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Create Top_K_Frequent.cpp
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‎Top_K_Frequent.cpp

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// Just Sorting
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// The easiest way to think of this problem and easy to implement.
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// Time complexity: O(nlogn), naive sort is o(nlogn)
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// Space complexity: O(n), for map and list
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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List<Map.Entry<String, Integer>> l = new LinkedList<>();
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for(Map.Entry<String, Integer> e:map.entrySet()){
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l.add(e);
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}
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Collections.sort(l, new MyComparator());//just use our Comparator to sort
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List<String> ans = new LinkedList<>();
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for(int i = 0;i<=k-1;i++){
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ans.add(l.get(i).getKey());
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}
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return ans;
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}
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}
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/*
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// Implement our own comparator for this problem, I will also use this Comparator in other methods(A little different in minHeap method).
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// We can also use anonymous Comparaotr or Lambda function.
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// */
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class MyComparator implements Comparator<Map.Entry<String, Integer>> {
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public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
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String word1 = e1.getKey();
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int freq1 = e1.getValue();
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String word2 = e2.getKey();
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int freq2 = e2.getValue();
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if(freq1!=freq2){
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return freq2-freq1;
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}
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else {
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return word1.compareTo(word2);
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}
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}
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}
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Max Heap
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Maintain a max heap and add all the words in it. Pop top K words to get the results.
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Time Complexity: O(nlogn + Klogn) = O(nlogn)
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Space Complexity: O(n), for heap
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>(new MyComparator());
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for(Map.Entry<String, Integer> e:map.entrySet()){
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pq.offer(e);
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}
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List<String> ans = new LinkedList<>();
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for(int i = 0;i<=k-1;i++){
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ans.add(pq.poll().getKey());
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}
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return ans;
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}
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}
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class MyComparator implements Comparator<Map.Entry<String, Integer>> {
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public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
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String word1 = e1.getKey();
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int freq1 = e1.getValue();
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String word2 = e2.getKey();
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int freq2 = e2.getValue();
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if(freq1!=freq2){
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return freq2-freq1;
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}
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else {
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return word1.compareTo(word2);
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}
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}
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}
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// Min Heap
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// Instead of using a max heap, we only store Top K Freqency word we have met so far in our min heap.
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// Time Complexity: O(nlogK), logK time for each word
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// Space Complexity: O(K), since the largest number of words in our minheap is K
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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MyComparator comparator = new MyComparator();
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PriorityQueue<Map.Entry<String, Integer>> pq = new PriorityQueue<>(comparator);
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for(Map.Entry<String, Integer> e:map.entrySet()){
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// If minHeap's size is smaller than K, we just add the entry
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if(pq.size()<k){
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pq.offer(e);
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}
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// Else, we compare the current entry with "min" entry in priority queue
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else {
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if(comparator.compare(e, pq.peek())>0){
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pq.poll();
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pq.offer(e);
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}
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}
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}
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List<String> ans = new LinkedList<>();
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for(int i = 0;i<=k-1;i++){
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ans.add(0, pq.poll().getKey());//the "smaller" entry poll out ealier
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}
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return ans;
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}
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}
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// The comparaotr is reversed as maxHeap
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class MyComparator implements Comparator<Map.Entry<String, Integer>> {
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public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
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String word1 = e1.getKey();
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int freq1 = e1.getValue();
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String word2 = e2.getKey();
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int freq2 = e2.getValue();
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if(freq1!=freq2){
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return freq1-freq2;
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}
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else {
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return word2.compareTo(word1);
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}
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}
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}
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// Bucket sort + Trie
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// 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.
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// The way to solve the tie problem is to use either trie or BST.
