Search for a Value in Linked List in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of searching for value 3 in this linked list?

Given the linked list: 1 -> 2 -> 3 -> 4 -> null, what will be the output of searching for value 3?

DSA Python

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

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

# Create linked list
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)

result = search(head, 3)
print(result)

A true

B False

C None

D True

💡 Hint

Check if the search function returns a boolean True or the string 'true'.

The search function returns the boolean value True when the target is found. Python booleans are capitalized: True.

❓ Predict Output

intermediate

What is the output when searching for a value not in the list?

Given the linked list: 5 -> 10 -> 15 -> null, what will be the output of searching for value 7?

DSA Python

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

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

# Create linked list
head = Node(5)
head.next = Node(10)
head.next.next = Node(15)

result = search(head, 7)
print(result)

A None

B True

C False

D 0

💡 Hint

What does the function return if the target is not found?

The function returns False if the target value is not found in the linked list.

🔧 Debug

advanced

What error does this code raise when searching for a value?

Consider this code snippet for searching a value in a linked list. What error will it raise when run?

DSA Python

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

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

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

print(search(head, 3))

A None

B IndexError

C AttributeError

D Infinite loop

💡 Hint

Check the loop condition and what happens when current is the last node.

The loop condition 'while current.next' skips checking the last node's value. When current becomes the last node, current.next is None, so the loop ends without checking the last node. But if the target is the last node's value, it will not be found. No error is raised, but the function returns False incorrectly.

🚀 Application

advanced

How many nodes are traversed when searching for value 4 in this list?

Given the linked list: 4 -> 7 -> 9 -> 12 -> null, how many nodes does the search function check when searching for value 4?

DSA Python

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

def search(head, target):
    current = head
    count = 0
    while current:
        count += 1
        if current.val == target:
            return count
        current = current.next
    return count

head = Node(4)
head.next = Node(7)
head.next.next = Node(9)
head.next.next.next = Node(12)

result = search(head, 4)
print(result)

A 1

B 2

C 3

D 4

💡 Hint

Think about where the target value is in the list.

The target value 4 is at the first node, so the search checks only one node before finding it.

🧠 Conceptual

expert

Which option best describes the time complexity of searching a value in a singly linked list?

What is the time complexity of searching for a value in a singly linked list with n nodes?

A O(1) - Constant time because we can jump to any node directly

B O(n) - Linear time because we may need to check each node

C O(log n) - Logarithmic time due to binary search on the list

D O(n^2) - Quadratic time because of nested loops

💡 Hint

Think about how nodes are connected and how you access them.

In a singly linked list, nodes are connected one after another. To find a value, you may need to check each node one by one, so the time complexity is O(n).

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