Get Length of Linked List in DSA Java - Time & Space Complexity

Time Complexity: Get Length of Linked List

O(n)

Understanding Time Complexity

We want to understand how the time to find the length of a linked list changes as the list grows.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int getLength(Node head) {
    int length = 0;
    Node current = head;
    while (current != null) {
        length++;
        current = current.next;
    }
    return length;
}
    

This code counts how many nodes are in a linked list by moving through each node one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving from one node to the next in the linked list.
• How many times: Once for each node until the end of the list (n times).

How Execution Grows With Input

As the list gets longer, the number of steps grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase in a straight line as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the length grows directly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Getting the length is instant or constant time because we just count once."

[OK] Correct: We must visit each node to count it, so the time grows with the list size.

Interview Connect

Knowing how to analyze simple linked list operations like this helps build confidence for more complex problems.

Self-Check

"What if the linked list stored its length as a variable updated on insertions? How would the time complexity change?"

Practice

1. What does the length of a linked list represent?

easy

A. The sum of all node values

B. The value of the first node

C. The memory size of the linked list

D. The number of nodes in the linked list

Solution

1. Step 1: Understand linked list structure

   A linked list is made of nodes connected one after another.

2. Step 2: Define length meaning

   The length is how many nodes are present in the list.

3. Final Answer:

   The number of nodes in the linked list -> Option D

4. Quick Check:

   Length = count of nodes [OK]

Hint: Length counts nodes, not values or memory [OK]

Common Mistakes:

• Confusing length with sum of values
• Thinking length is memory size
• Using first node value as length

2. Which of the following Java code snippets correctly initializes a variable to count the length of a linked list named head?

easy

A. int length = head; Node current = 0;

B. int length = 0; Node current = head;

C. Node length = 0; int current = head;

D. int length = null; Node current = head;

Solution

1. Step 1: Identify correct variable types

   Length should be an integer starting at 0; current should be a Node starting at head.

2. Step 2: Check each option

   int length = 0; Node current = head; correctly declares int length = 0; and Node current = head;. Others mix types or assign invalid values.

3. Final Answer:

   int length = 0; Node current = head; -> Option B

4. Quick Check:

   Initialize count as int 0 and pointer as Node head [OK]

Hint: Count is int zero; pointer is Node starting at head [OK]

Common Mistakes:

• Mixing variable types
• Assigning head to int
• Using null for int variable

3. What is the output of the following Java code snippet?

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

public class Test {
  public static int getLength(Node head) {
    int count = 0;
    Node current = head;
    while (current != null) {
      count++;
      current = current.next;
    }
    return count;
  }

  public static void main(String[] args) {
    Node n1 = new Node(5);
    Node n2 = new Node(10);
    Node n3 = new Node(15);
    n1.next = n2;
    n2.next = n3;
    System.out.println(getLength(n1));
  }
}

medium

A. 0

B. 2

C. 3

D. NullPointerException

Solution

1. Step 1: Trace linked list nodes

   Nodes are n1 -> n2 -> n3, so total 3 nodes linked.

2. Step 2: Count nodes in getLength

   Loop increments count for each node until current is null, so count = 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   3 nodes counted in list [OK]

Hint: Count increments once per node until null [OK]

Common Mistakes:

• Counting nodes incorrectly
• Expecting 0 for non-empty list
• Confusing null pointer errors

4. Identify the error in the following code that tries to get the length of a linked list:

public int getLength(Node head) {
  int count = 0;
  while (head.next != null) {
    count++;
    head = head.next;
  }
  return count;
}

medium

A. It does not count the last node

B. It causes infinite loop

C. It throws NullPointerException immediately

D. It counts nodes twice

Solution

1. Step 1: Analyze loop condition

   The loop runs while head.next != null, so it stops before last node.

2. Step 2: Check counting logic

   Since last node's next is null, it is never counted, so count misses last node.

3. Final Answer:

   It does not count the last node -> Option A

4. Quick Check:

   Loop condition excludes last node [OK]

Hint: Loop on current != null to count all nodes [OK]

Common Mistakes:

• Using head.next != null instead of head != null
• Missing last node count
• Confusing loop conditions

5. Given a linked list where each node contains an integer, write a method to find the length of the list but only count nodes with even values. Which approach is correct?

hard

A. Traverse all nodes, increment count only if node's data % 2 == 0

B. Count nodes until first odd value is found

C. Count all nodes and then subtract odd nodes count

D. Only count nodes if next node is even

Solution

1. Step 1: Understand problem requirement

   We must count nodes with even values only, ignoring odd nodes.

2. Step 2: Choose correct counting method

   Traverse entire list, check each node's data; increment count if even (data % 2 == 0).

3. Final Answer:

   Traverse all nodes, increment count only if node's data % 2 == 0 -> Option A

4. Quick Check:

   Count even nodes by checking each node [OK]

Hint: Check each node's value; count only if even [OK]

Common Mistakes:

• Stopping count at first odd node
• Subtracting odd count after full count
• Checking next node's value instead of current