Sort a Linked List Using Merge Sort in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle node of the linked list.

DSA Python

def get_middle(head):
    if head is None:
        return head
    slow = head
    fast = head.next
    while fast and fast.next:
        slow = slow.[1]
        fast = fast.next.next
    return slow

Drag options to blanks, or click blank then click option'

Arandom

Bprev

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes errors because singly linked lists don't have 'prev'.

Using 'fast.next' instead of 'fast.next.next' for the fast pointer.

The slow pointer moves one step at a time using the 'next' attribute to find the middle node.

2fill in blank

medium

Complete the code to merge two sorted linked lists.

DSA Python

def merge_sorted_lists(l1, l2):
    dummy = ListNode(0)
    tail = dummy
    while l1 and l2:
        if l1.val < l2.val:
            tail.next = l1
            l1 = l1.[1]
        else:
            tail.next = l2
            l2 = l2.next
        tail = tail.next
    tail.next = l1 or l2
    return dummy.next

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cchild

Drandom

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which does not exist in singly linked lists.

Using 'child' or 'random' which are not standard pointers.

To move forward in the linked list l1, we use l1.next.

3fill in blank

hard

Fix the error in the merge_sort function to correctly split the list.

DSA Python

def merge_sort(head):
    if not head or not head.next:
        return head
    middle = get_middle(head)
    next_to_middle = middle.next
    middle.[1] = None
    left = merge_sort(head)
    right = merge_sort(next_to_middle)
    sorted_list = merge_sorted_lists(left, right)
    return sorted_list

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cchild

Drandom

Attempts:

3 left

💡 Hint

Common Mistakes

Setting middle.prev which does not exist in singly linked lists.

Not breaking the list, causing infinite recursion.

To split the list, we set middle.next to None to break the list into two halves.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

DSA Python

words = ['apple', 'bat', 'carrot', 'dog', 'elephant']
lengths = {word: [1] for word in words if len(word) [2] 3}

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Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>' causing wrong filtering.

Using 'word' instead of 'len(word)' for dictionary values.

We use len(word) to get the length and '>' to filter words longer than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values greater than 0.

DSA Python

data = {'a': 1, 'b': 0, 'c': 3}
result = { [1] : [2] for k, v in data.items() if v [3] 0 }

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Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'k' instead of 'k.upper()' for keys.

Using '<' instead of '>' for filtering values.

Keys are converted to uppercase with k.upper(), values are v, and filter uses '>' to keep values greater than 0.