Merge k Sorted Lists

Here is a Python solution that merges k sorted linked-lists into one sorted linked-list:

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from queue import PriorityQueue

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        # Create a priority queue to hold the values of linked-lists
        q = PriorityQueue()

        # Iterate through each linked-list and add all the nodes into the priority queue
        for node in lists:
            current = node
            while current:
                q.put(current.val)
                current = current.next

        # Create a dummy node as a starting point
        dummy = ListNode(0)
        current = dummy

        # Retrieve the nodes from the priority queue in ascending order
        # and construct the merged linked-list
        while not q.empty():
            current.next = ListNode(q.get())
            current = current.next

        # Return the merged linked-list starting from the next node of the dummy
        return dummy.next

This solution leverages a priority queue to hold the values of all nodes, ensuring that they are sorted. By reconstructing the linked-list from the priority queue, we obtain a merged linked-list that is sorted in ascending order. This approach handles empty lists as well, returning an empty linked-list when no elements are present.

10 Prerequisite LeetCode Problems

“Merge k Sorted Lists” involves linked lists, sorting algorithms, and priority queues or heaps. Here are 10 problems to build these foundational skills:

  1. “Merge Two Sorted Lists” (LeetCode Problem #21): This problem introduces the basics of merging sorted lists, which is a key concept for “Merge k Sorted Lists”.

  2. “Linked List Cycle” (LeetCode Problem #141): This problem provides practice for basic linked list manipulation and understanding.

  3. “Reverse Linked List” (LeetCode Problem #206): This problem helps you practice manipulating linked list pointers, which is a useful skill for many linked list problems.

  4. “Sort List” (LeetCode Problem #148): This problem provides practice for sorting values in a linked list, which is a core component of the “Merge k Sorted Lists” problem.

  5. “Top K Frequent Elements” (LeetCode Problem #347): This problem introduces the use of a priority queue to solve problems, which is also used in “Merge k Sorted Lists”.

  6. “Kth Largest Element in an Array” (LeetCode Problem #215): This problem allows practice for heap usage in the context of finding Kth element, similar to merging K sorted lists.

  7. “Binary Heap Operations” (LeetCode Problem #1405): This problem introduces the basics of heap operations.

  8. “Swim in Rising Water” (LeetCode Problem #778): This problem provides a more complex usage of heaps/priority queues.

  9. “Flatten a Multilevel Doubly Linked List” (LeetCode Problem #430): This problem provides practice on complex linked list manipulations.

  10. “Insert into a Cyclic Sorted List” (LeetCode Problem #708): This problem gives practice on handling special cases in sorted list manipulations.

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def mergeKLists_heapq(self, lists):
	h = []
	head = tail = ListNode(0)
	for i in range(len(lists)):
		heapq.heappush(h, (lists[i].val, i, lists[i]))

	while h:
		node = heapq.heappop(h)
		node = node[2]
		tail.next = node
		tail = tail.next
		if node.next:
			i+=1
			heapq.heappush(h, (node.next.val, i, node.next))

	return head.next

Problem Classification

This problem can be classified as a Sorting and Merging problem in non-computer science terminology. The problem requires to handle multiple sorted sequences (which could represent any sort of ordered data, not necessarily ’lists’ in the computer science sense) and merge them into a single, sorted sequence.

Language Agnostic Coding Drills

This code is a Python function for merging k sorted linked lists using a priority queue provided by the heapq library. Here are the fundamental concepts and problem-solving steps it utilizes, listed in order of increasing difficulty:

  1. Understanding Variables and Basic Data Structures - Understanding how to declare variables, and use basic data structures like lists and tuples, is crucial. This code uses a list h as a priority queue and a tuple (lists[i].val, i, lists[i]) to store values from linked lists.

  2. Working with Lists - You need to know how to work with lists in Python including accessing elements, getting the length, and looping through them.

