DSA Tracker

Medium

Top K Frequent Words

A medium Heap problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.

Topic
Heap
Sheets
1
Core for
9 roles
Platform
LeetCode

The problem

Given an array of words and an integer k, return the k most frequent words. Ties are broken by alphabetical (lexicographical) order, with smaller words coming first.

Example 1

Input
words = ["i","love","leetcode","i","love","coding"], k = 2
Output
["i", "love"]
Why
"i" appears 2 times, "love" appears 2 times, "leetcode" and "coding" each appear once. Since "i" and "love" tie in frequency, "i" comes first alphabetically.

Example 2

Input
words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4
Output
["the", "is", "sunny", "day"]
Why
"the" appears 4 times, "is" appears 3 times, "sunny" appears 2 times, "day" appears once. The top 4 are ["the","is","sunny","day"] ordered by frequency then alphabetically for ties.

Constraints

  • 1 <= words.length <= 500
  • 1 <= words[i].length <= 10
  • words[i] consists of lowercase English letters
  • k is in the range [1, number of unique words]

How to think about it

Updated 2026-09-09

The comparison between two words is compound: higher frequency wins, but on a tie, lexicographically smaller wins. If you use a bounded heap of size k to drop the losers, the worst candidate in the heap is the one with lower frequency, or on tied frequency, the one that is lexicographically larger.

Approaches, worst first

  1. Sort all unique words

    time O(u log u * L) · space O(u * L)

    Count frequencies into a map. Extract unique words and sort with a custom comparator: frequency descending, then word ascending lexicographically. Slice the first k words.

  2. Bounded min-heap of size kWrite this one

    time O(u log k * L) · space O(u * L)

    Maintain a size-k heap where the root is the least desirable word: lower frequency, or higher lexicographical order on ties. When full, evict the root. Finally, pop and reverse to deliver the result in descending order.

Where people lose marks · 3
  • Reversing the tie-breaker inside the heap comparator. In a min-heap capped at k, the root must be the element you want to discard first, so tied frequencies must consider the lexicographically larger string as smaller in priority.
  • Returning the heap extraction order directly without reversing. Popping a min-heap produces increasing priority, whereas the final answer demands strictly decreasing priority.
  • Comparing strings with numerical subtraction instead of localeCompare or standard string operators.

The theory behind it

Heap — the ground this problem stands on. All Heap problems

What Heap is

A heap is a specialized tree that keeps only the single most extreme item at the very top. In a min-heap, every parent node is smaller than its children, so the smallest element in the entire collection sits immediately at the root. Unlike a binary search tree, a heap does not keep all items in full sorted order. It maintains only a partial order, making it fast at giving you the single smallest or largest item without spending time sorting everything else.

When to reach for it

Reach for a heap when a problem asks for the top k largest elements, the kth smallest value, or a running median from a stream of numbers. Signals include phrases like continuously finding the cheapest item, merging k sorted linked lists, or scheduling tasks with priorities. Whenever you need repeated access to the minimum or maximum value while items are added and removed dynamically, a priority heap is the tool.

How the pattern works

To find the k largest elements, keep a min-heap of fixed size k. Push incoming numbers into the heap; whenever the heap size grows past k, pop the top item, which is the smallest among them. After processing all elements, only the k largest remain. For a running median, balance two heaps: a max-heap holding the smaller half of numbers and a min-heap holding the larger half. In code, heaps are stored compactly as flat arrays where a node at index i has children at indices 2i plus 1 and 2i plus 2.

What each operation costs

OperationTime
read the minimum or maximum elementO(1)
insert a new element and sift into positionO(log n)
remove the top element and sift downO(log n)
build a heap from an array of n itemsO(n)
What usually goes wrong with Heap
  • Using a max-heap instead of a min-heap when keeping the k largest elements, causing the largest values to be evicted while small items stay behind.
  • Assuming that extracting elements by iterating over the backing array yields sorted order, without popping items from the heap one by one.
  • Forgetting that standard language libraries provide a min-heap by default, leading to wrong answers when a max-heap was required.

Which roles need this problem

Heap is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.

Secondary for 11 more roles, including SDE / Backend Engineer, Data Engineer, ML Engineer.

Track this in your role's order

Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.

Start free

More Heap problems

Problem set and role mapping as of .