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692. Top K Frequent Words

To help you guys have a better understanding of this problem, I tried to summary the methods I saw from others in this post. This is my first time to write a summary post, so correct me if there is any problem. Thanks!

To analyze Time Complexity and Space Complexity: n is the total number of words, m is the average length of each word

Just Sorting

The easiest way to think of this problem and easy to implement.
Time complexity: O(nlogn), naive sort is o(nlogn)
Space complexity: O(n), for map and list

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        List<Map.Entry<String, Integer>> l = new LinkedList<>();
        for(Map.Entry<String, Integer> e:map.entrySet()){
            l.add(e);
        }
        Collections.sort(l, new MyComparator());//just use our Comparator to sort
        List<String> ans = new LinkedList<>();
        for(int i = 0;i<=k-1;i++){
            ans.add(l.get(i).getKey());
        }
        return ans;
    }
}
/*
Implement our own comparator for this problem, I will also use this Comparator in other methods(A little different in minHeap method).
We can also use anonymous Comparaotr or Lambda function.
*/
class MyComparator implements Comparator<Map.Entry<String, Integer>> {

    public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
        String word1 = e1.getKey();
        int freq1 = e1.getValue();
        String word2 = e2.getKey();
        int freq2 = e2.getValue();
        if(freq1!=freq2){
            return freq2-freq1;
        }
        else {
            return word1.compareTo(word2);
        }
    }
}

Max Heap

Maintain a max heap and add all the words in it. Pop top K words to get the results.
Time Complexity: O(nlogn + Klogn) = O(nlogn)
Space Complexity: O(n), for heap

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>(new MyComparator());
        for(Map.Entry<String, Integer> e:map.entrySet()){
            pq.offer(e);
        }
        List<String> ans = new LinkedList<>();
        for(int i = 0;i<=k-1;i++){
            ans.add(pq.poll().getKey());
        }
        return ans;
    }
}
class MyComparator implements Comparator<Map.Entry<String, Integer>> {
    public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
        String word1 = e1.getKey();
        int freq1 = e1.getValue();
        String word2 = e2.getKey();
        int freq2 = e2.getValue();
        if(freq1!=freq2){
            return freq2-freq1;
        }
        else {
            return word1.compareTo(word2);
        }
    }
}

Min Heap

Instead of using a max heap, we only store Top K Freqency word we have met so far in our min heap.
Time Complexity: O(nlogK), logK time for each word
Space Complexity: O(K), since the largest number of words in our minheap is K

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        MyComparator comparator = new MyComparator();
        PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>(comparator);
        for(Map.Entry<String, Integer> e:map.entrySet()){
            // If minHeap's size is smaller than K, we just add the entry
            if(pq.size()<k){
                pq.offer(e);
            }
            // Else, we compare the current entry with "min" entry in priority queue
            else {
                if(comparator.compare(e, pq.peek())>0){
                    pq.poll();
                    pq.offer(e);
                }
            }
        }
        List<String> ans = new LinkedList<>();
        for(int i = 0;i<=k-1;i++){
            ans.add(0, pq.poll().getKey());//the "smaller" entry poll out ealier 
        }
        return ans;
    }    
}

// The comparaotr is reversed as maxHeap
class MyComparator implements Comparator<Map.Entry<String, Integer>> {
    public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
        String word1 = e1.getKey();
        int freq1 = e1.getValue();
        String word2 = e2.getKey();
        int freq2 = e2.getValue();
        if(freq1!=freq2){
            return freq1-freq2;
        }
        else {
            return word2.compareTo(word1);
        }
    }
}

Bucket sort + Trie

This method is derived from 347. Top K Frequent Elements. At 347, we use bucket sort(LinkedList in each bucket) to find top K frequency integers and we can choose any integer if there is a tie of frequency . But in this question, the problem is that when there is a tie of frequency, we need to compare the lexicographic order. Thus using bucket sort(LinkedList in each bucket) is not good.
The way to solve the tie problem is to use either trie or BST.
Time Complexity: O(nm) = O(n), m time to construct trie for each word and m is a constant
Space Complexity: O(nm) = O(n), m space for each bucket and m is a constant

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        
        Trie[] buckets = new Trie[words.length];
        for(Map.Entry<String, Integer> e:map.entrySet()){
            //for each word, add it into trie at its bucket
            String word = e.getKey();
            int freq = e.getValue();
            if(buckets[freq]==null){
                buckets[freq] = new Trie();
            }
            buckets[freq].addWord(word);
        }
        