  3. Object-Oriented Programming (OOP) - The code assumes that each element in the list lists is an object of a ListNode class that has attributes .val (for the value of the node) and .next (for the next node in the list). You need to understand how to create classes and objects, and how to access their properties.

  4. Heap Data Structure (Priority Queue) - Understanding the heap data structure is essential, as it is the primary tool used in this code for efficient sorting and retrieval of values from different lists. The functions heapq.heappush and heapq.heappop are used to add and remove elements from the heap, respectively.

  5. Linked Lists - This is a higher-level data structure used in this code. Each node in the list contains a value and a reference to the next node. You need to understand how to traverse linked lists and how to add nodes to them.

  6. Problem-Solving and Algorithms - The code is an implementation of a specific algorithm: merging k sorted lists. This involves understanding the problem requirements and constraints, devising a strategy to solve the problem efficiently, and then implementing the solution using the aforementioned programming concepts.

The general approach to the problem is as follows:

  1. Create a heap h and initialize head and tail pointers of the result list.

  2. Push the first node of each list in lists to h. Each element in the heap is a tuple containing the value of the node, its index in lists, and the node itself.

  3. While h is not empty, pop the node with the smallest value from h, append it to the result list, and push its next node to h if it exists.

  4. Continue this process until h is empty, at which point all nodes have been visited and the result list contains all nodes in ascending order.

  5. Finally, return the first node of the result list (ignoring the initial dummy node).

Targeted Drills in Python

  1. Understanding Variables and Basic Data Structures

Let’s declare some variables and basic data structures.

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# Variable declaration
x = 10
print(x)

# Declaration of a list
my_list = [1, 2, 3, 4, 5]
print(my_list)

# Declaration of a tuple
my_tuple = (6, 7, 8, 9, 10)
print(my_tuple)
  1. Working with Lists

Here we’ll go through some common list operations.

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my_list = [1, 2, 3, 4, 5]

# Accessing elements
print(my_list[0])  # prints 1

# Getting the length
print(len(my_list))  # prints 5

# Looping through a list
for i in my_list:
    print(i)  # prints 1, 2, 3, 4, 5 on separate lines
  1. Object-Oriented Programming (OOP)

Let’s define a simple ListNode class and create an object of this class.

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class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

# Creating an object of ListNode
node = ListNode(1)

print(node.val)  # prints 1
print(node.next)  # prints None
  1. Heap Data Structure (Priority Queue)

We’ll push some values into a heap and pop them off.

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import heapq

h = []

heapq.heappush(h, (3, "Apple"))
heapq.heappush(h, (1, "Banana"))
heapq.heappush(h, (2, "Cherry"))

print(heapq.heappop(h))  # prints (1, 'Banana')
print(heapq.heappop(h))  # prints (2, 'Cherry')
print(heapq.heappop(h))  # prints (3, 'Apple')
  1. Linked Lists

Let’s create a linked list and traverse through it.

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class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

# Creating a linked list 1 -> 2 -> 3
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)

node1.next = node2
node2.next = node3

# Traverse the linked list
current_node = node1
while current_node is not None:
    print(current_node.val)
    current_node = current_node.next
  1. Problem-Solving and Algorithms

Here’s the specific implementation for merging k sorted linked lists.

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from queue import PriorityQueue

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists):
    class Wrapper:
        def __init__(self, node):
            self.node = node
        def __lt__(self, other):
            return self.node.val < other.node.val

    head = point = ListNode(0)
    q = PriorityQueue()
    for l in lists:
        if l:
            q.put(Wrapper(l))
    while not q.empty():
        node = q.get().node
        point.next = node
        point = point.next
        node = node.next
        if node:
            q.put(Wrapper(node))
    return head.next

This is a bit more complex, but it’s basically using a priority queue to always pick the smallest node from the heads of the input lists, and appends that to the output list. It repeats this until all nodes from all lists have been visited.

