Merging K Sorted Lists Using Heapq Module In Python3

Problem:- merge k sorted lists. I want to solve this problem using min heap which can be implemented with heapq module in python. Below is the sample code of the function... heapq.

Solution 1:

When tuples are compared, their first elements are compared, and then any ties are broken using their second elements, their elements, and so on. For example, (2, "a") < (2, "b") will evaluate to True.

Here, you are inserting (node.val, node) tuples into your heap, which attempts to compare them. If there is a tie on the node value, it moves on to the second element of the tuples - the nodes themselves. These are just ListNode instances. Python does not know how to compare two ListNodes, hence the error.

To enable comparisons between ListNodes, you need to implement the rich comparison methods. A quick way to do it is to simply implement ListNode.__lt__ and then use the functools.total_ordering decorator:

import heapq
from functools import total_ordering


@total_orderingclassListNode:def__init__(self, value: float, label: str) -> None:self.value = value
        self.label = label

    def__lt__(self, other:'ListNode'):
        returnself.value <= other.value

    def__str__(self):
        return f"ListNode(label={self.label}, value={self.value})"



nodes = []
a =  ListNode(5, "A")
b = ListNode(3, "B")
c = ListNode(5, "C")
heapq.heappush(nodes, a)
heapq.heappush(nodes, b)
heapq.heappush(nodes, c)

whilenodes:
    x = heapq.heappop(nodes)
    print(x)

Here we say that comparing two ListNode objects is the same as just comparing their values. With comparisons defined, you don't even need to insert tuples at all. You can just insert ListNode objects directly, and rely on the comparison methods.

Solution 2:

I think you are talking about this: Leetcode Merge k Sorted Lists

I am sharing a working solution for you:

head = curr = ListNode(0)    # Creating a dummy node
    lst = []
    for l in lists:
        if l:
 # Here we need to first push val so that heap can know which is minimum and which is maximum. # If we insert directly node then heap can't work proper way (in this case heapq.heappop doesn't return minimum node).    
            lst.append((l.val,l))    

    
    heapq.heapify(lst)
    while lst:
        val,node = heapq.heappop(lst)
        curr.next = node
        curr = curr.next
        node = node.nextif node:
            heapq.heappush(lst,(node.val,node))
    return head.next