January 10, 2026

Finding Top-K Frequent Elements in Java: From Interview Solution to Production-Ready Code

Finding the top-K most frequent elements is a classic interview problem that looks simple—but is often used to evaluate how well a candidate understands time complexity, space optimization, and scalability trade-offs.

In this post, we’ll start with a common interview solution, analyze its shortcomings, and progressively evolve it into optimal and scalable approaches suitable for real-world systems.

Problem Statement

Given a list of strings, return the top K most frequent strings.

Example:

Input: ["def","abc","def","abc","def","ghi"]
K = 3
Output: def, abc, ghi

Baseline Interview Solution (Max Heap)

A very common approach is:

  1. Count frequencies using a HashMap

  2. Push all unique keys into a max-heap

  3. Poll k elements

Code

Map<String, Integer> map = new HashMap<>();
for (String s : list) {
    map.put(s, map.getOrDefault(s, 0) + 1);
}

PriorityQueue<String> pq =
    new PriorityQueue<>((a, b) -> map.get(b) - map.get(a));

pq.addAll(map.keySet());

for (int i = 0; i < k; i++) {
    System.out.println(pq.poll());
}

Complexity Analysis (Corrected)

Let:

  • N = total number of elements

  • U = number of unique elements

StepComplexity
Frequency countO(N)
Heap buildO(U log U)
Extract top KO(k log U)
SpaceO(U)

What Interviewers Notice

  • Heap size grows to all unique elements

  • Comparator uses subtraction → integer overflow risk

  • Inefficient when k is small and U is large

Improvement #1: Fix Comparator Overflow

This line is dangerous:

(a, b) -> map.get(b) - map.get(a)

If counts are large, subtraction can overflow.

Safer Version

(a, b) -> Integer.compare(map.get(b), map.get(a))

This is a frequent interview pitfall.

Improvement #2: Use a Min-Heap of Size K (Optimal)

Instead of storing all unique elements, we only keep the top K seen so far.

Key Idea

  • Use a min-heap

  • Remove smallest frequency whenever size exceeds k

Optimized Code

Map<String, Integer> freq = new HashMap<>();
for (String s : list) {
    freq.put(s, freq.getOrDefault(s, 0) + 1);
}

PriorityQueue<String> pq =
    new PriorityQueue<>((a, b) ->
        Integer.compare(freq.get(a), freq.get(b))
    );

for (String key : freq.keySet()) {
    pq.offer(key);
    if (pq.size() > k) {
        pq.poll();
    }
}

while (!pq.isEmpty()) {
    System.out.println(pq.poll());
}

Complexity

MetricValue
TimeO(N + U log K)
SpaceO(U + K)

This is the preferred interview solution.

Improvement #3: Avoid Repeated Map Lookups

A more efficient and cleaner approach is to store Map.Entry objects directly.

PriorityQueue<Map.Entry<String, Integer>> pq =
    new PriorityQueue<>(
        (a, b) -> Integer.compare(a.getValue(), b.getValue())
    );

for (Map.Entry<String, Integer> entry : freq.entrySet()) {
    pq.offer(entry);
    if (pq.size() > k) {
        pq.poll();
    }
}

while (!pq.isEmpty()) {
    System.out.println(pq.poll().getKey());
}

This reduces repeated hash lookups and improves readability.

Improvement #4: Bucket Sort (When Counts Are Bounded)

If frequencies are bounded by N, we can eliminate heap usage entirely.

Idea

  • Create buckets where index = frequency

  • Traverse from highest frequency down

Complexity

  • Time: O(N + U)

  • Space: O(N)

This is usually mentioned verbally in interviews to demonstrate algorithmic range.

Handling Massive Data (1B+ Records)

For very large datasets:

  • Holding all strings in memory is not feasible

  • HashMap alone can exhaust heap

Real-World Approach

Mentioning this shows system design awareness, not just coding skill.

Key Interview Takeaways

What this problem really tests:

Final Thoughts

A solution that works is not the same as a solution that scales.

If you can:

  • Start with the basic heap

  • Optimize to O(U log K)

  • Discuss bucket sort and streaming

You’ll stand out as someone who writes production-grade code, not just interview code.

If you want, I can:

Just say the word.

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