Merge k Sorted Lists

1. Clarify the problem:

Before we begin, let's clarify the problem statement. The "Merge k Sorted Lists" problem asks us to merge k sorted linked lists into a single sorted linked list and return the head of the merged list. Each linked list is sorted in ascending order.

2. Analyze the problem:

Let's analyze the problem to identify the input, output, and constraints.

Input: An array of k linked lists, where each linked list has nodes containing integer values. The linked lists are already sorted in ascending order.

Output: The head of a single merged linked list that contains all the nodes from the input lists in sorted order.

Constraints:

The number of linked lists, k, can be up to 10^4.
The total number of nodes across all linked lists can be up to 10^5.

3. Design an algorithm:

To solve this problem efficiently, we can use a divide-and-conquer approach. Here's the general outline of our algorithm:

Divide the given array of k linked lists into two halves recursively until we have only one or two linked lists.
Merge the two linked lists using a helper function that merges two sorted linked lists into a single sorted linked list.
Repeat the merge operation for pairs of merged linked lists until we have a single merged linked list.

4. Explain your approach:

Our approach involves using a divide-and-conquer strategy to merge the linked lists efficiently. We divide the array of linked lists into halves recursively and then merge the resulting linked lists. By merging pairs of linked lists repeatedly, we can obtain the final merged linked list.

5. Write clean and readable code:

Let's implement the algorithm in Python:

python

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


def mergeKLists(lists):
    if not lists:
        return None

    while len(lists) > 1:
        merged_lists = []

        for i in range(0, len(lists), 2):
            l1 = lists[i]
            l2 = lists[i + 1] if i + 1 < len(lists) else None
            merged = mergeTwoLists(l1, l2)
            merged_lists.append(merged)

        lists = merged_lists

    return lists[0]


def mergeTwoLists(l1, l2):
    dummy = ListNode(0)
    current = dummy

    while l1 and l2:
        if l1.val < l2.val:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next

        current = current.next

    current.next = l1 if l1 else l2

    return dummy.next

6. Test your code:

Let's test the code with some example test cases and explain why we chose them:

python

# Example test case
list1 = ListNode(1, ListNode(4, ListNode(5)))
list2 = ListNode(1, ListNode(3, ListNode(4)))
list3 = ListNode(2, ListNode(6))
print(mergeKLists([list1, list2, list3]))  # Expected: 1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6

We chose this test case to cover the scenario of merging multiple sorted linked lists. The expected output is the merged linked list in sorted order.

7. Optimize if necessary:

The algorithm we implemented has a time complexity of O(N log k), where N is the total number of nodes across all linked lists and k is the number of linked lists. This is because we divide the lists into halves log k times, and each merge operation takes O(N) time.

The space complexity of the algorithm is O(1) since we only use a constant amount of extra space to store pointers and variables.

Thus, the algorithm is already efficient and doesn't require further optimization.

8. Handle error cases:

Our code assumes that the input will be a valid array of linked lists or an empty array. However, it doesn't handle cases where the input is not a list or contains invalid elements.

To handle these error cases gracefully, we can add some input validation at the beginning of our code:

python

def mergeKLists(lists):
    if not isinstance(lists, list):
        return None

    for linked_list in lists:
        if not isinstance(linked_list, ListNode):
            return None

    # Rest of the code

    return merged_list

This modification ensures that the function returns None for invalid inputs.

9. Discuss complexity analysis:

As mentioned earlier, the time complexity of the algorithm is O(N log k), where N is the total number of nodes across all linked lists and k is the number of linked lists. This is because we divide the lists into halves log k times, and each merge operation takes O(N) time.

The space complexity is O(1) since we only use a constant amount of extra space to store pointers and variables.

The algorithm's performance is efficient and can handle a large number of linked lists and nodes efficiently without exceeding memory limitations.

Python