L1 · FoundationsList basics~1 min · 4 tests

Count Occurrences of an Element

Count how many times a value appears in a Python list with a loop, and compare with list.count(). Linear search basics with live tests in your browser.

The problem

Return how many times target appears in nums, without list.count().

Examples

Example 1
Input
```
count_of([1, 2, 2, 3, 2], 2)
```
Expected output
```
3
```
Example 2
Input
```
count_of([1, 2, 3], 9)
```
Expected output
```
0
```

+ 2 hidden tests on Submit.

Edge cases to ask about

Not present
Empty list

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · count_of

def count_of(nums, target):
    """How many times target appears, without .count()."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Scan and count	O(n)	O(1)

Walkthrough of the optimal approach (try it yourself first)

One pass, one counter. If you need counts for many different targets, build a frequency dict once (O(n)) instead of scanning again for each one — that is the bridge to L2.

Complexity: O(n) time, O(1) space. Every element is compared once.

Reveal the reference solution

def count_of(nums, target):
    count = 0
    for x in nums:
        if x == target:
            count += 1
    return count

Follow-ups interviewers ask

Counts for every distinct value at once.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Count Occurrences of an Element in Python?

The optimal solution runs in O(n) time and O(1) auxiliary space. Every element is compared once.

What follow-up questions do interviewers ask about Count Occurrences of an Element?

Counts for every distinct value at once.