A file is given to us. which contains words. Our objective is to find k word which occurred most in the file.
There are multiple ways to solve this.
Simplest Way(HashMap): Get all the word and store in the hashmap as key and value would frequency of the word, if word is already present in the HashMap then increment the count.Once processing of the file is done then traverse through the hash map and return the k words with maximum counts.
Trie and Min Heap : Another approach is, to create a Trie to store the word and maintain a Min Heap of size k. Trie is to search the String but can also stores count of occurrences of words.
Trie and Min Heap are linked with each other by storing an additional field in Trie ‘indexMinHeap’ and a pointer "MyTrieNode" in Min Heap.
Initially indexMinHeap is -1, means word is not present in the Min Heap Otherwise it shows the index in Min Heap.
How to fill Min Heap:
TrieNode will look like:
class MyTrieNode {
char data;
boolean is_end_of_string;
Map<Character, MyTrieNode> nodes;
int frequency = 0;
int minHeapIndex = -1;
public MyTrieNode(char data) {
this.data = data;
is_end_of_string = false;
nodes = new HashMap<Character, MyTrieNode>();
}
public MyTrieNode children(char data) {
return nodes.get(data);
}
public boolean isChildExist(char c) {
return children(c) != null;
}
}
Min Heap Class Code
class MinHeap {
int size;
int capacity;
MyNode[] nodes;
}
class MyNode {
String word;
int frequency;
//This is extra pointer to point to Trie
MyTrieNode node;
}
TRIE Class:
class MyTrie {
MyTrieNode root;
MinHeap minHeap;
public MyTrie(int frequency) {
root = new MyTrieNode(' ');
this.minHeap = new MinHeap();
this.minHeap.nodes = new MyNode[frequency];
this.minHeap.capacity = frequency;
}
/*
* This method to insert the node into the TRIE
* */
public void insert(String s) {
if (s == null || s.trim().length() == 0) {
return;
}
MyTrieNode current = root;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!current.isChildExist(c)) {
MyTrieNode node = new MyTrieNode(c);
current.nodes.put(c, node);
}
current = current.children(c);
}
if (current.is_end_of_string) {
current.frequency++;
} else {
current.frequency = 1;
current.is_end_of_string = true;
}
insertInMinHeap(s, current);
}
/*
* This method is to insert the word into Min Heap using below rule.
* 1. If word is already is present then, increment the count in MinHeap and call heapify method.
* 2. If Min Heap is not full(not contains k element), and word is not present then add the node to Min Heap and update its minHeapIndex, and call heapify.
* 3.If word is not present and Min Heap is full and new word frequency is more than minimum head then replace the top element(index=0) with new word and update the minHeapIndexOf both the words.
* */
private void insertInMinHeap(String s, MyTrieNode current) {
if (current.minHeapIndex != -1) {
this.minHeap.nodes[current.minHeapIndex].frequency++;
minheapify(current.minHeapIndex);
} else if (this.minHeap.size < this.minHeap.capacity) {
++this.minHeap.size;
MyNode node = new MyNode();
node.word = s;
node.frequency = current.frequency;
node.node = current;
node.node.minHeapIndex = this.minHeap.size - 1;
this.minHeap.nodes[this.minHeap.size - 1] = node;
buildHeap();
} else if (current.frequency > this.minHeap.nodes[0].frequency) {
this.minHeap.nodes[0].node.minHeapIndex = -1;
this.minHeap.nodes[0].node = current;
this.minHeap.nodes[0].frequency = current.frequency;
this.minHeap.nodes[0].word = s;
current.minHeapIndex = 0;
minheapify(0);
}
}
private void buildHeap() {
for (int i = (this.minHeap.size - 1) / 2; i >= 0; i--) {
minheapify(i);
}
}
/*
* To search any word is the Trie
* */
public boolean search(String s) {
if (s == null || s.trim().length() == 0) {
return false;
}
MyTrieNode current = root;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!current.isChildExist(c)) {
return false;
}
current = current.children(c);
}
return current.is_end_of_string;
}
/*
* This method is to heapify the given and make sure it satisfies the property of the node
* */
public void minheapify(int node) {
