This approach uses a hash map to count the frequency of each element. We then use a min-heap to keep track of the top k elements.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) for storing frequencies.
1using System;
2using System.Collections.Generic;
3using System.Linq;
4
5public class Solution {
6 public int[] TopKFrequent(int[] nums, int k) {
7 var freqMap = new Dictionary<int, int>();
8 foreach (var num in nums) {
9 if (!freqMap.ContainsKey(num))
10 freqMap[num] = 0;
11 freqMap[num]++;
12 }
13
14 return freqMap.OrderByDescending(x => x.Value).Take(k).Select(x => x.Key).ToArray();
15 }
16
17 public static void Main(string[] args) {
18 int[] nums = new int[] {1, 1, 1, 2, 2, 3};
19 int k = 2;
20 Solution sol = new Solution();
21 int[] result = sol.TopKFrequent(nums, k);
22 Console.WriteLine(string.Join(", ", result));
23 }
24}
25We use a dictionary to count frequencies, then order by descending frequency and take the top k elements.
This approach involves using bucket sort where we create buckets for frequency counts and then extract the top k frequent elements.
Time Complexity: O(n + k).
Space Complexity: O(n).
1function topKFrequent(nums, k) {
2 const freqMap = {};
3 nums.forEach(num => {
4 freqMap[num] = (freqMap[num] || 0) + 1;
5 });
6
7 const buckets = Array(nums.length + 1).fill().map(() => []);
8 Object.keys(freqMap).forEach(num => {
9 buckets[freqMap[num]].push(parseInt(num));
10 });
11
12 const result = [];
13 for (let i = buckets.length - 1; i >= 0 && result.length < k; i--) {
14 if (buckets[i].length > 0) {
15 result.push(...buckets[i]);
16 }
17 }
18 return result.slice(0, k);
19}
20
21const nums = [1, 1, 1, 2, 2, 3];
22const k = 2;
23console.log(topKFrequent(nums, k));
24In JavaScript, elements are counted via a map and ordered into buckets representing frequencies, making it possible to fetch the most frequent.