Top K Frequent Words using heaps in Python

Question

I'm trying to solve the Top K Frequent Words Leetcode problem in O(N log K) time and am getting an undesirable result. My Python3 code and console output are below:

from collections import Counter
import heapq

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        
        counts = Counter(words)
        print('Word counts:', counts)
        
        result = []
        for word in counts:
            print('Word being added:', word)
            if len(result) < k:
                heapq.heappush(result, (-counts[word], word))
                print(result)
            else:
                heapq.heappushpop(result, (-counts[word], word))
        result = [r[1] for r in result]
        
        return result

----------- Console output -----------

Word counts: Counter({'the': 3, 'is': 3, 'sunny': 2, 'day': 1})
Word being added: the
[(-3, 'the')]
Word being added: day
[(-3, 'the'), (-1, 'day')]
Word being added: is
[(-3, 'is'), (-1, 'day'), (-3, 'the')]
Word being added: sunny
[(-3, 'is'), (-2, 'sunny'), (-3, 'the'), (-1, 'day')]

When I run the test case ["the", "day", "is", "sunny", "the", "the", "sunny", "is", "is"] with K = 4, I find that the word the gets shifted to the end of the list (after day) once is is added even though they both have a count of 3. This makes sense since the parent need only be <= the children and the children are not ordered in any way. Since (-2, 'sunny') and (-3, 'the') are both > (-3, 'is'), the heap invariant is, in fact, maintained even though (-3, 'the') < (-2, 'sunny') and is the right child of (-3, 'is'). The expected result is ["is","the","sunny","day"] while the output of my code is ["is","sunny","the","day"].

Should I be using heaps to solve this problem in O(N log K) time, and if so, how can I modify my code to achieve the desired result?

Answer 1

You're on the right track with using heapq and Counter you just need to make a slight modification in how you are using them in relation to k: (you need to iterate the whole of counts before adding anything to result):

from collections import Counter
import heapq

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        counts = collections.Counter(words)
        max_heap = []
        for key, val in counts.items():
            heapq.heappush(max_heap, (-val, key))
        
        result = []
        while k > 0:
            result.append(heapq.heappop(max_heap)[1])
            k -= 1
        
        return result

---

Did not read the requirement of being O(N log k) before, here is a modification to the above solution to achieve that:

from collections import Counter, deque
import heapq

class WordWithFrequency(object):
    def __init__(self, word, frequency):
        self.word = word
        self.frequency = frequency

    def __lt__(self, other):
        if self.frequency == other.frequency:
            return lt(other.word, self.word)
        else:
            return lt(self.frequency, other.frequency)

class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:    
        counts = collections.Counter(words)
        
        max_heap = []
        for key, val in counts.items():
            heapq.heappush(max_heap, WordWithFrequency(key, val))
            if len(max_heap) > k:
                heapq.heappop(max_heap)
        
        result = deque([]) # can also use a list and just reverse at the end
        while k > 0:
            result.appendleft(heapq.heappop(max_heap).word)
            k -= 1
        
        return list(result)

Answer 2

You don't need to bother with a heap. Counter() already has a method to return the most common elements.

>>> c = Counter(["the", "day", "is", "sunny", "the", "the", "sunny", "is", "is"])
>>> c.most_common(4)
[('the', 3), ('is', 3), ('sunny', 2), ('day', 1)]

Answer 3

Using Counter() and sort(), this'd also pass on O(N Log N):

class Solution:
    def topKFrequent(self, words, k):
        words_countmap = collections.Counter(words)
        items = list(words_countmap.items())
        items.sort(key=lambda item: (-item[1], item[0]))
        return [item[0] for item in items[0:k]]

Here is a Java Solution using PriorityQueue:

class Solution {
    public static final List<String> topKFrequent(
        final String[] words,
        final int k
    ) {
        LinkedList<String> frequentWords = new LinkedList<>();
        HashMap<String, Integer> wordsMap = new HashMap<>();

        for (int i = 0; i < words.length; i++) {
            wordsMap.put(words[i], wordsMap.getOrDefault(words[i], 0) + 1);
        }

        PriorityQueue<Map.Entry<String, Integer>> wordCounterQueue = new PriorityQueue<>(
            (a, b) -> a.getValue() == b.getValue() ? b.getKey().compareTo(a.getKey()) : a.getValue() -
            b.getValue()
        );

        for (Map.Entry<String, Integer> key : wordsMap.entrySet()) {
            wordCounterQueue.offer(key);

            if (wordCounterQueue.size() > k) {
                wordCounterQueue.poll();
            }
        }

        while (!wordCounterQueue.isEmpty()) {
            frequentWords.add(0, wordCounterQueue.poll().getKey());
        }

        return frequentWords;
    }
}

Similarly with C++:

// Most of headers are already included;
// Can be removed;
#include <iostream>
#include <cstdint>
#include <vector>
#include <string>
#include <unordered_map>
#include <queue>
#include <utility>

// The following block might slightly improve the execution time;
// Can be removed;
static const auto __optimize__ = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();




struct Solution {
    using ValueType = std::uint_fast16_t;
    using Pair = std::pair<std::string, ValueType>;
    static const std::vector<std::string> topKFrequent(
        const std::vector<std::string>& words,
        const ValueType k
    ) {
        std::unordered_map<std::string, ValueType> word_counts;

        for (const auto& word : words) {
            ++word_counts[word];
        }

        std::priority_queue<Pair, std::vector<Pair>, Comparator> word_freqs;

        for (const auto& word_count : word_counts) {
            word_freqs.push(word_count);


            if (std::size(word_freqs) > k) {
                word_freqs.pop();
            }
        }

        std::vector<std::string> k_frequent;

        while (!word_freqs.empty()) {
            k_frequent.emplace_back(word_freqs.top().first);
            word_freqs.pop();
        }

        std::reverse(std::begin(k_frequent), std::end(k_frequent));

        return k_frequent;
    }

  private:
    struct Comparator {
        Comparator() {}
        ~Comparator() {}
        const bool operator()(
            const Pair& a,
            const Pair& b
        ) {
            return a.second > b.second || (a.second == b.second && a.first < b.first);
        }
    };
};

int main() {
    std::vector<std::string> words = {"i", "love", "leetcode", "i", "love", "coding"};
    std::vector<std::string> top_k_frequents = Solution().topKFrequent(words, 2);

    for (const auto& word : top_k_frequents) {
        std::cout << word << "\n";
    }
}