From cbc858b667d63dbe9f4b5c47a93cf05e4b3ba046 Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Fri, 2 Oct 2026 12:42:01 -0400
Subject: [PATCH] feat(work): add 290.word-pattern solution

---
 work/1/Easy/Hash Table/290.word-pattern.py | 83 ++++++++++++++++++++++
 1 file changed, 83 insertions(+)
 create mode 100644 work/1/Easy/Hash Table/290.word-pattern.py

diff --git a/work/1/Easy/Hash Table/290.word-pattern.py b/work/1/Easy/Hash Table/290.word-pattern.py
new file mode 100644
index 0000000..4d12e33
--- /dev/null
+++ b/work/1/Easy/Hash Table/290.word-pattern.py	
@@ -0,0 +1,83 @@
+"""
+290. Word Pattern
+Difficulty: Easy
+https://leetcode.com/problems/word-pattern/
+
+──────────────────────────────────────────────────
+
+Given a pattern and a string s, find if s follows the same pattern.
+
+Here follow means a full match, such that there is a bijection
+between a letter in pattern and a non-empty word in s. Specifically:
+
+        • Each letter in pattern maps to exactly one unique word in s.
+
+        • Each unique word in s maps to exactly one letter in pattern.
+
+• No two letters map to the same word, and no two words map to the
+same letter.
+
+
+
+Example 1:
+
+Input: pattern = "abba", s = "dog cat cat dog"
+
+Output: true
+
+Explanation:
+
+The bijection can be established as:
+
+        • 'a' maps to "dog".
+
+        • 'b' maps to "cat".
+
+Example 2:
+
+Input: pattern = "abba", s = "dog cat cat fish"
+
+Output: false
+
+Example 3:
+
+Input: pattern = "aaaa", s = "dog cat cat dog"
+
+Output: false
+
+
+
+Constraints:
+
+        • 1 <= pattern.length <= 300
+
+        • pattern contains only lower-case English letters.
+
+        • 1 <= s.length <= 3000
+
+        • s contains only lowercase English letters and spaces ' '.
+
+        • s does not contain any leading or trailing spaces.
+
+        • All the words in s are separated by a single space.
+"""
+
+
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        if len(pattern) != len(s.split()):
+            return False
+
+        words = s.split()
+        map = {}
+
+        for i in range(len(pattern)):
+            c = pattern[i]
+            w = words[i]
+
+            if c in map and map[c] != w:
+                return False
+            else:
+                map[c] = w
+
+        return len(set(pattern)) == len(set(words))