        List<String> ans = new LinkedList<>();

        for(int i = buckets.length-1;i>=0;i--){
        //for trie in each bucket, get all the words with same frequency in lexicographic order. Compare with k and get the result
            if(buckets[i]!=null){
                List<String> l = new LinkedList<>();                               
                buckets[i].getWords(buckets[i].root, l);
                if(l.size()<k){
                    ans.addAll(l);
                    k = k - l.size(); 
                }
                else {
                   for(int j = 0;j<=k-1;j++){
                       ans.add(l.get(j));
                   } 
                    break;
                }
            }
        }
        return ans;
    }
}

class TrieNode {
    TrieNode[] children = new TrieNode[26];
    String word = null;
}

class Trie {
    TrieNode root = new TrieNode();
    public void addWord(String word){
        TrieNode cur = root;
        for(char c:word.toCharArray()){
            if(cur.children[c-'a']==null){
                cur.children[c-'a'] = new TrieNode();
            }
            cur = cur.children[c-'a'];
        }
        cur.word = word;
    }
    
    public void getWords(TrieNode node, List<String> ans){
        //use DFS to get lexicograpic order of all the words with same frequency
        if(node==null){
            return;
        }
        if(node.word!=null){
            ans.add(node.word);
        }
        for(int i = 0;i<=25;i++){
            if(node.children[i]!=null){
                getWords(node.children[i], ans);
            }
        }
        
    }
}

Bucket sort + BST

The reason we use Trie is to break the tie of same word frequency. Thus we can easily use BST to replace Trie(In Java, we can use TreeMap or TreeSet)
Time Complexity: O(n), not sure
Space Complexity: O(n), not sure

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        TreeMap<String, Integer>[] buckets = new TreeMap[words.length];
        for(Map.Entry<String, Integer> e:map.entrySet()){
            String word = e.getKey();
            int freq = e.getValue();
            if(buckets[freq]==null){
                buckets[freq] = new TreeMap<>((a, b)->{
                    return a.compareTo(b);
                });
            }
            buckets[freq].put(word, freq);
        }
        
        List<String> ans = new LinkedList<>();
        for(int i = buckets.length-1;i>=0;i--){
            if(buckets[i]!=null){
                TreeMap<String, Integer> temp = buckets[i];
                if(temp.size()<k){
                    k = k - temp.size();
                    while(temp.size()>0){
                        ans.add(temp.pollFirstEntry().getKey());
                    }  
                }
                else {
                    while(k>0){
                        ans.add(temp.pollFirstEntry().getKey());
                        k--;
                    }
                    break;
                }
            }
        }     
        return ans;
    }
}

Quick select

If the question is to find Kth frequency word, quick select is a good solution and only cost O(n), for this question, after getting Top K frequency words by using quick select, we also need to do a sort to make sure they are in the right order.
Time Complexity: O(n+KlogK), n time for quick select and KlogK time for sort
Space Complexity: O(n)

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class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> map = new HashMap<>();
        for(String word:words){
            map.put(word, map.getOrDefault(word, 0)+1);
        }
        
        Map.Entry<String, Integer>[] entrys = new Map.Entry[map.size()];
        int index = 0;
        for(Map.Entry<String, Integer> e:map.entrySet()){
            entrys[index] = e;
            index++;
        }
        //do quick select
        int start = 0;
        int end = entrys.length-1;
        int mid = 0;
        while(start<=end){
            mid = partition(entrys, start, end);
            if(mid == k-1){
                break;
            }
            else if(mid<k-1){
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }

        List<String> ans = new LinkedList<>();
        List<Map.Entry<String, Integer>> l = new LinkedList<>();
        for(int i = 0;i<=mid;i++){
            l.add(entrys[i]);
        }
        //still need to sort these K words, because we only know they are in result, but not in right order
        Collections.sort(l, new MyComparator());
        for(Map.Entry<String, Integer> e:l){
            ans.add(e.getKey());
        }
        return ans;
    }
    
    private int partition(Map.Entry<String, Integer>[] entrys, int start, int end){
        int pivot = start;
        int left = start + 1;
        int right = end;
        MyComparator myComparator = new MyComparator();
        while(true){
            while(left<=end){
                if(myComparator.compare(entrys[left], entrys[pivot])<=0){
                    left++;
                }
                else {
                    break;
                }
            }
            
            while(right>=start+1){
                if(myComparator.compare(entrys[right], entrys[pivot])>0){
                    right--;
                }
                else {
                    break;
                }
            }
            if(left>right){
                break;
            }
            swap(entrys, left, right);
        }
        swap(entrys, pivot, right);
        return right;
    }

    private void swap(Map.Entry<String, Integer>[] entrys, int i, int j){
        Map.Entry<String, Integer> a = entrys[i];
        entrys[i] = entrys[j];
        entrys[j] = a;
    }
}

class MyComparator implements Comparator<Map.Entry<String, Integer>> {

    public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
        String word1 = e1.getKey();
        int freq1 = e1.getValue();
        String word2 = e2.getKey();
        int freq2 = e2.getValue();
        if(freq1!=freq2){
            return freq2-freq1;
        }
        else {
            return word1.compareTo(word2);
        }
    }
}
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