LeetCode Everyday: 347. Top K Frequent Elements

</pre>https://leetcode.com/problems/top-k-frequent-elements/Given a non-empty array of integers, return the <span style="box-sizing: border-box; font-weight: 700;"><span style="box-sizing: border-box;">k</span></span> most frequent elements.</p><p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;">For example,<br style="box-sizing: border-box;" />Given <code style="box-sizing: border-box; font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 12.6px; padding: 2px 4px; color: rgb(199, 37, 78); border-radius: 4px; background-color: rgb(249, 242, 244);">[1,1,1,2,2,3]</code> and k = 2, return <code style="box-sizing: border-box; font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 12.6px; padding: 2px 4px; color: rgb(199, 37, 78); border-radius: 4px; background-color: rgb(249, 242, 244);">[1,2]</code>.</p><p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"><span style="box-sizing: border-box; font-weight: 700;">Note: </span><br style="box-sizing: border-box;" /></p><ul style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"><li style="box-sizing: border-box;">You may assume <span style="box-sizing: border-box;">k</span> is always valid, 1 ≤ <span style="box-sizing: border-box;">k</span> ≤ number of unique elements.</li></ul><p></p><ul style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;"><li style="box-sizing: border-box;">Your algorithm's time complexity <span style="box-sizing: border-box; font-weight: 700;">must be</span> better than O(<span style="box-sizing: border-box;">n</span> log <span style="box-sizing: border-box;">n</span>), where <span style="box-sizing: border-box;">n</span> is the array's size.</li></ul><div><span style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;color:#333333;"><span style="font-size: 14px; line-height: 30px;">Method 1: count sort. O(n)</span></span></div><div><span style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;color:#333333;"><span style="font-size: 14px; line-height: 30px;"></span></span><pre name="code" class="java">public class Solution {    public List<Integer> topKFrequent(int[] nums, int k) {        HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();        int max = 0;        for(int i:nums){            int n = hm.getOrDefault(i,0)+1;            max = Math.max(max,n);            hm.put(i,n);        }        ArrayList<Integer>[] arr = new ArrayList[max+1];        for(int i=0;i<arr.length; i++){            arr[i] = new ArrayList<Integer>();        }        ArrayList<Integer> ret = new ArrayList<Integer>();        for(int i: hm.keySet()){            arr[hm.get(i)].add(i);        }        int add = k;        while(k>0){            ret.addAll(arr[max]);            k-=arr[max].size();            max--;        }        return ret;    }}

Method 2: 

Idea is very straightforward:

• build a counter map that maps a num to its frequency
• build a heap/priority queue that keeps track of k most significant entries
• iterate through the final heap and get the keys
• The total time complexity I think should be O(N+k+(N-k)lgk), where N is the time spent on traversing over the map, k is the time to build heap, and (N-k)lgk is the time to insert the remaining (N-k)elements.
  

Code in Java:

public List<Integer> topKFrequent(int[] nums, int k) {    Map<Integer, Integer> counterMap = new HashMap<>();    for(int num : nums) {        int count = counterMap.getOrDefault(num, 0);        counterMap.put(num, count+1);    }    PriorityQueue<Map.Entry<Integer, Integer>> pq = new PriorityQueue<>((a, b) -> a.getValue()-b.getValue());    for(Map.Entry<Integer, Integer> entry : counterMap.entrySet()) {        pq.offer(entry);        if(pq.size() > k) pq.poll();    }    List<Integer> res = new LinkedList<>();    while(!pq.isEmpty()) {        res.add(0, pq.poll().getKey());    }    return res;}