feat: refactored docs structure

apps/docs/content/(01-linear-structures)/hash-table/(problems)/205_isomorphic-strings.mdx

@@ -1,52 +0,0 @@
----
-title: '205. Isomorphic Strings'
-description: Given two strings s and t, determine if they are isomorphic
-icon: dot
-topics:
-  - { name: "Hash Table", slug: "hash-table" }
-  - { name: "String", slug: "string" }
-issue: "https://github.com/prdlk/leetcode/issues/33"
-sidebar:
-  label: 'Isomorphic Strings'
-  badge: 'Easy'
----
-
-### 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.
-
-## Solution
-
-```py
-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))
-```