Top K Frequent Elements

 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]

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
       

        HashMap<Integer,Integer> map= new HashMap<>();
        for(int num:nums)
        {
            map.put(num,map.getOrDefault(num,0)+1);
        }
        PriorityQueue<Map.Entry<Integer,Integer>> heap= new PriorityQueue<>((a,b)->a.getValue()-b.getValue());

        for(Map.Entry<Integer,Integer> entry:map.entrySet())
        {
            heap.offer(entry);

            if(heap.size()>k)
            {
                heap.poll();
            }
        }
            int [] topk=new int[k];
            int i=0;
            while(!heap.isEmpty())
            {
                topk[i++]=heap.poll().getKey();
            }
           
       
         return topk;
    }
}




If frequencies are equal, how can I pick the key which has a larger value?”

This means:
Among elements with the same frequency, prefer the one with the larger number (key).

Normally, you sort by frequency:

(a, b) -> a.getValue() - b.getValue()

To prefer larger key (number) when frequency is equal, change comparator to

(a, b) -> {

    if (a.getValue().equals(b.getValue())) {

        return b.getKey() - a.getKey();  // prefer bigger key

    }

    return a.getValue() - b.getValue(); // smaller frequency first

}



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