Q&A

Why there are test cases such as [[], []] and not a word about possible inputs expected output?

The presence of test cases like [[], []] or other edge cases serves multiple purposes:

  1. Edge Cases: These cases are typically edge cases that test the lower or upper bounds of the input constraints. They help verify if the solution can handle all kinds of input and doesn’t break or produce incorrect output.

  2. Robustness: Handling such cases correctly makes your algorithm more robust. It must not only work for standard cases but also for corner cases.

  3. Specifications: The problem statement usually specifies the constraints and allowable inputs for a problem. In this problem, the constraints allow for k to be as low as 0 and the length of each list to be as low as 0, making [[], []] a valid test case.

  4. Explicitness: Even if the problem doesn’t explicitly mention that empty lists can be part of the input, the constraints make it implicitly allowed. Therefore, your code should be able to handle it.

  5. Real-World Scenarios: In actual applications, you might encounter such empty structures. Your code should be resilient enough to handle them gracefully rather than crashing or producing incorrect output.

As for not mentioning the expected output for such edge cases, it’s generally understood that an edge case like [[], []] would naturally result in an empty list since there are no elements to merge. Providing expected output for standard test cases is common, but for edge cases like this, the expected output is often considered self-explanatory given the problem’s constraints and requirements.

Python…

I assume that everyone realizes that without using ListNode that the solution to this problem is one line of code. Flatten the lists to one list and sort the list.

Why are there so many problems that have to be answered using ListNodes? It’s a complete waste of time in a language such as python. Shouldn’t the simplest solution be the best solution?

The use of ListNode or linked lists in algorithmic problems isn’t necessarily about finding the most efficient solution in Python or any specific language. Rather, it’s about teaching or testing certain concepts. Here are a few reasons:

  1. Conceptual Understanding: Using ListNode often helps in understanding how to manually manipulate pointers and nodes, which is crucial for a deep understanding of data structures.

  2. Language Agnostic: These problems are designed to be language-agnostic. While Python has easy list manipulation, languages like C or C++ do not have such built-in capabilities. The problem is set to be fair to everyone, regardless of the programming language used.

  3. Algorithm Complexity: Some problems require an in-depth understanding of algorithm complexity. Using built-in functions like sort can sometimes give a false sense of simplicity when, in fact, the actual task of merging the lists might be more complex.

  4. Resource Utilization: In some cases, especially in embedded systems or other resource-constrained environments, understanding how to work with nodes and pointers is essential for efficient memory usage.

  5. Skill Assessment: These problems are often used in interviews to assess understanding of fundamental computer science concepts, beyond syntactic sugar provided by specific programming languages.

  6. Real-world Use Cases: Not all data comes in simple, sorted, array-like structures. Linked lists, trees, and graphs are data structures commonly used in real-world applications. Understanding them is key to being a well-rounded developer.

  7. Optimal Substructure: Problems that mandate the use of data structures like linked lists often involve an optimal substructure that makes for an elegant solution, thereby teaching you problem-solving patterns that can be applied elsewhere.

  8. Consistency: If a problem was initially posed in a computer science paper or textbook using ListNodes, it’s often kept that way to maintain consistency with classic algorithm and data structure literature.

So while it may seem unnecessary or verbose, using ListNode provides a broader educational benefit, ensuring that you understand both high-level language features and low-level data structure manipulation.

Can someone help clarify why Divide & Conquer is O(N log k) instead of O(Nk)?

How I got to O(N k): Assuming k lists, the first round we merge k/2, then k/4, k/8, etc..

This series k/2 + k/4 + k/8 + … is a known series that sums to k.

Since there’s this many merges and each merge is O(N), I thought runtime would be O(Nk)

The divide-and-conquer approach for merging k sorted linked-lists is often O(N log k) due to the way the merge operations are organized. Here’s why:

  1. Divide Stage: You have k sorted linked-lists, and you divide them into pairs. You keep doing this until each pair is merged into a single sorted list. The total number of lists is halved in each iteration. The number of times you can halve k until you get to 1 is log(k) (base 2).

