Search for a Value in Linked List in DSA Rust - 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 one.

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.

intermediate

What is the time complexity of searching a linked list?

The time complexity is O(n), where n is the number of nodes. You might have to check each node once in the worst case.

intermediate

Why can't we do faster search like binary search on a linked list?

Because linked lists do not allow direct access to the middle element. You must follow links one by one, unlike arrays where you can jump to any index.

beginner

Show a simple Rust function to search a value in a singly linked list.

```rust
struct Node {
    data: i32,
    next: Option<Box<Node>>,
}

fn search(head: &Option<Box<Node>>, target: i32) -> bool {
    let mut current = head;
    while let Some(node) = current {
        if node.data == target {
            return true;
        }
        current = &node.next;
    }
    false
}
```

What does each node in a linked list contain?

A Only a link to the next node

B Only data

C Data and a link to the next node

D Data and a link to the previous node only

Each node holds data and a link to the next node, forming the chain.

What is the first step when searching for a value in a linked list?

A Check the last node

B Jump to the middle node

C Sort the list first

D Check the first node

You start searching from the first node and move forward.

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

A Return true

B Return false after checking all nodes

C Throw an error

D Restart the search

If the value is not found after checking all nodes, return false.

Why is binary search not suitable for linked lists?

A Because linked lists do not allow direct access to middle elements

B Because linked lists are always sorted

C Because linked lists have no data

D Because linked lists are faster than arrays

Linked lists require sequential access, so you cannot jump to the middle element directly.

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

A O(n)

B O(log n)

C O(n log n)

D O(1)

You may need to check every node once, so the time complexity is O(n).

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

Think of walking through a chain checking each link.

Write a simple Rust function to search for an integer value in a singly linked list and explain how it works.

Use Option<Box<Node>> and while let syntax.

Practice

1. What is the main purpose of searching for a value in a linked list?

easy

A. To find if a specific value exists in the list

B. To add a new node at the end of the list

C. To delete the first node of the list

D. To reverse the order of nodes in the list

Solution

1. Step 1: Understand the search operation

   Searching in a linked list means checking each node to see if it contains the value we want.

2. Step 2: Identify the goal of searching

   The goal is to find whether the value exists anywhere in the list, not to modify the list.

3. Final Answer:

   To find if a specific value exists in the list -> Option A

4. Quick Check:

   Search means find value [OK]

Hint: Search means check each node for the value [OK]

Common Mistakes:

• Confusing search with insertion or deletion
• Thinking search changes the list
• Assuming search reverses the list

2. Which Rust code snippet correctly starts a search for a value in a linked list node called head?

easy

A. let current = head.as_mut();

B. let current = head.unwrap();

C. let mut current = head.unwrap();

D. let mut current = head.as_ref();

Solution

1. Step 1: Understand linked list node type in Rust

   Linked list nodes are often wrapped in Option and Box, so head is Option.

2. Step 2: Choose safe way to start traversal

   Using head.as_ref() gives an Option reference to the node without moving ownership, allowing safe traversal.

3. Final Answer:

   let mut current = head.as_ref(); -> Option D

4. Quick Check:

   Use as_ref() to borrow node safely [OK]

Hint: Use as_ref() to get a reference without moving ownership [OK]

Common Mistakes:

• Using unwrap() which can panic if list is empty
• Using as_mut() when only reading values
• Not using mut when needing to traverse

3. What will be the output of this Rust code searching for value 5 in the linked list 1 -> 3 -> 5 -> 7?

struct Node {
    value: i32,
    next: Option,
}

fn search(head: &Option, target: i32) -> bool {
    let mut current = head.as_ref();
    while let Some(node) = current {
        if node.value == target {
            return true;
        }
        current = node.next.as_ref();
    }
    false
}

fn main() {
    let list = Some(Box::new(Node { value: 1, next: Some(Box::new(Node { value: 3, next: Some(Box::new(Node { value: 5, next: Some(Box::new(Node { value: 7, next: None })) })) })) }));
    println!("{}", search(&list, 5));
}

medium

A. false

B. true

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Trace the search function

   The function starts at head (value 1), checks 1 != 5, moves to 3, checks 3 != 5, moves to 5, finds 5 == 5, returns true.

2. Step 2: Confirm output of main

   Main prints the boolean returned by search, which is true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Value found in list = true [OK]

Hint: Follow nodes until value matches or list ends [OK]

Common Mistakes:

• Assuming search returns false without checking all nodes
• Confusing Option types causing compile errors
• Forgetting to use as_ref() when traversing

4. Identify the bug in this Rust function that searches for a value in a linked list:

fn search(head: &Option, target: i32) -> bool {
    let mut current = head.unwrap();
    while current.is_some() {
        if current.value == target {
            return true;
        }
        current = current.next.unwrap();
    }
    false
}

medium

A. Incorrect comparison operator used

B. Missing mut keyword for current variable

C. Using unwrap() on Option without checking can cause panic

D. Function does not return a boolean

Solution

1. Step 1: Analyze unwrap() usage

   Calling unwrap() on Option without checking if it is Some can cause the program to panic if the list is empty or next is None.

2. Step 2: Explain safe alternative

   Using as_ref() and pattern matching with while let Some(node) = current avoids panics by safely handling None cases.

3. Final Answer:

   Using unwrap() on Option without checking can cause panic -> Option C

4. Quick Check:

   Unsafe unwrap causes panic [OK]

Hint: Never unwrap Option without checking Some first [OK]

Common Mistakes:

• Ignoring possibility of None causing panic
• Not using while let for safe traversal
• Assuming unwrap always safe

5. Given a linked list where each node contains an integer, how would you modify the search function to return the position (0-based index) of the first occurrence of a target value, or -1 if not found? Choose the correct Rust code snippet.

struct Node {
    value: i32,
    next: Option,
}

fn search_position(head: &Option, target: i32) -> i32 {
    // Fill in the code
}

hard

A. let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } -1

B. let mut current = head.unwrap(); let mut index = 0; while current.is_some() { if current.value == target { return index; } current = current.next.unwrap(); index += 1; } -1

C. let mut current = head.as_ref(); let mut index = 1; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } 0

D. let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value != target { return index; } current = node.next.as_ref(); index += 1; } -1

Solution

1. Step 1: Initialize traversal and index

   Start with current as a reference to head and index at 0 to count positions from start.

2. Step 2: Traverse nodes checking values

   Use while let Some(node) to safely traverse. If node.value equals target, return current index. Otherwise, move to next node and increment index.

3. Step 3: Return -1 if not found

   If loop ends without finding target, return -1 to indicate absence.

4. Final Answer:

   let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } -1 -> Option A

5. Quick Check:

   Return index if found, else -1 [OK]

Hint: Return index on match, else -1 after full traversal [OK]

Common Mistakes:

• Starting index at 1 instead of 0
• Returning index when value does not match
• Using unwrap() causing panic on empty list