Linear Search Algorithm in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of linear search index return

What is the output of this Python code that performs a linear search for the value 7 in the list?

DSA Python

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

result = linear_search([3, 5, 7, 9, 11], 7)
print(result)

A 2

B 3

C -1

D 1

💡 Hint

Remember that list indexes start at 0 and linear search returns the first matching index.

The value 7 is at index 2 in the list. The function returns the index where the target is found.

🧠 Conceptual

intermediate

Number of comparisons in worst case linear search

If a linear search is performed on a list of 10 elements and the target is not present, how many comparisons will be made?

A 9

B 10

C 11

D 0

💡 Hint

In worst case, linear search checks every element once.

If the target is not found, linear search compares with all 10 elements once each.

🔧 Debug

advanced

Identify the error in this linear search code

What error will this code produce when searching for 4 in the list?

DSA Python

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

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

A IndexError

B -1

C 0

D TypeError

💡 Hint

Check the loop condition and how indexes are accessed.

The loop condition allows i to reach len(arr), which is out of range for indexing.

❓ Predict Output

advanced

Output of linear search with multiple occurrences

What is the output of this code searching for 5 in the list with duplicates?

DSA Python

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

print(linear_search([1, 5, 3, 5, 7], 5))

A 3

B 4

C -1

D 1

💡 Hint

Linear search returns the first index where the target is found.

The first 5 is at index 1, so the function returns 1 immediately.

🧠 Conceptual

expert

Time complexity of linear search in average case

What is the average time complexity of linear search on a list of n elements?

A O(n log n)

B O(log n)

C O(n)

D O(1)

💡 Hint

Think about how many elements are checked on average before finding the target.

On average, linear search checks half the elements, which is proportional to n, so O(n).

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