feat: refactored docs structure

docs/(01-linear-structures)/hash-table/(problems)/128_longest-consecutive-sequence.mdx

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+---
+title: '128. Longest Consecutive Sequence'
+description: Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Union-Find", slug: "union-find" }
+issue: "https://github.com/prdlk/leetcode/issues/34"
+sidebar:
+  label: 'Longest Consecutive Sequence'
+  badge: 'Medium'
+---
+
+::::warning
+You must write an algorithm that runs in O(n) time.
+::::
+
+### Example 1:
+- Input: `nums = [100,4,200,1,3,2]`
+- Output: `4`
+- Explanation: The longest consecutive elements sequence is [1, `2`, `3`, 4]. Therefore its length is `4`.
+
+### Example 2:
+- Input: `nums = [0,3,7,2,5,8,4,6,0,1]`
+- Output: `9`
+
+### Example 3:
+- Input: `nums = [1,0,1,2]`
+- Output: `3`
+
+### Constraints:
+
+- `0 <= nums.length <= 10^5`
+- `-10^9 <= nums[i] <= 10^9`
+
+## Solution
+
+```py
+class Solution:
+    def longestConsecutive(self, nums: list[int]) -> int:
+        seen = set(nums)
+        best = 0
+
+        for n in seen:
+            if n - 1 in seen:
+                continue
+
+            length = 1
+            while length + n in seen:
+                length += 1
+
+            best = max(length, best)
+
+        return best
+```

docs/(01-linear-structures)/hash-table/(problems)/1832_check-if-the-sentence-is-pangram.mdx

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+---
+title: '1832. Check if the Sentence Is Pangram'
+description: A pangram is a sentence where every letter of the English alphabet appears at least once
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+sidebar:
+  label: 'Check if the Sentence Is Pangram'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `sentence = "thequickbrownfoxjumpsoverthelazydog"`
+- Output: `true`
+- Explanation: `sentence` contains at least one of every letter of the English alphabet.
+
+### Example 2:
+- Input: `sentence = "leetcode"`
+- Output: `false`
+
+### Constraints:
+
+- `1 <= sentence.length <= 1000`
+- `sentence` consists of lowercase English letters.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["checkIfPangram(sentence)"]) --> I["freq = {}"]
+  I --> L{"more char c?"}
+  L -- yes --> C["freq[c] = (freq[c] or 0) + 1"]
+  C --> L
+  L -- no --> Q{"distinct keys in freq == 26?"}
+  Q -- yes --> R(["return true"])
+  Q -- no --> F(["return false"])
+```
+
+## Solution
+
+```js
+/**
+ * @param {string} sentence
+ * @return {boolean}
+ */
+var checkIfPangram = function(sentence) {
+    const freq = {};
+    for (let c of sentence) freq[c] = (freq[c] || 0) + 1;
+    
+    if (Object.keys(freq).length === 26){
+        return true;
+    }
+    return false;
+};
+```

docs/(01-linear-structures)/hash-table/(problems)/1_two-sum.mdx

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+---
+title: '1. Two Sum'
+description: You are given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+video: "KLlXCFG5TnA"
+issue: "https://github.com/prdlk/leetcode/issues/29"
+sidebar:
+  label: 'Two Sum'
+  badge: 'Easy'
+---
+
+::::warning
+Can you come up with an algorithm that is less than O(n^2) time complexity?
+::::
+
+### Example 1:
+- Input: `nums = [2,7,11,15], target = 9`
+- Output: `[0,1]`
+- Explanation: Because `nums[0] + nums[1]` == `9`, we return [0, 1].
+
+### Example 2:
+- Input: `nums = [3,2,4], target = 6`
+- Output: `[1,2]`
+
+### Example 3:
+- Input: `nums = [3,3], target = 6`
+- Output: `[0,1]`
+
+### Constraints:
+
+- `2 <= nums.length <= 10^4`
+- `-10^9 <= nums[i] <= 10^9`
+- `-10^9 <= target <= 10^9`
+- Only one valid answer exists.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["twoSum(nums, target)"]) --> I["seen = {} — value to index"]
+  I --> L{"more (i, num) in nums?"}
+  L -- no --> E(["return []"])
+  L -- yes --> C["complement = target - num"]
+  C --> H{"complement in seen?"}
+  H -- yes --> R(["return [seen[complement], i]"])
+  H -- no --> W["seen[num] = i"]
+  W --> L
+```
+
+## Solution
+
+```py
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        seen = {}
+        for i, num in enumerate(nums):
+            complement = target - num
+            if complement in seen:
+                return [seen[complement], i]
+            seen[num] = i
+        return []
+```

