23. Merge k Sorted Lists

Hard

---

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 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 won't 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]:
        min_heap = []
        count = 0 # This is not useful(only used for comparison) but create so that we don't get error in heap
        # Push First K elements each from one of the K list
        for index, node in enumerate(lists):
            if node:
                heapq.heappush(min_heap, (node.val, count, node))
                count+=1
        head = dummy = ListNode()
        # Pick min from heap and put the next element of that list to which this element belongs
        while min_heap:
            val, _ ,node = heapq.heappop(min_heap)
            if node:
                head.next = node
                head = node
                node = node.next
                if node:
                    heapq.heappush(min_heap, (node.val, count, node))
                    count+=1
        return dummy.next