Top K Frequent Words asks you to find the k most common words from a list of strings. Each string is a word, and words may repeat many times. You need to return the top k words sorted by frequency, with an important tie-break rule: if two words have the same frequency, the word that is lexicographically smaller (alphabetically earlier) should come first.
That tie-break is not a small detail. It is the main place where people get tripped up, because it forces you to think about ordering beyond simple counts. You are not just finding the most frequent words. You are also producing them in a specific, predictable order.
This problem is common in interviews because it combines counting, sorting logic, and data structure trade-offs in a way that mirrors real-world tasks like log analysis, search ranking, and trend detection.
Why counting alone is not enough
The first step is obvious: you need to know how many times each word appears. A frequency map is the natural tool for this. You scan the list once, and for every word, you increment its count.
But once you have counts, you still have to select the top k words and return them in the correct order. If you simply sort all words by frequency and then take the first k, you will get the right result, but it may not be the most efficient approach when the list is large or when k is small compared to the number of unique words.
Interviewers usually accept full sorting as a baseline, then ask how you would optimize selection.
The core idea behind the solution
This problem has two tasks that must work together.
First, build a frequency map so you know how common each word is.
Second, extract the top k words using an ordering rule that prioritizes higher frequency, and for ties, prioritizes alphabetical order.
Once you state those two tasks clearly, the rest is choosing the right tool to perform the extraction efficiently.
A clean and common approach: sorting by a custom rule
After building the frequency map, you can convert the unique words into a list and sort them using a custom comparator.
The comparator sorts primarily by frequency in descending order. If two words have the same frequency, it sorts them lexicographically in ascending order.
After sorting, the first k words in the sorted list are your answer.
This approach is easy to explain and easy to get right, which is why it is often the most interview-friendly solution when constraints are moderate.
Want to explore more coding problem solutions? Check out the All Paths From Source to Target and Binary Tree Paths.
A more scalable approach: using a heap for top-k selection
When the number of unique words is large, sorting everything can be unnecessary. A heap allows you to keep only the best candidates.
The idea is to maintain a heap of size k that stores the current top words seen so far. You define the heap ordering so that the “least desirable” word among the top k sits at the top of the heap. That makes it easy to remove when a better candidate arrives.
“Least desirable” here means lower frequency, and when frequencies are equal, lexicographically larger, because it should be pushed out first.
You push words into the heap and pop when the size exceeds k. At the end, the heap contains exactly the top k words, but you may need to output them in the correct order, since heap order is not the final sorted order.
This approach reduces work because you avoid sorting all unique words. It is especially valuable when k is small.
Why the tie-break rule changes the ordering logic
Many candidates handle frequency correctly but get ties wrong.
If two words appear the same number of times, the lexicographically smaller word must rank higher.
That means your comparison logic must be consistent everywhere: when sorting, when pushing into the heap, and when extracting results. If you get the tie-break direction reversed in any step, your output will be incorrect even if the counts are correct.
Interviewers often probe this because it reveals whether you can translate a problem statement into a reliable ordering rule.
Performance in simple terms
Building the frequency map is linear in the number of input words.
Sorting all unique words costs more, but it is simple and often acceptable.
Using a heap keeps the extraction step efficient when k is small, because you limit how many elements you actively manage at once.
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