A bounded heap retains the best k candidates without sorting every item. Its root should represent the easiest retained item to replace.
Before you start
You should know arrays, loops and basic complexity notation. Start with a small input you can trace by hand. State what each variable means and which invariant is maintained, then use that invariant to explain correctness before discussing performance or writing a more compact solution.
The practical goal is to reason through this situation: A min-heap of size k keeps the k largest values while scanning. Read the walkthrough first, then try the interview exercise before opening its answer. The important part is explaining the decision and its consequences, rather than remembering a definition alone.
Step-by-step walkthrough
Step 1: Choose the retained set
Keep the k largest candidates seen so far.
Step 2: Use the correct root
A min-heap exposes the smallest retained value for replacement.
Step 3: Bound heap size
Insert until k, then replace only when the new value is larger.
Worked scenario
A min-heap of size k keeps the k largest values while scanning.
For k=2 and [5,1,9,3], retain 5 and 1 initially. Nine replaces 1; three is discarded because it is smaller than the retained minimum five. Work is O(n log k) with O(k) storage for positive k, subject to initialization and output-order requirements.
Common mistake
Using the wrong heap direction discards the desired extremes.
Verify the behavior
Test k=0, k greater than input size, ties and negative values.
Interview exercise
Analyze the method.
Answer and reasoning
Each insertion or replacement costs O(log k), giving O(n log k) time and O(k) working storage.
Continue learning
Compare the scenario with the Data Structures and Algorithms interview questions and test your understanding with the Data Structures and Algorithms MCQs. For terminology and implementation details, consult the reference material.