Linear Search Algorithm in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the index of the target in the list or -1 if not found.

DSA Python

def linear_search(arr, target):
    for i in range(len(arr)):
        if arr[i] == [1]:
            return i
    return -1

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

Alen(arr)

Btarget

Carr

Di

Attempts:

3 left

💡 Hint

Common Mistakes

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

Comparing arr[i] to the whole list 'arr' instead of the target.

We compare each element arr[i] with the target value to find a match.

2fill in blank

medium

Complete the code to return True if the target is found, otherwise False.

DSA Python

def contains(arr, target):
    for element in arr:
        if element == [1]:
            return True
    return False

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

Atarget

Barr

Celement

DFalse

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing element to itself instead of the target.

Returning False too early inside the loop.

We check if each element equals the target to decide if it exists in the list.

3fill in blank

hard

Fix the error in the code to correctly perform linear search and return the index or -1.

DSA Python

def search(arr, target):
    i = 0
    while i < len(arr):
        if arr[i] == [1]:
            return i
        i += 1
    return -1

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

Alen(arr)

Bi

Carr

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[i] to the index variable 'i' instead of the target.

Comparing arr[i] to the whole list 'arr'.

The comparison must be between arr[i] and the target value to find a match.

4fill in blank

hard

Fill both blanks to create a dictionary of elements and their indices for elements greater than 5.

DSA Python

result = { [1]: [2] for [2], [1] in enumerate(arr) if [1] > 5 }

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

Ax

Bi

Carr[x]

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up keys and values in the dictionary comprehension.

Using undefined variables in the condition.

We use 'x' as the element variable and 'i' as the index variable to create a dictionary of elements (keys) and their indices (values) for elements greater than 5.

5fill in blank

hard

Fill all three blanks to create a dictionary of elements and their indices for elements less than 10.

DSA Python

result = { [1]: [2] for [2], [1] in enumerate(arr) if [3] < 10 }

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

Aval

Bidx

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up keys and values in the dictionary comprehension.

Using undefined variables in the condition.

We use 'val' as the element variable and 'idx' as the index variable. The condition checks if 'val' is less than 10.

Practice

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

easy

A. Sort the list first then search using binary steps

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

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

D. Use a hash table to find the target instantly

Solution

1. Step 1: Understand linear search process

   Linear search checks each element in order, one by one.

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 C

4. Quick Check:

   Linear search = sequential check [OK]

Hint: Linear search checks items one by one [OK]

Common Mistakes:

• Confusing linear search with binary search
• Thinking linear search requires sorted list
• Assuming linear search is faster than binary search always

2. Which of the following is the correct syntax to start a linear search loop over a list arr in Python?

easy

A. for i to len(arr):

B. for i in range(arr):

C. for i in arr:

D. for i in range(len(arr)):

Solution

1. Step 1: Identify how to loop over indices

   To access elements by index, use range(len(arr)).

2. Step 2: Check each option's correctness

   for i in arr: loops over elements, not indices; B is invalid because range needs an integer; D is invalid syntax.

3. Final Answer:

   for i in range(len(arr)): -> Option D

4. Quick Check:

   Loop indices = range(len(arr)) [OK]

Hint: Use range(len(arr)) to loop indices [OK]

Common Mistakes:

• Using range(arr) instead of range(len(arr))
• Trying to loop with invalid syntax like 'for i to len(arr)'
• Confusing looping over elements vs indices

3. What will be the output of the following code?

def linear_search(arr, target):
    for i in range(len(arr)):
        if arr[i] == target:
            return i
    return -1

print(linear_search([4, 2, 7, 1, 3], 7))

medium

A. 2

B. 3

C. -1

D. 7

Solution

1. Step 1: Trace the loop and compare elements

   Check elements: arr[0]=4 (no), arr[1]=2 (no), arr[2]=7 (yes, matches target 7).

2. Step 2: Return index when found

   Function returns index 2 immediately when target found.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Index of 7 in list = 2 [OK]

Hint: Return index when target matches during loop [OK]

Common Mistakes:

• Returning element value instead of index
• Returning -1 even when target is present
• Confusing index with element value

4. Find the error in this linear search function:

def linear_search(arr, target):
    for i in range(len(arr)):
        if arr[i] = target:
            return i
    return -1

medium

A. Using '=' instead of '==' in the if condition

B. Missing colon ':' after for loop

C. Returning -1 instead of None when not found

D. Looping over range(len(arr)) instead of arr

Solution

1. Step 1: Identify syntax error in condition

   The if condition uses '=' which is assignment, not comparison.

2. Step 2: Correct operator for comparison

   Use '==' to compare values in if condition.

3. Final Answer:

   Using '=' instead of '==' in the if condition -> Option A

4. Quick Check:

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

Hint: Use '==' for comparison, '=' is assignment [OK]

Common Mistakes:

• Using single '=' in if condition
• Confusing assignment and comparison operators
• Ignoring syntax errors causing runtime failure

5. You want to find all positions of a target number in a list using linear search. Which modification to the code below will achieve this?

def linear_search_all(arr, target):
    positions = []
    for i in range(len(arr)):
        if arr[i] == target:
            # What to do here?
    return positions

print(linear_search_all([1, 3, 7, 3, 9, 3], 3))

hard

A. Return i immediately when target is found

B. Append index i to positions list inside the if block

C. Break the loop after first match

D. Use positions = target instead of list

Solution

1. Step 1: Understand requirement to find all matches

   We must collect all indices where target appears, not stop at first.

2. Step 2: Append matching indices to list

   Inside if, add positions.append(i) to store all found positions.

3. Final Answer:

   Append index i to positions list inside the if block -> Option B

4. Quick Check:

   Collect all indices by appending [OK]

Hint: Append all matching indices to list [OK]

Common Mistakes:

• Returning immediately on first match
• Breaking loop early
• Assigning target to positions instead of appending