Add 290

Author: trung-hn

Readme.md

@@ -35,6 +35,7 @@
 | 264 | Ugly Number II | Medium | O(1) | O(1) | Heap, Dynamic Programming | | |
 | 270 | Closest Binary Search Tree Value | Easy | O(H) | O(1) | Binary Search, Tree | | 🔒 |
 | 274 | H-Index | Medium | O(NlogN) | O(N) | Hash Table, Sort | There is a better solution | |
+| 290 | Word Pattern | Easy | O(N) | O(N) | Hash Table | | |
 | 295 | Find Median from Data Stream | Hard | O(logN) | O(N) | Design, Hard, Heap | | |
 | 319 | Bulb Switcher | Medium | O(1) | O(1) | Math, Tricky | | |
 | 328 | Odd Even Linked List | Medium | O(N) | O(1) | Linked List | | |

inputs/290

@@ -0,0 +1,10 @@
+""abc"
+"b c a"
+"abc"
+"dog cat dog"
+"abba"
+"dog dog dog dog"
+"abba"
+"dog cat cat dog"
+""
+"beef""

src/290.word-pattern.py

@@ -0,0 +1,45 @@
+#
+# @lc app=leetcode id=290 lang=python3
+#
+# [290] Word Pattern
+#
+
+# @lc code=start
+# TAGS: Hash Table
+
+import itertools
+class Solution:
+    # 36 ms, 66 % pythonic code using set
+    def wordPattern1(self, pattern: str, str: str) -> bool:
+        t = str.split()
+        p = pattern
+        return len(set(zip(p,t))) == len(set(p)) == len(set(t)) and len(p) == len(t)
+    
+    # 32 ms, 60.84 %. Time and Space: O(N)
+    def wordPattern(self, pattern: str, str: str) -> bool:
+        S = str.split()
+        if len(S) != len(pattern): return False
+        match1 = {}
+        match2 = {}
+        for c, word in itertools.zip_longest(pattern, S):
+            if c in match1 and word in match2:
+                if match1[c] != word: return False        
+                if match2[word] != c: return False
+            elif c not in match1 and word not in match2:
+                match1[c] = word
+                match2[word] = c
+            else:
+                return False
+        return True
+
+    # 32 ms 89 %. Time and Space: O(N)
+    def wordPattern(self, pattern: str, str: str) -> bool:
+        S = str.split()
+        if len(S) != len(pattern) or len(set(S)) != len(set(pattern)): return False
+        D = {}
+        for i, c in enumerate(pattern):
+            if c in D and S[i] != D[c]: return False
+            D[c] = S[i]
+        return True
+            
+# @lc code=end