  2. Conquer (Merge) Stage: In each of these log(k) stages, you are performing merge operations. Each merge operation for two lists with a total of N elements would be O(N).

  3. Total Complexity: The total time complexity for these operations is, therefore, log(k) stages of O(N) merge operations, resulting in a time complexity of O(N log k).

The reason it’s not O(Nk) is that you’re not merging N elements k times. The division process reduces the number of lists you have to merge at each step, and the height of this division is log(k), not k. This division step is a key aspect of divide-and-conquer algorithms, and it’s what gives you the log(k) factor.

So, in short, the O(N log k) time complexity arises from merging N elements log(k) times, not k times.

Is it possible to solve this question under the same time complexity without implementing the priority queue?

Yes, you can solve the problem with the same O(N log k) time complexity without using a priority queue. One alternative is to use a divide-and-conquer approach to merge the lists two at a time until only one list remains. Here’s how:

Divide-and-Conquer without Priority Queue

  1. Divide Stage: Pair up k sorted linked lists and merge each pair. After the first round of merging, you will have k/2 lists.

  2. Repeat: Continue this process until you have only one list remaining. This will take log(k) rounds.

  3. Conquer (Merge) Stage: Each merge operation for two lists with a total of N elements will have a time complexity of O(N).

The division step will be performed log(k) times, and each time you have to merge all N elements, so the overall complexity remains O(N log k).

Pseudocode

Here’s some simple pseudocode to demonstrate the idea:

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def mergeTwoLists(l1, l2):
    # Merge two sorted lists l1 and l2, return the merged sorted list
    # Complexity: O(N)

def mergeKLists(lists):
    length = len(lists)
    
    # Base case
    if length == 0:
        return None
    if length == 1:
        return lists[0]
    
    # Initialize new list to store the merged pairs
    new_lists = []
    
    # Merge lists in pairs
    for i in range(0, length, 2):
        l1 = lists[i]
        l2 = lists[i + 1] if i + 1 < length else None
        new_lists.append(mergeTwoLists(l1, l2))
    
    # Recursively call mergeKLists with the newly merged lists
    return mergeKLists(new_lists)

The function mergeKLists will be called log(k) times, and each call will involve merging N elements, giving you a time complexity of O(N log k).

I think my code’s complexity is also O(nlogk) and not using heap or priority queue, n means the total elements and k means the size of list.

The mergeTwoLists functiony in my code comes from the problem Merge Two Sorted Lists whose complexity obviously is O(n), n is the sum of length of l1 and l2.

To put it simpler, assume the k is 2^x, So the progress of combination is like a full binary tree, from bottom to top. So on every level of tree, the combination complexity is n, beacause every level have all n numbers without repetition. The level of tree is x, ie logk. So the complexity is O(nlogk).

for example, 8 ListNode, and the length of every ListNode is x1, x2, x3, x4, x5, x6, x7, x8, total is n.

on level 3: x1+x2, x3+x4, x5+x6, x7+x8 sum: n

on level 2: x1+x2+x3+x4, x5+x6+x7+x8 sum: n

on level 1: x1+x2+x3+x4+x5+x6+x7+x8 sum: n

total 3n, nlog8

You’re correct in your analysis. The code you’re describing does indeed have a time complexity of O(N log k) using the divide-and-conquer approach without using a heap or priority queue. The full binary tree analogy aptly captures the process.

Each “level” of your binary tree represents a round of merging lists, and the number of levels is log(k). Since you’re merging all elements at each level (totaling N), the overall complexity becomes N multiplied by log(k), or O(N log k).

Your understanding aligns with the complexities typically found in divide-and-conquer algorithms, where you break down the problem into smaller parts and then combine them. It seems like your implementation is both efficient and correct based on your description.