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

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+---
+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))
+```

docs/(01-linear-structures)/hash-table/(problems)/217_contains-duplicate.mdx

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+---
+title: '217. Contains Duplicate'
+description: Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Sorting", slug: "sorting" }
+video: "3OamzN90kPg"
+issue: "https://github.com/prdlk/leetcode/issues/27"
+sidebar:
+  label: 'Contains Duplicate'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `nums = [1,2,3,1]`
+- Output: `true`
+- Explanation: The element `1` occurs at the indices `0` and `3`.
+
+### Example 2:
+- Input: `nums = [1,2,3,4]`
+- Output: `false`
+- Explanation: All elements are distinct.
+
+### Example 3:
+- Input: `nums = [1,1,1,3,3,4,3,2,4,2]`
+- Output: `true`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^5`
+- `-10^9 <= nums[i] <= 10^9`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["containsDuplicate(nums)"]) --> I["freq = {}"]
+  I --> L{"more n in nums?"}
+  L -- no --> E(["return false"])
+  L -- yes --> C["freq[n] = (freq[n] or 0) + 1"]
+  C --> Q{"freq[n] >= 2?"}
+  Q -- yes --> R(["return true"])
+  Q -- no --> L
+```
+
+## Solution
+
+```py
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(set(nums)) != len(nums)
+```

docs/(01-linear-structures)/hash-table/(problems)/242_valid-anagram.mdx

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+---
+title: '242. Valid Anagram'
+description: Given two strings s and t, return true if t is an anagram of s, and false otherwise
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sorting", slug: "sorting" }
+video: "9UtInBqnCgA"
+issue: "https://github.com/prdlk/leetcode/issues/28"
+sidebar:
+  label: 'Valid Anagram'
+  badge: 'Easy'
+---
+
+::::warning
+What if the inputs contain Unicode characters? How would you adapt your solution to such a case?
+::::
+
+### Example 1:
+- Input: `s = "anagram", t = "nagaram"`
+- Output: `true`
+
+### Example 2:
+- Input: `s = "rat", t = "car"`
+- Output: `false`
+
+### Constraints:
+
+- `1 <= s.length, t.length <= 5 * 10^4`
+- `s` and `t` consist of lowercase English letters.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["isAnagram(s, t)"]) --> I["freq: count every char of s"]
+  I --> L{"more char c in t?"}
+  L -- no --> E(["return true"])
+  L -- yes --> Q{"freq[c] missing or 0?"}
+  Q -- yes --> R(["return false — t has a char s cannot cover"])
+  Q -- no --> D["freq[c] -= 1"]
+  D --> L
+```
+
+## Solution
+
+```js
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isAnagram = function(s, t) {
+  let freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+  
+  for (let c of t) {
+    if(!freq[c] || freq[c] === 0){
+      return false;
+    }
+    freq[c] = freq[c] - 1;
+  }
+  return true;
+};
+```

docs/(01-linear-structures)/hash-table/(problems)/290_word-pattern.mdx

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+---
+title: '290. Word Pattern'
+description: Given a pattern and a string s, find if s follows the same pattern
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+issue: "https://github.com/prdlk/leetcode/issues/35"
+sidebar:
+  label: 'Word Pattern'
+  badge: 'Easy'
+---
+
+### 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.
+
+## Solution
+
+```py
+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))
+```

docs/(01-linear-structures)/hash-table/(problems)/347_top-k-frequent-elements.mdx

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+---
+title: '347. Top K Frequent Elements'
+description: Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Bucket Sort", slug: "bucket-sort" }
+  - { name: "Counting", slug: "counting" }
+  - { name: "Quickselect", slug: "quickselect" }
+video: "YPTqKIgVk-k"
+issue: "https://github.com/prdlk/leetcode/issues/31"
+sidebar:
+  label: 'Top K Frequent Elements'
+  badge: 'Medium'
+---
+
+::::warning
+Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
+::::
+
+### Example 1:
+- Input: `nums = [1,1,1,2,2,3], k = 2`
+- Output: `[1,2]`
+
+### Example 2:
+- Input: `nums = [1], k = 1`
+- Output: `[1]`
+
+### Example 3:
+- Input: `nums = [1,2,1,2,1,2,3,1,3,2], k = 2`
+- Output: `[1,2]`
+
+### Constraints:
+
+- `1 <= nums.length <= 10^5`
+- `-10^4 <= nums[i] <= 10^4`
+- `k` is in the range `[1, the number of unique elements in the array]`.
+- It is guaranteed that the answer is unique.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["topKFrequent(nums, k)"]) --> F["freq: count every value in nums"]
+  F --> K["keys of freq, cast to Number"]
+  K --> O["sort by freq[b] - freq[a] — most frequent first"]
+  O --> R(["return slice(0, k)"])
+```
+
+## Solution
+
+```js
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function (nums, k) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  return Object.keys(freq)
+    .map(Number)
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+};
+```

docs/(01-linear-structures)/hash-table/(problems)/387_first-unique-character-in-a-string.mdx

