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.
1function topKFrequent(nums, k) {
2 const freqMap = {};
3 for (const num of nums) {
4 freqMap[num] = (freqMap[num] || 0) + 1;
5 }
6
7 const minHeap = Object.keys(freqMap).sort((a, b) => freqMap[b] - freqMap[a]).slice(0, k);
8
9 return minHeap.map(Number);
10}
11
12const nums = [1, 1, 1, 2, 2, 3];
13const k = 2;
14console.log(topKFrequent(nums, k));
15We use an object to count frequencies, sort keys by frequency, and take the top k.
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).
1from collections import Counter
2
3def topKFrequent(nums, k):
4 count = Counter(nums)
5 buckets = [[] for _ in range(len(nums) + 1)]
6
7 for num, freq in count.items():
8 buckets[freq].append(num)
9
10 result = []
11 for i in range(len(buckets) - 1, 0, -1):
12 for num in buckets[i]:
13 result.append(num)
14 if len(result) == k:
15 return result
16
17if __name__ == '__main__':
18 nums = [1, 1, 1, 2, 2, 3]
19 k = 2
20 print(topKFrequent(nums, k))
21Even with Python, we leverage frequency count with buckets to directly access and collect top k frequent elements.