Medium
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?
import java.util.SortedSet
import java.util.TreeSet
@Suppress("NAME_SHADOWING")
class Solution {
fun topKFrequent(words: Array<String>, k: Int): List<String> {
var k = k
val map: MutableMap<String, Int> = HashMap()
for (word in words) {
map[word] = map.getOrDefault(word, 0) + 1
}
val sortedset: SortedSet<Map.Entry<String, Int>> = TreeSet(
java.util.Comparator { (key, value): Map.Entry<String, Int>, (key1, value1): Map.Entry<String, Int> ->
return@Comparator if (value != value1) {
value1 - value
} else {
key.compareTo(key1, ignoreCase = true)
}
},
)
sortedset.addAll(map.entries)
val result: MutableList<String> = ArrayList()
val iterator: Iterator<Map.Entry<String, Int>> = sortedset.iterator()
while (iterator.hasNext() && k-- > 0) {
result.add(iterator.next().key)
}
return result
}
}