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LeetCode pattern lab · bounded priority

Keep the few.Reveal the best.

Stream every candidate through a capacity-K heap. Watch the boundary decide who enters, who leaves, and why sorting everything is unnecessary.

Keeper capacity

K = 2

Min-heap · keep largest

Problem deck

Choose what “best” means

Array valuesCapacity K

Prediction modeRejected dock

Current mission

Kth Largest Element in an Array

Find the kth largest value without sorting the full array.

Medium

Steps14

Checkpoints5

Capacity2

Playback console

Control the stream

1 / 14

Speed720ms

Choose a mission, set capacity K, then inspect the keeper heap.

Bounded heap arena

Configure the keeper heap

A min-heap keeps the weakest retained candidate visible at the root.

Capacity2

Stored0

Remaining6

Candidate streamprocessed 0 / 6

3value 3

2value 2

1value 1

5value 5

6value 6

4value 4

Keeper chamber

0 of 2 slots occupied

Heap waiting for candidates

RejectedNothing has crossed the boundary yet.

Live decision inspector

Which heap belongs at the boundary?

Min-heap · capacity 2

root = smallest kept

›

Next action

Stream candidates one at a time

The root must expose the easiest retained candidate to replace.

Invariant

The heap never stores more than k = 2 candidates.

Execution trace

Code in motion

def findKthLargest(nums, k):
    heap = []
    for value in nums:
        heappush(heap, value)
        if len(heap) > k:
            heappop(heap)
    return heap[0]

Why does this work?

Invariant laboratory

Predict which candidate crosses the boundary, then reveal the induction argument.

Local mastery

Your pattern record

Kth Largest Element in an Array0 runs0%

Top K Frequent Elements0 runs0%

K Closest Points to Origin0 runs0%

Completion and best checkpoint accuracy remain in this browser.

Stream in progress

The boundary is still deciding.

Complete the scan to record this mission. Prediction mastery requires at least 70% accuracy.

Recognize the pattern

When should you reach for a bounded heap?

Look for “largest,” “smallest,” “closest,” “most frequent,” or “best K” when K is much smaller than the full input. A heap avoids the cost and disruption of fully sorting candidates you will never return.

◇Can I score each candidate?

◇Do I need only K results?

◇Which retained item is weakest?

◇Can that weakest item live at the root?

Decision recipe

01Define the score and whether larger or smaller is better.
02Choose a root that exposes the weakest retained item.
03Keep heap size at most K; replace only when the boundary improves.
04Return the root for kth order statistics or all retained nodes for Top K.