docs(docs): add MDX page for Merge k Sorted Lists problem

apps/docs/content/(array)/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
+sidebar:
+  label: 'Kth Largest Element in an Array'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Heap / Priority Queue</Badge>
+
+### 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`
+
+## 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)
+```

apps/docs/content/(array)/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)
+sidebar:
+  label: 'K Closest Points to Origin'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Heap / Priority Queue</Badge>
+
+### 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`
+
+## 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
+```

apps/docs/content/(linked-list)/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
+sidebar:
+  label: 'Merge k Sorted Lists'
+  badge: 'Hard'
+---
+
+<Badge variant="accent">Heap / Priority Queue</Badge>
+
+### 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.
+
+## 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
+```

work/1/Hard/Linked List/23.merge-k-sorted-lists.py

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+"""
+23. Merge k Sorted Lists
+Difficulty: Hard
+https://leetcode.com/problems/merge-k-sorted-lists/
+
+──────────────────────────────────────────────────
+
+You are given an array of k linked-lists lists, each linked-list is
+sorted in ascending order.
+
+Merge all the linked-lists into one sorted linked-list and return it.
+
+
+
+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.
+"""
+
+# 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

work/1/Medium/Array/215.kth-largest-element-in-an-array.py

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+"""
+215. Kth Largest Element in an Array
+Difficulty: Medium
+https://leetcode.com/problems/kth-largest-element-in-an-array/
+
+──────────────────────────────────────────────────
+
+Given an integer array nums and an integer k, return the k^th largest
+element in the array.
+
+Note that it is the k^th largest element in the sorted order, not the
+k^th distinct element.
+
+Can you solve it without sorting?
+
+
+
+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
+"""
+
+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)

work/1/Medium/Array/973.k-closest-points-to-origin.py

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+"""
+973. K Closest Points to Origin
+Difficulty: Medium
+https://leetcode.com/problems/k-closest-points-to-origin/
+
+──────────────────────────────────────────────────
+
+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).
+
+The distance between two points on the X-Y plane is the Euclidean
+distance (i.e., √(x1 - x2)^2 + (y1 - y2)^2).
+
+You may return the answer in any order. The answer is guaranteed to
+be unique (except for the order that it is in).
+
+
+
+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
+"""
+
+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