int left = (node << 1) + 1;
int right = (node << 1) + 2;
int smallest = node;
if (left < this.minHeap.size
&& this.minHeap.nodes[smallest].frequency > this.minHeap.nodes[left].frequency) {
smallest = left;
}
if (right < this.minHeap.size
&& this.minHeap.nodes[smallest].frequency > this.minHeap.nodes[right].frequency) {
smallest = right;
}
if (smallest != node) {
int index = this.minHeap.nodes[smallest].node.minHeapIndex;
this.minHeap.nodes[smallest].node.minHeapIndex = this.minHeap.nodes[node].node.minHeapIndex;
this.minHeap.nodes[node].node.minHeapIndex = index;
MyNode temp = this.minHeap.nodes[smallest];
this.minHeap.nodes[smallest] = this.minHeap.nodes[node];
this.minHeap.nodes[node] = temp;
minheapify(smallest);
}
}
/*
* Traverse through Min Heap and show all words and their frequency
* */
public void display() {
for (int i = 0; i < this.minHeap.size; i++) {
System.out.println("word\t:\t" + this.minHeap.nodes[i].word
+ "\t\t\t\tfrequency\t:\t" + this.minHeap.nodes[i].frequency);
}
}
}
Main/Test Class
public class TrieForOccurenceOfString {
public static void main(String[] args) throws IOException {
File file=new File("Path of input file"/input.txt");
BufferedReader reader=new BufferedReader(new FileReader(file));
int k=5;
MyTrie t = new MyTrie(k);
String ss=null;
while((ss=reader.readLine())!=null){
String[] array=ss.split(" ");
for (int i = 0; i < array.length; i++) {
t.insert(array[i]);
}
}
reader.close();
t.display();
}
}
There are multiple ways to solve this.
Simplest Way(HashMap): Get all the word and store in the hashmap as key and value would frequency of the word, if word is already present in the HashMap then increment the count.Once processing of the file is done then traverse through the hash map and return the k words with maximum counts.
Trie and Min Heap : Another approach is, to create a Trie to store the word and maintain a Min Heap of size k. Trie is to search the String but can also stores count of occurrences of words.
Trie and Min Heap are linked with each other by storing an additional field in Trie ‘indexMinHeap’ and a pointer "MyTrieNode" in Min Heap.
Initially indexMinHeap is -1, means word is not present in the Min Heap Otherwise it shows the index in Min Heap.
How to fill Min Heap:
- If word is already is present then, increment the count in MinHeap and call heapify method.
- If Min Heap is not full(not contains k element), and word is not present then add the node to Min Heap and update its minHeapIndex, and call heapify.
- If word is not present and Min Heap is full and new word frequency is more than minimum head then replace the top element(index=0) with new word and update the minHeapIndexOf both the words.
TrieNode will look like:
class MyTrieNode {
char data;
boolean is_end_of_string;
Map<Character, MyTrieNode> nodes;
int frequency = 0;
int minHeapIndex = -1;
public MyTrieNode(char data) {
this.data = data;
is_end_of_string = false;
nodes = new HashMap<Character, MyTrieNode>();
}
public MyTrieNode children(char data) {
return nodes.get(data);
}
public boolean isChildExist(char c) {
return children(c) != null;
}
}
Min Heap Class Code
class MinHeap {
int size;
int capacity;
MyNode[] nodes;
}
class MyNode {
String word;
int frequency;
//This is extra pointer to point to Trie
MyTrieNode node;
}
TRIE Class:
class MyTrie {
MyTrieNode root;
MinHeap minHeap;
public MyTrie(int frequency) {
root = new MyTrieNode(' ');
this.minHeap = new MinHeap();
this.minHeap.nodes = new MyNode[frequency];
this.minHeap.capacity = frequency;
}
/*
* This method to insert the node into the TRIE
* */
public void insert(String s) {
if (s == null || s.trim().length() == 0) {
return;
}
MyTrieNode current = root;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!current.isChildExist(c)) {
MyTrieNode node = new MyTrieNode(c);
current.nodes.put(c, node);
}
current = current.children(c);
}
if (current.is_end_of_string) {
current.frequency++;
} else {
current.frequency = 1;
current.is_end_of_string = true;
}
insertInMinHeap(s, current);
}
/*
* This method is to insert the word into Min Heap using below rule.
