feat(work): add Isomorphic Strings solution

work/1/Easy/Hash Table/205.isomorphic-strings.py

@@ -0,0 +1,76 @@
+"""
+205. Isomorphic Strings
+Difficulty: Easy
+https://leetcode.com/problems/isomorphic-strings/
+
+──────────────────────────────────────────────────
+
+Given two strings s and t, determine if they are isomorphic.
+
+Two strings s and t are isomorphic if the characters in s can be
+replaced to get t.
+
+All occurrences of a character must be replaced with another
+character while preserving the order of characters. No two characters
+may map to the same character, but a character may map to itself.
+
+
+
+Example 1:
+
+Input: s = "egg", t = "add"
+
+Output: true
+
+Explanation:
+
+The strings s and t can be made identical by:
+
+        • Mapping 'e' to 'a'.
+
+        • Mapping 'g' to 'd'.
+
+Example 2:
+
+Input: s = "f11", t = "b23"
+
+Output: false
+
+Explanation:
+
+The strings s and t can not be made identical as '1' needs to be
+mapped to both '2' and '3'.
+
+Example 3:
+
+Input: s = "paper", t = "title"
+
+Output: true
+
+
+
+Constraints:
+
+        • 1 <= s.length <= 5 * 10^4
+
+        • t.length == s.length
+
+        • s and t consist of any valid ascii character.
+"""
+
+
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        N = len(t)
+        seen = {}
+
+        for i in range(N):
+            c1, c2 = s[i], t[i]
+            if c1 in seen and seen[c1] != c2:
+                return False
+            elif c1 in seen and seen[c1] == c2:
+                continue
+            else:
+                seen[c1] = c2
+
+        return len(set(s)) == len(set(t))