feat: refactored docs structure

docs/(03-hierarchical)/heap-priority-queue/(problems)/215_kth-largest-element-in-an-array.mdx

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+---
+title: '215. Kth Largest Element in an Array'
+description: Given an integer array nums and an integer k, return the k^th largest element in the array
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Quickselect", slug: "quickselect" }
+video: "XEmy13g1Qxc"
+issue: "https://github.com/prdlk/leetcode/issues/118"
+sidebar:
+  label: 'Kth Largest Element in an Array'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `nums = [3,2,1,5,6,4], k = 2`
+- Output: `5`
+
+### Example 2:
+- Input: `nums = [3,2,3,1,2,4,5,5,6], k = 4`
+- Output: `4`
+
+### Constraints:
+
+- `1 <= k <= nums.length <= 10^5`
+- `-10^4 <= nums[i] <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["findKthLargest(nums, k)"]) --> F{"more i in range(len(nums))?"}
+  F -- yes --> N["nums[i] = -nums[i] — negate so the min-heap acts as a max-heap"]
+  N --> F
+  F -- no --> H["heapq.heapify(nums)"]
+  H --> W{"more _ in range(k - 1)?"}
+  W -- yes --> P["heapq.heappop(nums) — discard the k-1 largest"]
+  P --> W
+  W -- no --> E(["return -heapq.heappop(nums)"])
+```
+
+## Solution
+
+```py
+import heapq
+
+
+class Solution:
+    def findKthLargest(self, nums: List[int], k: int) -> int:
+        for i in range(len(nums)):
+            nums[i] = -nums[i]
+
+        heapq.heapify(nums)
+
+        for _ in range(k - 1):
+            heapq.heappop(nums)
+
+        return -heapq.heappop(nums)
+```

docs/(03-hierarchical)/heap-priority-queue/(problems)/23_merge-k-sorted-lists.mdx

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+---
+title: '23. Merge k Sorted Lists'
+description: You are given an array of k linked-lists lists, each linked-list is sorted in ascending order
+icon: dot
+topics:
+  - { name: "Linked List", slug: "linked-list" }
+  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Merge Sort", slug: "merge-sort" }
+  - { name: "Tournament Sort", slug: "tournament-sort" }
+video: "q5a5OiGbT6Q"
+issue: "https://github.com/prdlk/leetcode/issues/120"
+sidebar:
+  label: 'Merge k Sorted Lists'
+  badge: 'Hard'
+---
+
+### Example 1:
+- Input: `lists = [[1,4,5],[1,3,4],[2,6]]`
+- Output: `[1,1,2,3,4,4,5,6]`
+- Explanation: The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted linked list: 1->1->2->3->4->4->5->6
+
+### Example 2:
+- Input: `lists = []`
+- Output: `[]`
+
+### Example 3:
+- Input: `lists = [[]]`
+- Output: `[]`
+
+### Constraints:
+
+- `k == lists.length`
+- `0 <= k <= 10^4`
+- `0 <= lists[i].length <= 500`
+- `-10^4 <= lists[i][j] <= 10^4`
+- `lists[i]` is sorted in ascending order.
+- The sum of lists[i].length will not exceed 10^4.
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["mergeKLists(lists)"]) --> I["heap = []"]
+  I --> F{"more i, node in enumerate(lists)?"}
+  F -- yes --> C{"node?"}
+  C -- yes --> P["heapq.heappush(heap, (node.val, i, node)) — i breaks ties so nodes are never compared"]
+  P --> F
+  C -- no --> F
+  F -- no --> D["D = ListNode(), cur = D"]
+  D --> W{"heap?"}
+  W -- no --> E(["return D.next"])
+  W -- yes --> O["val, i, node = heapq.heappop(heap) — the smallest head across all lists"]
+  O --> L["cur.next = node, cur = node, node = node.next"]
+  L --> N{"node?"}
+  N -- yes --> Q["heapq.heappush(heap, (node.val, i, node))"]
+  Q --> W
+  N -- no --> W
+```
+
+## Solution
+
+```py
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+import heapq
+
+
+class Solution:
+    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
+        heap = []
+
+        # K log K
+        for i, node in enumerate(lists):
+            if node:
+                heapq.heappush(heap, (node.val, i, node))
+
+        D = ListNode()
+        cur = D
+
+        # n log k
+        while heap:
+            val, i, node = heapq.heappop(heap)
+            cur.next = node
+            cur = node
+            node = node.next
+
+            if node:
+                heapq.heappush(heap, (node.val, i, node))
+
+        # Time: O(N log k)
+        # Space: O(n)
+        return D.next
+```

docs/(03-hierarchical)/heap-priority-queue/(problems)/973_k-closest-points-to-origin.mdx

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+---
+title: '973. K Closest Points to Origin'
+description: Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0)
+icon: dot
+topics:
+  - { name: "Array", slug: "array" }
+  - { name: "Math", slug: "math" }
+  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
+  - { name: "Geometry", slug: "geometry" }
+  - { name: "Sorting", slug: "sorting" }
+  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
+  - { name: "Quickselect", slug: "quickselect" }
+  - { name: "K-D Tree", slug: "k-d-tree" }
+video: "rI2EBUEMfTk"
+issue: "https://github.com/prdlk/leetcode/issues/119"
+sidebar:
+  label: 'K Closest Points to Origin'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `points = [[1,3],[-2,2]], k = 1`
+- Output: `[[-2,2]]`
+- Explanation: The distance between `(1`, `3`) and the origin is sqrt(10). The distance between `(-2`, `2`) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), `(-2`, `2`) is closer to the origin. We only want the closest `k = 1 points` from the origin, so the answer is just [[-2,2]].
+
+### Example 2:
+- Input: `points = [[3,3],[5,-1],[-2,4]], k = 2`
+- Output: `[[3,3],[-2,4]]`
+- Explanation: The answer [[-2,4],[3,3]] would also be accepted.
+
+### Constraints:
+
+- `1 <= k <= points.length <= 10^4`
+- `-10^4 <= xi, yi <= 10^4`
+
+## Approach
+
+```mermaid
+flowchart TD
+  S(["kClosest(points, k)"]) --> I["minHeap = []"]
+  I --> F{"more x, y in points?"}
+  F -- yes --> D["dist = x**2 + y**2 — squared distance orders the same as sqrt"]
+  D --> A["minHeap.append([dist, x, y])"]
+  A --> F
+  F -- no --> H["heapq.heapify(minHeap), res = [] — O(n) build, smallest dist on top"]
+  H --> W{"k > 0?"}
+  W -- yes --> P["dist, x, y = heapq.heappop(minHeap)"]
+  P --> R["res.append([x, y]), k -= 1"]
+  R --> W
+  W -- no --> E(["return res"])
+```
+
+## Solution
+
+```py
+import heapq
+
+
+class Solution:
+    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
+        minHeap = []
+        for x, y in points:
+            dist = (x**2) + (y**2)
+            minHeap.append([dist, x, y])
+
+        heapq.heapify(minHeap)
+        res = []
+
+        while k > 0:
+            dist, x, y = heapq.heappop(minHeap)
+            res.append([x, y])
+            k -= 1
+
+        return res
+```

docs/(03-hierarchical)/heap-priority-queue/index.mdx

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+---
+title: Heap / Priority Queue
+description: 'The largest or smallest item is always at the top, and it is removed in logarithmic time.'
+hidden: true
+---