Description#
Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105-104 <= nums[i] <= 104k is in the range [1, the number of unique elements in the array].- It is guaranteed that the answer is unique.
Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
Solutions#
Solution 1#
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| class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
cnt = Counter(nums)
return [v[0] for v in cnt.most_common(k)]
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| class Solution {
public int[] topKFrequent(int[] nums, int k) {
Map<Integer, Long> frequency = Arrays.stream(nums).boxed().collect(
Collectors.groupingBy(Function.identity(), Collectors.counting()));
Queue<Map.Entry<Integer, Long>> queue = new PriorityQueue<>(Map.Entry.comparingByValue());
for (var entry : frequency.entrySet()) {
queue.offer(entry);
if (queue.size() > k) {
queue.poll();
}
}
return queue.stream().mapToInt(Map.Entry::getKey).toArray();
}
}
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| using pii = pair<int, int>;
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
for (int v : nums) ++cnt[v];
priority_queue<pii, vector<pii>, greater<pii>> pq;
for (auto& [num, freq] : cnt) {
pq.push({freq, num});
if (pq.size() > k) {
pq.pop();
}
}
vector<int> ans(k);
for (int i = 0; i < k; ++i) {
ans[i] = pq.top().second;
pq.pop();
}
return ans;
}
};
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| func topKFrequent(nums []int, k int) []int {
cnt := map[int]int{}
for _, v := range nums {
cnt[v]++
}
h := hp{}
for v, freq := range cnt {
heap.Push(&h, pair{v, freq})
if len(h) > k {
heap.Pop(&h)
}
}
ans := make([]int, k)
for i := range ans {
ans[i] = heap.Pop(&h).(pair).v
}
return ans
}
type pair struct{ v, cnt int }
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].cnt < h[j].cnt }
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
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| function topKFrequent(nums: number[], k: number): number[] {
let hashMap = new Map();
for (let num of nums) {
hashMap.set(num, (hashMap.get(num) || 0) + 1);
}
let list = [...hashMap];
list.sort((a, b) => b[1] - a[1]);
let ans = [];
for (let i = 0; i < k; i++) {
ans.push(list[i][0]);
}
return ans;
}
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| use std::collections::HashMap;
impl Solution {
pub fn top_k_frequent(nums: Vec<i32>, k: i32) -> Vec<i32> {
let mut map = HashMap::new();
let mut max_count = 0;
for &num in nums.iter() {
let val = map.get(&num).unwrap_or(&0) + 1;
map.insert(num, val);
max_count = max_count.max(val);
}
let mut k = k as usize;
let mut res = vec![0; k];
while k > 0 {
let mut next = 0;
for key in map.keys() {
let val = map[key];
if val == max_count {
res[k - 1] = *key;
k -= 1;
} else if val < max_count {
next = next.max(val);
}
}
max_count = next;
}
res
}
}
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Solution 2#
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| class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
cnt = Counter(nums)
hp = []
for num, freq in cnt.items():
heappush(hp, (freq, num))
if len(hp) > k:
heappop(hp)
return [v[1] for v in hp]
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| class Solution {
public int[] topKFrequent(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
for (int v : nums) {
cnt.put(v, cnt.getOrDefault(v, 0) + 1);
}
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[1] - b[1]);
for (var e : cnt.entrySet()) {
pq.offer(new int[] {e.getKey(), e.getValue()});
if (pq.size() > k) {
pq.poll();
}
}
int[] ans = new int[k];
for (int i = 0; i < k; ++i) {
ans[i] = pq.poll()[0];
}
return ans;
}
}
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| function topKFrequent(nums: number[], k: number): number[] {
const map = new Map<number, number>();
let maxCount = 0;
for (const num of nums) {
map.set(num, (map.get(num) ?? 0) + 1);
maxCount = Math.max(maxCount, map.get(num));
}
const res = [];
while (k > 0) {
for (const key of map.keys()) {
if (map.get(key) === maxCount) {
res.push(key);
k--;
}
}
maxCount--;
}
return res;
}
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