Linear Search Algorithm in DSA Python - Time & Space Complexity

Time Complexity: Linear Search Algorithm

O(n)

Understanding Time Complexity

We want to understand how the time taken by linear search changes as the list size grows.

How many steps does it take to find an item or decide it's not there?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code looks through each item in the list one by one to find the target value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking each element in the list one by one.
• How many times: Up to n times, where n is the number of items in the list.

How Execution Grows With Input

As the list gets bigger, the number of checks grows roughly the same as the list size.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The number of steps grows directly with the size of the list.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the target grows linearly with the number of items.

Common Mistake

[X] Wrong: "Linear search always takes the same time no matter the list size."

[OK] Correct: The time depends on where the target is or if it is missing; bigger lists usually mean more checks.

Interview Connect

Understanding linear search time helps you compare it with faster methods and shows you can analyze simple algorithms clearly.

Self-Check

"What if the list was sorted? How would the time complexity change if you used a different search method?"

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