Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2]

Example 2:

Input: nums = [1], k = 1 Output: [1]

Note:

  • You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
  • Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.
  • It’s guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
  • You can return the answer in any order.

Solution:

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class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int num : nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }

        Queue<Integer> q = new PriorityQueue<>(Comparator.comparingInt(freq::get));

        for (int num : freq.keySet()) {
            q.add(num);
            if (q.size() > k) {
                q.remove();
            }
        }
        int[] result = new int[k];
        int i = q.size() - 1;
        for (int num : q) {
            result[i--] = num;
        }
        return result;
    }
}

Solution 2021-10-31

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class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> freq = new HashMap<>();
        
        for (int num: nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        // 1 2 3 4 
        PriorityQueue<Pair<Integer, Integer>> heap = new PriorityQueue<>((a, b) -> {
            return a.getValue() - b.getValue();
        });
        
        for (var entry: freq.entrySet()) {
            heap.add(new Pair<>(entry.getKey(), entry.getValue()));
            
            if (heap.size() > k) {
                heap.poll();
            }
        }
        
        int[] res = new int[k];
        
        int i = 0;
        while (heap.size() > 0) {
            res[i] = heap.poll().getKey();
            i = i + 1;
        }
        return res;
    }
}