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// Time Complexity: O(nm) = O(n), m time to construct trie for each word and m is a constant
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// Space Complexity: O(nm) = O(n), m space for each bucket and m is a constant
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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Trie[] buckets = new Trie[words.length];
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for(Map.Entry<String, Integer> e:map.entrySet()){
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//for each word, add it into trie at its bucket
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String word = e.getKey();
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int freq = e.getValue();
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if(buckets[freq]==null){
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buckets[freq] = new Trie();
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}
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buckets[freq].addWord(word);
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}
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List<String> ans = new LinkedList<>();
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for(int i = buckets.length-1;i>=0;i--){
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//for trie in each bucket, get all the words with same frequency in lexicographic order. Compare with k and get the result
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if(buckets[i]!=null){
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List<String> l = new LinkedList<>();
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buckets[i].getWords(buckets[i].root, l);
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if(l.size()<k){
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ans.addAll(l);
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k = k - l.size();
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}
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else {
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for(int j = 0;j<=k-1;j++){
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ans.add(l.get(j));
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}
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break;
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}
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}
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}
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return ans;
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}
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}
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class TrieNode {
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TrieNode[] children = new TrieNode[26];
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String word = null;
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}
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class Trie {
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TrieNode root = new TrieNode();
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public void addWord(String word){
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TrieNode cur = root;
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for(char c:word.toCharArray()){
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if(cur.children[c-'a']==null){
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cur.children[c-'a'] = new TrieNode();
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}
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cur = cur.children[c-'a'];
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}
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cur.word = word;
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}
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public void getWords(TrieNode node, List<String> ans){
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//use DFS to get lexicograpic order of all the words with same frequency
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if(node==null){
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return;
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}
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if(node.word!=null){
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ans.add(node.word);
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}
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for(int i = 0;i<=25;i++){
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if(node.children[i]!=null){
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getWords(node.children[i], ans);
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}
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}
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}
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}
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// Bucket sort + BST
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// 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)
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// Time Complexity: O(n), not sure
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// Space Complexity: O(n), not sure
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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TreeMap<String, Integer>[] buckets = new TreeMap[words.length];
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for(Map.Entry<String, Integer> e:map.entrySet()){
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String word = e.getKey();
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int freq = e.getValue();
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if(buckets[freq]==null){
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buckets[freq] = new TreeMap<>((a, b)->{
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return a.compareTo(b);
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});
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}
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buckets[freq].put(word, freq);
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}
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List<String> ans = new LinkedList<>();
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for(int i = buckets.length-1;i>=0;i--){
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if(buckets[i]!=null){
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TreeMap<String, Integer> temp = buckets[i];
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if(temp.size()<k){
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k = k - temp.size();
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while(temp.size()>0){
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ans.add(temp.pollFirstEntry().getKey());
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}
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}
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else {
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while(k>0){
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ans.add(temp.pollFirstEntry().getKey());
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k--;
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}
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break;
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}
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}
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}
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return ans;
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}
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}
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// Quick select
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// 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.
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// Time Complexity: O(n+KlogK), n time for quick select and KlogK time for sort
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// Space Complexity: O(n)
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class Solution {
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public List<String> topKFrequent(String[] words, int k) {
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Map<String, Integer> map = new HashMap<>();
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for(String word:words){
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map.put(word, map.getOrDefault(word, 0)+1);
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}
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Map.Entry<String, Integer>[] entrys = new Map.Entry[map.size()];
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int index = 0;
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for(Map.Entry<String, Integer> e:map.entrySet()){
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entrys[index] = e;
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index++;
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}
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//do quick select
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int start = 0;
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int end = entrys.length-1;
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int mid = 0;
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while(start<=end){
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mid = partition(entrys, start, end);
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if(mid == k-1){
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break;
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}
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else if(mid<k-1){
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start = mid + 1;
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}
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else {
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end = mid - 1;
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}
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}
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List<String> ans = new LinkedList<>();
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List<Map.Entry<String, Integer>> l = new LinkedList<>();
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for(int i = 0;i<=mid;i++){
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l.add(entrys[i]);
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}
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//still need to sort these K words, because we only know they are in result, but not in right order
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Collections.sort(l, new MyComparator());
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for(Map.Entry<String, Integer> e:l){
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ans.add(e.getKey());
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}
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return ans;
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}
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private int partition(Map.Entry<String, Integer>[] entrys, int start, int end){
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int pivot = start;
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int left = start + 1;
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int right = end;
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MyComparator myComparator = new MyComparator();
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while(true){
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while(left<=end){
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if(myComparator.compare(entrys[left], entrys[pivot])<=0){
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left++;
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}
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else {
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break;
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}
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}
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while(right>=start+1){
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if(myComparator.compare(entrys[right], entrys[pivot])>0){
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right--;
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}
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else {
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break;
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}
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}
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if(left>right){
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break;
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}
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swap(entrys, left, right);
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}
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swap(entrys, pivot, right);
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return right;
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}
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private void swap(Map.Entry<String, Integer>[] entrys, int i, int j){
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Map.Entry<String, Integer> a = entrys[i];
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entrys[i] = entrys[j];
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entrys[j] = a;
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}
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}
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class MyComparator implements Comparator<Map.Entry<String, Integer>> {
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public int compare(Map.Entry<String, Integer> e1, Map.Entry<String, Integer> e2){
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String word1 = e1.getKey();
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int freq1 = e1.getValue();
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String word2 = e2.getKey();
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int freq2 = e2.getValue();
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if(freq1!=freq2){
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return freq2-freq1;
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}
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else {
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return word1.compareTo(word2);
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}
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}
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}

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