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 <= 5001 <= words[i].length <= 10words[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 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 × log n), and the space complexity is O(n). Here, n is the number of words.
PythonJavaC++GoTypeScript
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](code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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]
}(code-box)
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);
}(code-box)
