Search for a Value in Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start traversing the linked list from the head node.

DSA Python

current = [1]

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

Atail

Bhead

CNone

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Starting traversal from tail instead of head.

Assigning current to None initially.

We start searching from the head of the linked list, so we assign current = head.

2fill in blank

medium

Complete the code to move to the next node in the linked list during traversal.

DSA Python

current = current[1]

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

A.next

B.head

C.prev

D.tail

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which is for doubly linked lists.

Using head or tail which are not node attributes.

To move forward in a singly linked list, we assign current = current.next.

3fill in blank

hard

Fix the error in the condition to check if the current node's value matches the target.

DSA Python

if current[1] == target:

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

A.data

B.value

C.val

D.key

Attempts:

3 left

💡 Hint

Common Mistakes

Using val or value when the node uses data.

Using key which is not typical for linked list nodes.

In this linked list, the node's value is stored in the data attribute, so we use current.data.

4fill in blank

hard

Fill both blanks to complete the while loop that searches the linked list for the target value.

DSA Python

while current is not [1] and current.[2] != target:
    current = current.next

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

ANone

Bdata

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using head instead of None.

Checking current.value when the attribute is data.

The loop continues while current is not None (end of list) and the current node's data is not equal to the target.

5fill in blank

hard

Fill all three blanks to return True if the target is found, otherwise False after the loop.

DSA Python

if current is not [1] and current.[2] == [3]:
    return True
else:
    return False

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

ANone

Bdata

Ctarget

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with head instead of None.

Using wrong attribute name instead of data.

Comparing with a wrong variable instead of target.

If current is not None and its data equals target, we found the value and return True. Otherwise, return False.

Practice

1. What does searching for a value in a linked list involve?

easy

A. Sorting the list first then searching for the value

B. Jumping directly to the middle node to find the value

C. Checking each node one by one until the value is found or list ends

D. Using an index to access the value directly

Solution

1. Step 1: Understand linked list traversal

   Linked lists do not have indexes, so we must check nodes one by one from the start.

2. Step 2: Search each node for the value

   We compare the target value with each node's data until we find a match or reach the end.

3. Final Answer:

   Checking each node one by one until the value is found or list ends -> Option C

4. Quick Check:

   Search = Sequential check [OK]

Hint: Traverse nodes sequentially to find the value [OK]

Common Mistakes:

• Assuming direct access by index
• Trying to jump to middle node without traversal
• Sorting before searching in linked list

2. Which of the following is the correct way to start searching for a value in a singly linked list named head?

easy

A. current = head.next

B. current = head

C. current = None

D. current = head.prev

Solution

1. Step 1: Identify the starting node

   Search starts from the first node, which is referenced by head.

2. Step 2: Assign current pointer correctly

   Set current = head to begin traversal from the first node.

3. Final Answer:

   current = head -> Option B

4. Quick Check:

   Start at head node [OK]

Hint: Always start search from head node [OK]

Common Mistakes:

• Starting from head.next skips first node
• Setting current to None causes no traversal
• Using head.prev in singly linked list (no prev pointer)

3. What will be the output of the following code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def search(head, target):
    current = head
    while current:
        if current.data == target:
            return True
        current = current.next
    return False

# Create linked list: 5 -> 10 -> 15 -> None
head = Node(5)
head.next = Node(10)
head.next.next = Node(15)

print(search(head, 10))
print(search(head, 20))

medium

A. True\nFalse

B. False\nTrue

C. True\nTrue

D. False\nFalse

Solution

1. Step 1: Trace search for 10

   Nodes checked: 5 (no), 10 (yes) -> returns True.

2. Step 2: Trace search for 20

   Nodes checked: 5, 10, 15 (none match) -> returns False.

3. Final Answer:

   True\nFalse -> Option A

4. Quick Check:

   10 found, 20 not found [OK]

Hint: Return True immediately when value found [OK]

Common Mistakes:

• Returning False too early
• Not moving to next node correctly
• Confusing True/False outputs

4. Find the error in this search function for a linked list:

def search(head, target):
    current = head
    while current.next:
        if current.data == target:
            return True
        current = current.next
    return False

medium

A. The loop condition should be while current, not while current.next

B. The return False should be inside the loop

C. The function should return the node, not True

D. The current pointer should start at head.next

Solution

1. Step 1: Analyze loop condition

   Using while current.next skips checking the last node.

2. Step 2: Correct loop to include last node

   Change to while current to check all nodes including last.

3. Final Answer:

   The loop condition should be while current, not while current.next -> Option A

4. Quick Check:

   Loop must check all nodes [OK]

Hint: Use while current to include last node in search [OK]

Common Mistakes:

• Stopping before last node
• Returning False too early
• Starting current at head.next skips first node

5. You have a linked list with nodes containing values: 3 -> 7 -> 3 -> 9 -> None. You want to find if value 3 exists and also count how many times it appears. Which approach correctly combines search and count?

hard

A. Use recursion to find first 3, then stop

B. Search for 3 once, then traverse again to count occurrences

C. Sort the list first, then count consecutive 3s

D. Traverse the list once, increment count each time node data equals 3, then check if count > 0

Solution

1. Step 1: Traverse list once counting matches

   Check each node; if data is 3, increase count.

2. Step 2: After traversal, check if count > 0

   If count is more than zero, value 3 exists and count shows occurrences.

3. Final Answer:

   Traverse the list once, increment count each time node data equals 3, then check if count > 0 -> Option D

4. Quick Check:

   Single pass count and check [OK]

Hint: Count while searching in one pass [OK]

Common Mistakes:

• Doing two separate traversals
• Sorting linked list unnecessarily
• Stopping after first find without counting all