Top K Frequent Words - Solution & Explanation

Explanation:

To solve this problem, we can use a hash map (or dictionary) to count the frequency of each word in the given array. Once we have the frequencies, we can then create a list of words and sort them based on their frequency in descending order. An additional condition for sorting is that if two words have the same frequency, we sort them alphabetically.

This approach simplifies the solution since sorting (bound by the number of unique words) helps easily extract the top k frequent words.

In this C solution, we use an array of Node structs to record words and their counts. We iterate through the input array, updating frequency counts. After processing the input, sorting is done via qsort using a custom comparator that sorts first by frequency, then lexicographically. Finally, we collect the first k elements as the result.

Explanation:

This method takes advantage of a min-heap to efficiently determine the top k frequent words. Instead of sorting the entire set of words, which incurs an O(n log n) cost, we can maintain a min-heap of size k. As we iterate through the frequency map, we push elements into the heap. If heap size exceeds k, we remove the smallest element (i.e., least frequent word or alphabetically greater if frequencies are tied).

This is guided by the Follow-up question, showcasing an alternative solution bounded by O(n log k) time complexity.

In this C++ example, we create a custom comparator to maintain a priority queue (min-heap) based on the frequency and lexicon criteria. By limiting the heap to size k, we efficiently manage only the most critical elements, producing a final list once extraneous words are removed from the heap.