔medium/0890.Find-and-Replace-Pattern/find_and_replace_pattern.py)         |           |
 | [0894](https://leetcode.com/problems/all-possible-full-binary-trees/)                        | All Possible Full Binary Trees                        | medium       | [python](🤔medium/0894.All-Possible-Full-Binary-Trees/all_possible_full_binary_trees.py)         |           |
 | [0896](https://leetcode.com/problems/monotonic-array/)                        | Monotonic Array                        | easy       | [python](🙂easy/0896.Monotonic-Array/monotonic_array.py)         |           |
 | [0905](https://leetcode.com/problems/sort-array-by-parity/)                                  | Sort Array By Parity                      | easy       | golang         |           |

🤔medium/0890.Find-and-Replace-Pattern/find_and_replace_pattern.py

@@ -0,0 +1,29 @@
+# 890. Find and Replace Pattern
+# https://leetcode.com/problems/find-and-replace-pattern/
+
+from typing import List
+from collections import defaultdict
+
+
+class Solution:
+    # pattern 에 맞는 word 를 찾아 리스트로 리턴하는 문제다.
+    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
+
+        # 딕셔너리 2개를 이용해서 1:1 매칭 되게 만들어주면 된다.
+        res = []
+        for word in words:
+            d1 = defaultdict(str)
+            d2 = defaultdict(str)
+            for w, p in zip(word, pattern):
+                if w not in d1:
+                    d1[w] = p
+                if p not in d2:
+                    d2[p] = w
+
+                # 1:1 매칭이 안되면 종료
+                if d1[w] != p or d2[p] != w:
+                    break
+            else:
+                res.append(word)
+
+        return res