Linear Search Algorithm in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the index of the target if found, else return None.

DSA Rust

fn linear_search(arr: &[i32], target: i32) -> Option<usize> {
    for (index, &value) in arr.iter().enumerate() {
        if value == [1] {
            return Some(index);
        }
    }
    None
}

Drag options to blanks, or click blank then click option'

Avalue

Bindex

Ctarget

Darr

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'index' instead of 'target' in the comparison.

Comparing 'value' to 'arr' which is the whole array.

We compare each element 'value' with the 'target' to find a match.

2fill in blank

medium

Complete the code to return None if the target is not found after checking all elements.

DSA Rust

fn linear_search(arr: &[i32], target: i32) -> Option<usize> {
    for (index, &value) in arr.iter().enumerate() {
        if value == target {
            return Some(index);
        }
    }
    [1]
}

Drag options to blanks, or click blank then click option'

ASome(0)

BNone

CSome(target)

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning Some(0) which incorrectly indicates the target is at index 0.

Returning 0 which is not an Option type.

If the target is not found, the function should return None to indicate absence.

3fill in blank

hard

Fix the error in the loop to correctly iterate over the array with index and value.

DSA Rust

fn linear_search(arr: &[i32], target: i32) -> Option<usize> {
    for ([1], &value) in arr.iter().enumerate() {
        if value == target {
            return Some(index);
        }
    }
    None
}

Drag options to blanks, or click blank then click option'

Aidx

Bvalue

Ci

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name in the loop but returning 'index'.

Using 'value' as the index variable.

The variable 'index' is used later to return the position, so it must be named correctly in the loop.

4fill in blank

hard

Fill both blanks to complete the function that searches for target and returns its index or None.

DSA Rust

fn linear_search(arr: &[i32], target: i32) -> Option<usize> {
    for ([1], &value) in arr.iter().enumerate() {
        if value [2] target {
            return Some(index);
        }
    }
    None
}

Drag options to blanks, or click blank then click option'

Aindex

B==

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causing wrong logic.

Using a different variable name than 'index' causing errors.

The loop variable must be 'index' and the comparison operator must be '==' to check equality.

5fill in blank

hard

Fill all three blanks to create a linear search that returns the index of the target or None.

DSA Rust

fn linear_search(arr: &[i32], target: i32) -> Option<usize> {
    for ([1], &[2]) in arr.iter().enumerate() {
        if [3] == target {
            return Some(index);
        }
    }
    None
}

Drag options to blanks, or click blank then click option'

Aindex

Bvalue

Didx

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Using 'idx' instead of 'index' causing errors.

The index variable is 'index', the value variable is 'value', and the comparison uses 'value'.

Practice

1. What is the main idea behind the Linear Search Algorithm?

easy

A. Divide the list into halves repeatedly to find the target

B. Use a hash table to find the target instantly

C. Sort the list first then search using binary steps

D. Check each element one by one until the target is found or list ends

Solution

1. Step 1: Understand linear search process

   Linear search checks each element in order from start to end.

2. Step 2: Compare with other search methods

   Binary search divides list, hash uses extra memory, linear is simple sequential check.

3. Final Answer:

   Check each element one by one until the target is found or list ends -> Option D

4. Quick Check:

   Linear search = sequential check [OK]

Hint: Linear search means checking items one by one [OK]

Common Mistakes:

• Confusing linear search with binary search
• Thinking linear search requires sorted data
• Assuming linear search uses extra memory

2. Which of the following is the correct Rust syntax to start a linear search loop over a vector arr?

easy

A. while i < arr.len() { }

B. for i in arr { }

C. for i in 0..arr.len() { }

D. loop i in arr.len() { }

Solution

1. Step 1: Identify correct Rust for loop syntax

   Rust uses for i in 0..arr.len() to loop over indices.

2. Step 2: Check other options for syntax errors

   for i in arr iterates over values, not indices; while needs initialization; loop i in is invalid.

3. Final Answer:

   for i in 0..arr.len() { } -> Option C

4. Quick Check:

   Rust for loop over range = for i in 0..len() [OK]

Hint: Use for i in 0..arr.len() to loop indices [OK]

Common Mistakes:

• Using for i in arr to get indices instead of values
• Writing invalid loop syntax like loop i in ...
• Forgetting to initialize loop variable in while loops

3. What will be the output of this Rust code snippet?

let arr = vec![3, 5, 7, 9];
let target = 7;
let mut index = -1;
for i in 0..arr.len() {
    if arr[i] == target {
        index = i as i32;
        break;
    }
}
println!("{}", index);

medium

A. 0

B. 2

C. 3

D. -1

Solution

1. Step 1: Trace the loop to find target index

   Loop checks arr[0]=3, arr[1]=5, arr[2]=7 matches target, sets index=2 and breaks.

2. Step 2: Print the index value

   After break, index is 2, so printed output is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Target 7 found at index 2 [OK]

Hint: Stop loop at first match, print that index [OK]

Common Mistakes:

• Not breaking loop after finding target
• Printing index before loop ends
• Confusing index type casting

4. Identify the error in this Rust linear search code and choose the fix:

let arr = vec![1, 2, 3];
let target = 4;
let mut found = false;
for i in 0..arr.len() {
    if arr[i] = target {
        found = true;
        break;
    }
}
println!("{}", found);

medium

A. Use '==' for comparison instead of '='

B. Change 'found' to integer type

C. Remove the break statement

D. Use while loop instead of for loop

Solution

1. Step 1: Check the if condition syntax

   The code uses '=' which is assignment, not comparison. It should be '==' to compare values.

2. Step 2: Verify other options

   Changing 'found' type or removing break won't fix syntax error. Using while loop is valid but not necessary.

3. Final Answer:

   Use '==' for comparison instead of '=' -> Option A

4. Quick Check:

   Comparison needs '==' not '=' [OK]

Hint: Use '==' to compare values, '=' assigns [OK]

Common Mistakes:

• Using '=' instead of '==' in if conditions
• Removing break causing unnecessary iterations
• Changing variable types unnecessarily

5. You want to find all indices of a target value in a vector arr using linear search in Rust. Which approach correctly collects all matching indices?

hard

A. Use a loop over indices, push index to a vector when element equals target

B. Break the loop at first match and return that index

C. Sort the vector first, then use binary search for all matches

D. Use a hash map to store element counts and find indices

Solution

1. Step 1: Understand requirement to find all indices

   We must check every element and collect indices where element equals target.

2. Step 2: Evaluate options for correctness

   Breaking loop stops early, sorting changes order, hash map counts but doesn't store indices. Only pushing indices in loop works.

3. Final Answer:

   Use a loop over indices, push index to a vector when element equals target -> Option A

4. Quick Check:

   Collect all matches by looping and storing indices [OK]

Hint: Collect indices by checking all elements without breaking [OK]

Common Mistakes:

• Breaking loop too early to find all matches
• Sorting vector which changes original order
• Using hash map counts instead of indices