Search for a Value in Linked List in DSA Rust - Execution Trace

Concept Flow - Search for a Value in Linked List

Start at head node

↓

Check if current node is None?

Yes→Value not found, stop

No↓

Compare current node value with target

↓

Found value, stop

↩Back to check None

Start from the head node, check each node's value. If it matches the target, stop. Otherwise, move to the next node until the list ends.

Execution Sample

DSA Rust

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

This code checks each node in the linked list for the target value and returns true if found, false otherwise.

Execution Table

Step	Operation	Current Node Value	Pointer Changes	Visual State
1	Start at head	3	current -> head (3)	3 -> 7 -> 10 -> null
2	Compare current value with target (7)	3	current -> next (7)	3 -> 7 -> 10 -> null
3	Compare current value with target (7)	7	no pointer change (value matches)	3 -> 7 -> 10 -> null
4	Value matches target (7)?	7	Stop search	3 -> 7 -> 10 -> null
5	Return true	-	-	-

💡 Value 7 found at step 4, search stops.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
current	head (3)	node with value 3	node with value 7	node with value 7	node with value 7	search stopped
found	false	false	false	true	true	true

Key Moments - 3 Insights

Why do we check if current node is None before comparing values?

Because if current is None, it means we've reached the end of the list and the value is not found. This is shown in step 2 where we move current to next node and check again.

Why do we stop the search immediately when the value matches?

Because the goal is to find if the value exists. Once found (step 4), continuing would be unnecessary and inefficient.

What happens if the list is empty (head is None)?

The loop never runs because current is None at start, so the function returns false immediately, meaning value not found.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of current node at step 3?

A 3

B 7

C 10

D None

💡 Hint

Check the 'Current Node Value' column at step 3 in the execution table.

At which step does the search stop because the value is found?

A Step 4

B Step 3

C Step 2

D Step 5

💡 Hint

Look for the step where 'Value matches target' and 'Stop search' appear in the execution table.

If the target value was 10 instead of 7, at which step would the search stop?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Imagine the pointer moving through nodes until it reaches the node with value 10, then stops.

Concept Snapshot

Search in Linked List:
- Start at head node
- Check if current node is None (end)
- Compare node value with target
- If match, stop and return true
- Else move to next node
- Return false if not found

Full Transcript

To search for a value in a linked list, start at the head node. Check if the current node is None, which means the list ended without finding the value. If not None, compare the current node's value with the target. If they match, return true immediately. Otherwise, move to the next node and repeat. If the list is empty, the search returns false immediately. This process continues until the value is found or the list ends.

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