* 1. If word is already is present then, increment the count in MinHeap and call heapify method.
* 2. If Min Heap is not full(not contains k element), and word is not present then add the node to Min Heap and update its minHeapIndex, and call heapify.
* 3.If word is not present and Min Heap is full and new word frequency is more than minimum head then replace the top element(index=0) with new word and update the minHeapIndexOf both the words.
* */
private void insertInMinHeap(String s, MyTrieNode current) {
if (current.minHeapIndex != -1) {
this.minHeap.nodes[current.minHeapIndex].frequency++;
minheapify(current.minHeapIndex);
} else if (this.minHeap.size < this.minHeap.capacity) {
++this.minHeap.size;
MyNode node = new MyNode();
node.word = s;
node.frequency = current.frequency;
node.node = current;
node.node.minHeapIndex = this.minHeap.size - 1;
this.minHeap.nodes[this.minHeap.size - 1] = node;
buildHeap();
} else if (current.frequency > this.minHeap.nodes[0].frequency) {
this.minHeap.nodes[0].node.minHeapIndex = -1;
this.minHeap.nodes[0].node = current;
this.minHeap.nodes[0].frequency = current.frequency;
this.minHeap.nodes[0].word = s;
current.minHeapIndex = 0;
minheapify(0);
}
}
private void buildHeap() {
for (int i = (this.minHeap.size - 1) / 2; i >= 0; i--) {
minheapify(i);
}
}
/*
* To search any word is the Trie
* */
public boolean search(String s) {
if (s == null || s.trim().length() == 0) {
return false;
}
MyTrieNode current = root;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (!current.isChildExist(c)) {
return false;
}
current = current.children(c);
}
return current.is_end_of_string;
}
/*
* This method is to heapify the given and make sure it satisfies the property of the node
* */
public void minheapify(int node) {
int left = (node << 1) + 1;
int right = (node << 1) + 2;
int smallest = node;
if (left < this.minHeap.size
&& this.minHeap.nodes[smallest].frequency > this.minHeap.nodes[left].frequency) {
smallest = left;
}
if (right < this.minHeap.size
&& this.minHeap.nodes[smallest].frequency > this.minHeap.nodes[right].frequency) {
smallest = right;
}
if (smallest != node) {
int index = this.minHeap.nodes[smallest].node.minHeapIndex;
this.minHeap.nodes[smallest].node.minHeapIndex = this.minHeap.nodes[node].node.minHeapIndex;
this.minHeap.nodes[node].node.minHeapIndex = index;
MyNode temp = this.minHeap.nodes[smallest];
this.minHeap.nodes[smallest] = this.minHeap.nodes[node];
this.minHeap.nodes[node] = temp;
minheapify(smallest);
}
}
/*
* Traverse through Min Heap and show all words and their frequency
* */
public void display() {
for (int i = 0; i < this.minHeap.size; i++) {
System.out.println("word\t:\t" + this.minHeap.nodes[i].word
+ "\t\t\t\tfrequency\t:\t" + this.minHeap.nodes[i].frequency);
}
}
}
Main/Test Class
public class TrieForOccurenceOfString {
public static void main(String[] args) throws IOException {
File file=new File("Path of input file"/input.txt");
BufferedReader reader=new BufferedReader(new FileReader(file));
int k=5;
MyTrie t = new MyTrie(k);
String ss=null;
while((ss=reader.readLine())!=null){
String[] array=ss.split(" ");
for (int i = 0; i < array.length; i++) {
t.insert(array[i]);
}
}
reader.close();
t.display();
}
}
What is the time complexity of this algorithm ?
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