Python 3.11+
The code uses standard Python syntax and containers supported by the browser Judge. The Linear Search proof does not depend on a minor Python release.
Verify in Python docs(opens in a new tab)Inspect candidates sequentially when no exploitable ordering or index is available. Learn the algorithm's preconditions, state transition, correctness argument, Python cost model, and transfer from one focused drill to a full Interview Problem.
Recognize it when
Consider Linear Search when the prompt's constraints and required operations match this shape: Inspect candidates sequentially when no exploitable ordering or index is available.

Sign in to save your code and progress across devices. The lesson and problem statement remain public.
Linear search inspects items in order and stops as soon as the requested occurrence is determined. It needs no ordering or preprocessing.
First, last, any, and all matches are different contracts. First match returns immediately; last match must scan the entire input while updating a candidate index.
def linear_trace(values,target):
trace=[]
for index,value in enumerate(values):
matched=value==target;trace.append([index,value,matched])
if matched:break
return traceImplement linear_trace(values,target). Return [index,value,matched] for inspected items through the first match.
Loading interactive editor.If it does not appear, the reference and starter code remain readable.Reload editor
Choose a missing result that cannot be confused with a valid answer, commonly -1 or None. Do not use truthiness for index zero.
def find_first(values,target):
for index,value in enumerate(values):
if value==target:return index
return -1Implement find_first(values,target). Return the first matching index, or -1.
Loading interactive editor.If it does not appear, the reference and starter code remain readable.Reload editor
Use a scan for one or few queries over unsorted data. Build a dictionary, set, or sorted index when many queries justify O(n) preprocessing and extra space. A single scan is O(n) time and O(1) auxiliary space.
Say: “I inspect in input order, and returning here proves this is the first match. If the loop ends, no item satisfies the predicate.” Cover empty input, duplicates, equality semantics, and the missing sentinel.
Python 3.11+
The code uses standard Python syntax and containers supported by the browser Judge. The Linear Search proof does not depend on a minor Python release.
Verify in Python docs(opens in a new tab)Interview bounds depend on the stated representation, input model, and real Python operations.
| Operation | Average | Worst | Interview note |
|---|---|---|---|
| Linear Search complete workflow | O(n) | O(n) | Lowercase each character and count membership in the five-vowel set. |
O(1) for the focused Count Vowels implementation.
These are the mistakes most likely to survive a happy-path example and fail a boundary case.
Verify algorithm preconditions such as sorted input, nonnegative weights, acyclicity, or admissible heuristics before applying it.
Prevent it: State and verify this precondition before coding: The search space and equality or monotone-boundary contract are explicit.
Avoid
def count_vowels(text):
passUse instead
def count_vowels(text):
return sum(character.lower() in 'aeiou' for character in text)Where you will hit this: Count Vowels(opens in a new tab)
This implementation ignores uppercase vowels and undercounts mixed-case input.
Prevent it: Preserve this proof obligation: Every discarded candidate or interval is excluded by a direct comparison or monotone predicate.
Avoid
def count_vowels(text):
return sum(character in 'aeiou' for character in text)Use instead
def count_vowels(text):
return sum(character.lower() in 'aeiou' for character in text)Where you will hit this: Count Vowels(opens in a new tab)
Include visited state, boundary cases, recursion depth, and hidden copying costs in correctness and complexity analysis.
Prevent it: Count every sort, slice, copy, membership check, heap update, and recursive frame before claiming O(n).
Avoid
def count_vowels(text):
return sum(character in 'aeiou' for character in text)Use instead
def count_vowels(text):
return sum(character.lower() in 'aeiou' for character in text)Where you will hit this: First and Last Target Position(opens in a new tab)
Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.
python-docs · checked 2026-07-27
python-docs · checked 2026-07-12