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+---
+title: '387. First Unique Character in a String'
+description: Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Queue", slug: "queue" }
+  - { name: "Counting", slug: "counting" }
+video: "rBENYgWy3xU"
+sidebar:
+  label: 'First Unique Character in a String'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `s = "leetcode"`
+- Output: `0`
+- Explanation: The character `'l'` at index `0` is the first character that does not occur at any other index.
+
+### Example 2:
+- Input: `s = "loveleetcode"`
+- Output: `2`
+
+### Example 3:
+- Input: `s = "aabb"`
+- Output: `-1`
+
+### Constraints:
+
+- `1 <= s.length <= 10^5`
+- `s` consists of only lowercase English letters.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["firstUniqChar(s)"]) --> F["Pass 1 — freq: count every char of s"]
+  F --> L{"Pass 2 — more i in 0..len-1?"}
+  L -- no --> E(["return -1"])
+  L -- yes --> Q{"freq[s[i]] == 1?"}
+  Q -- yes --> R(["return i — leftmost wins because the scan is in order"])
+  Q -- no --> L
+```
+
+## Solution
+
+```js
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var firstUniqChar = function (s) {
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  for (let i = 0; i < s.length; i++) {
+    if (freq[s[i]] === 1) {
+      return i;
+    }
+  }
+  return -1;
+};
+```

docs/(01-linear-structures)/hash-table/(problems)/451_sort-characters-by-frequency.mdx

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+---
+title: '451. Sort Characters By Frequency'
+description: Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string
+icon: dot
+topics:
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Bucket Sort", slug: "bucket-sort" }
+  - { name: "Counting", slug: "counting" }
+video: "OXdXc9HTrIg"
+sidebar:
+  label: 'Sort Characters By Frequency'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `s = "tree"`
+- Output: `"eert"`
+- Explanation: `'e'` appears twice while `'r'` and `'t'` both appear once. So `'e'` must appear before both `'r'` and `'t'`. Therefore `"eetr"` is also a valid answer.
+
+### Example 2:
+- Input: `s = "cccaaa"`
+- Output: `"aaaccc"`
+- Explanation: Both `'c'` and `'a'` appear three times, so both `"cccaaa"` and `"aaaccc"` are valid answers. Note that `"cacaca"` is incorrect, as the same characters must be together.
+
+### Example 3:
+- Input: `s = "Aabb"`
+- Output: `"bbAa"`
+- Explanation: `"bbaA"` is also a valid answer, but `"Aabb"` is incorrect. Note that `'A'` and `'a'` are treated as two different characters.
+
+### Constraints:
+
+- `1 <= s.length <= 5 * 10^5`
+- `s` consists of uppercase and lowercase English letters and digits.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["frequencySort(s)"]) --> F["freq: count every char of s"]
+  F --> C["split s and sort with a two-key comparator"]
+  C --> K1["primary: freq[b] - freq[a] — most frequent first"]
+  C --> K2["tie-break: a.localeCompare(b) — alphabetical"]
+  K1 --> R(["join and return"])
+  K2 --> R
+```
+
+## Solution
+
+```js
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var frequencySort = function (s) {
+  // count the frequency of each character
+  const freq = {};
+  for (let c of s) freq[c] = (freq[c] || 0) + 1;
+
+  // sort the characters by frequency
+  return s
+    .split("")
+    .sort((a, b) => freq[b] - freq[a] || a.localeCompare(b))
+    .join("");
+};
+```

docs/(01-linear-structures)/hash-table/(problems)/49_group-anagrams.mdx

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+---
+title: '49. Group Anagrams'
+description: Given an array of strings strs, group the anagrams together. You can return the answer in any order
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Hash Table", slug: "hash-table" }
+  - { name: "String", slug: "string" }
+  - { name: "Sorting", slug: "sorting" }
+video: "vzdNOK2oB2E"
+issue: "https://github.com/prdlk/leetcode/issues/30"
+sidebar:
+  label: 'Group Anagrams'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `strs = ["eat","tea","tan","ate","nat","bat"]`
+- Output: `[["bat"],["nat","tan"],["ate","eat","tea"]]`
+- Explanation: There is no string in `strs` that can be rearranged to form `"bat"`. The strings `"nat"` and `"tan"` are anagrams as they can be rearranged to form each other. The strings `"ate"`, `"eat"`, and `"tea"` are anagrams as they can be rearranged to form each other.
+
+### Example 2:
+- Input: `strs = [""]`
+- Output: `[[""]]`
+
+### Example 3:
+- Input: `strs = ["a"]`
+- Output: `[["a"]]`
+
+### Constraints:
+
+- `1 <= strs.length <= 10^4`
+- `0 <= strs[i].length <= 100`
+- `strs[i]` consists of lowercase English letters.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["groupAnagrams(strs)"]) --> K["sorted[i] = letters of strs[i], sorted — the anagram key"]
+  K --> I["anagrams = {}"]
+  I --> L{"more i in 0..n-1?"}
+  L -- no --> E(["return Object.values(anagrams)"])
+  L -- yes --> Q{"anagrams has sorted[i]?"}
+  Q -- no --> N["anagrams[sorted[i]] = [strs[i]]"]
+  Q -- yes --> P["anagrams[sorted[i]].push(strs[i])"]
+  N --> L
+  P --> L
+```
+
+## Solution
+
+```js
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+var groupAnagrams = function(strs) {
+  let sorted = strs.map(str => str.split("").sort().join(""));
+  let anagrams = {};
+
+  for (let i = 0; i < strs.length; i++){
+    if(!anagrams[sorted[i]]){
+      anagrams[sorted[i]] = [strs[i]]
+    }else{
+      anagrams[sorted[i]].push(strs[i])
+    }
+  }
+
+  return Object.values(anagrams);
+};
+```