Search for a Value in Linked List in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a linked list?

A linked list is a chain of nodes where each node holds data and a link to the next node. It is like a treasure hunt where each clue leads to the next.

beginner

How do you search for a value in a linked list?

Start at the first node and check its value. If it matches, stop. If not, move to the next node. Repeat until you find the value or reach the end.

beginner

What happens if the value is not found in the linked list?

You reach the end of the list (null) without finding the value. This means the value is not in the list.

intermediate

Why can't we directly access a node in a linked list like in an array?

Because linked lists are connected by links, not by index. You must follow the links from the start to reach a node.

intermediate

What is the time complexity of searching a value in a linked list?

It is O(n), where n is the number of nodes, because you may need to check each node once.

What do you check first when searching for a value in a linked list?

A The value of the current node

B The value of the last node

C The total number of nodes

D The index of the node

You always check the value of the current node before moving on.

If the value is not found, where does the search end in a linked list?

A At the first node

B At the last node

C At the middle node

D At null (end of the list)

The search ends when you reach null, meaning no more nodes.

What is the main reason linked lists do not support direct access by index?

A Nodes have no values

B Nodes are not stored in continuous memory

C Nodes are sorted

D Nodes are stored in arrays

Linked list nodes are scattered in memory and connected by links.

What is the worst-case time complexity to find a value in a linked list with n nodes?

A O(1)

B O(log n)

C O(n)

D O(n log n)

You may need to check all n nodes in the worst case.

Which of these is NOT a step in searching a linked list?

A Jump directly to the middle node

B Check the node's value

C Start at the head node

D Move to the next node if no match

You cannot jump directly; you must follow links one by one.

Explain how to search for a value in a linked list step-by-step.

Think of walking through a chain of friends asking if they have a message.

Why is searching a linked list slower than searching an array by index?

Imagine looking for a book in a pile versus a shelf with numbered slots.

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