692. Top K Frequent Words

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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: Hash Table + Sorting

We can use a hash table $\textit{cnt}$ to record the frequency of each word. Then, we sort the key-value pairs in the hash table by value, and if the values are the same, we sort by key.

Finally, we take the first $k$ keys.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of words.

Python3

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]

Java

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> cnt = new HashMap<>();
        for (String w : words) {
            cnt.merge(w, 1, Integer::sum);
        }
        Arrays.sort(words, (a, b) -> {
            int c1 = cnt.get(a), c2 = cnt.get(b);
            return c1 == c2 ? a.compareTo(b) : c2 - c1;
        });
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < words.length && ans.size() < k; ++i) {
            if (i == 0 || !words[i].equals(words[i - 1])) {
                ans.add(words[i]);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> cnt;
        for (const auto& w : words) {
            ++cnt[w];
        }
        vector<string> ans;
        for (const auto& [w, _] : cnt) {
            ans.push_back(w);
        }
        ranges::sort(ans, [&](const string& a, const string& b) {
            return cnt[a] > cnt[b] || (cnt[a] == cnt[b] && a < b);
        });
        ans.resize(k);
        return ans;
    }
};

Go

func topKFrequent(words []string, k int) (ans []string) {
	cnt := map[string]int{}
	for _, w := range words {
		cnt[w]++
	}
	for w := range cnt {
		ans = append(ans, w)
	}
	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]
}

TypeScript

function topKFrequent(words: string[], k: number): string[] {
    const cnt: Map<string, number> = new Map();
    for (const w of words) {
        cnt.set(w, (cnt.get(w) || 0) + 1);
    }
    const ans: string[] = Array.from(cnt.keys());
    ans.sort((a, b) => {
        return cnt.get(a) === cnt.get(b) ? a.localeCompare(b) : cnt.get(b)! - cnt.get(a)!;
    });
    return ans.slice(0, k);
}