From 24312437ebc3be470015efbb18279e7e57717557 Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Mon, 14 Sep 2026 12:49:55 -0400
Subject: [PATCH] =?UTF-8?q?feat(work):=20add=20solution=20for=20LeetCode?=
 =?UTF-8?q?=2023=E2=80=AFMerge=E2=80=AFk=E2=80=AFSorted=E2=80=AFLists?=
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---
 .../Linked List/23.merge-k-sorted-lists.py    | 87 +++++++++++++++++++
 1 file changed, 87 insertions(+)
 create mode 100644 work/1/Hard/Linked List/23.merge-k-sorted-lists.py

diff --git a/work/1/Hard/Linked List/23.merge-k-sorted-lists.py b/work/1/Hard/Linked List/23.merge-k-sorted-lists.py
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+++ b/work/1/Hard/Linked List/23.merge-k-sorted-lists.py	
@@ -0,0 +1,87 @@
+"""
+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