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692. Top K Frequent Words

Description

Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.

 

Example 1:

Input: words = ["i","love","leetcode","i","love","coding"], k = 2
Output: ["i","love"]
Explanation: "i" and "love" are the two most frequent words.
Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:

Input: words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4
Output: ["the","is","sunny","day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.

 

Constraints:

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

 

Follow-up: Could you solve it in O(n log(k)) time and O(n) extra space?

Solutions

Solution 1

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class Solution:
    def topKFrequent(self, words: List[str], k: int) -> List[str]:
        cnt = Counter(words)
        return sorted(cnt, key=lambda x: (-cnt[x], x))[:k]

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> cnt = new HashMap<>();
        for (String v : words) {
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
        }
        PriorityQueue<String> q = new PriorityQueue<>((a, b) -> {
            int d = cnt.get(a) - cnt.get(b);
            return d == 0 ? b.compareTo(a) : d;
        });
        for (String v : cnt.keySet()) {
            q.offer(v);
            if (q.size() > k) {
                q.poll();
            }
        }
        LinkedList<String> ans = new LinkedList<>();
        while (!q.isEmpty()) {
            ans.addFirst(q.poll());
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> cnt;
        for (auto& v : words) ++cnt[v];
        vector<string> ans;
        for (auto& [key, _] : cnt) ans.emplace_back(key);
        sort(ans.begin(), ans.end(), [&](const string& a, const string& b) -> bool {
            return cnt[a] == cnt[b] ? a < b : cnt[a] > cnt[b];
        });
        ans.erase(ans.begin() + k, ans.end());
        return ans;
    }
};

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func topKFrequent(words []string, k int) []string {
    cnt := map[string]int{}
    for _, v := range words {
        cnt[v]++
    }
    ans := []string{}
    for v := range cnt {
        ans = append(ans, v)
    }
    sort.Slice(ans, func(i, j int) bool {
        a, b := ans[i], ans[j]
        return cnt[a] > cnt[b] || cnt[a] == cnt[b] && a < b
    })
    return ans[:k]
}

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