Count the number of words with given prefix using Trie
Prerequisite: Trie
Given a list of string str[] and a prefix string pre. The task is to count the number of words in the list of string with a given prefix using trie.
Examples:
Input: str = [ “apk”, “app”, “apple”, “arp”, “array” ], pre = “ap”
Output: 3
Explanation:
Below is the representation of trie from using above string.
The words in str having prefix “ap” are apk, app and apple.
So, the count is 3Input: str = [ “gee”, “geek”, “geezer”, “w3wiki”, “geekiness”, “geekgod” ], pre = “geek”
Output: 4
Approach:
To solve this problem Trie Data Structure is used and each node of this Trie contains the following three fields:
- children: This field is used for mapping from a character to the next level trie node
- isEndOfWord: This field is used to distinguish the node as the end of the word node
- num: This field is used to count the number of times a node is visited during insertion in trie
Steps:
- Insert a list of string in trie such that every string in the list is inserted as an individual trie node.
- During inserting update all the fields in every node of the trie
- For a given prefix, traverse the trie till we reach the last character of the given prefix pre.
- The value of the num field in the last node of string pre is the count of prefix in the given list of string.
Below is the implementation of the above approach:
C++14
// C++ implementation of counting the // number of words in a trie with a // given prefix #include "bits/stdc++.h" using namespace std; // Trie Node struct TrieNode { // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient map< char , TrieNode*> children; // If isEndOfWord is true, then // node represents end of word bool isEndOfWord; // num represents number of times // a character has appeared during // insertion of the words in the // trie map< char , int > num; }; // Declare root node struct TrieNode* root; // Function to create New Trie Node struct TrieNode* getNewTrieNode() { struct TrieNode* pNode = new TrieNode; pNode->isEndOfWord = false ; return pNode; } // Function to insert a string in trie void insertWord(string word) { // To hold the value of root struct TrieNode* current = root; // To hold letters of the word char s; // Traverse through strings in list for ( int i = 0; i < word.length(); i++) { s = word[i]; // If s is not present in the // character field of current node if (current->children.find(s) == current->children.end()) { // Get new node struct TrieNode* p = getNewTrieNode(); // Insert s in character // field of current node // with reference to node p (current->children)[s] = p; // Insert s in num field // of current node with // value 1 (current->num)[s] = 1; } else { // Increment the count // corresponding to the // character s current->num[s] = (current->num)[s] + 1; } // Go to next node current = (current->children)[s]; } current->isEndOfWord = true ; } // Function to count the number of // words in trie with given prefix int countWords(vector<string>& words, string prefix) { root = getNewTrieNode(); // Size of list of string int n = words.size(); // Construct trie containing // all the words for ( int i = 0; i < n; i++) { insertWord(words[i]); } struct TrieNode* current = root; char s; // Initialize the wordCount = 0 int wordCount = 0; for ( int i = 0; prefix[i]; i++) { s = prefix[i]; // If the complete prefix is // not present in the trie if (current->children.find(s) == current->children.end()) { // Make wordCount 0 and // break out of loop wordCount = 0; break ; } // Update the wordCount wordCount = (current->num)[s]; // Go to next node current = (current->children)[s]; } return wordCount; } // Driver Code int main() { // input list of words vector<string> words; words = { "apk" , "app" , "apple" , "arp" , "array" }; // Given prefix to find string prefix = "ap" ; // Print the number of words with // given prefix cout << countWords(words, prefix); return 0; } |
Java
// Java implementation of counting the // number of words in a trie with a // given prefix import java.util.HashMap; import java.util.Map; import java.util.Vector; // Trie Node class TrieNode { // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient Map<Character, TrieNode> children; // If isEndOfWord is true, then // node represents end of word boolean isEndOfWord; // num represents number of times // a character has appeared during // insertion of the words in the // trie Map<Character, Integer> num; public TrieNode() { this .children = new HashMap<>(); this .isEndOfWord = false ; this .num = new HashMap<>(); } } public class Main { // Declare root node static TrieNode root; // Function to create New Trie Node static TrieNode getNewTrieNode() { TrieNode pNode = new TrieNode(); pNode.isEndOfWord = false ; return pNode; } // Function to insert a string in trie static void insertWord(String word) { // To hold the value of root TrieNode current = root; // To hold letters of the word char s; // Traverse through strings in list for ( int i = 0 ; i < word.length(); i++) { s = word.charAt(i); // If s is not present in the // character field of current node if (!current.children.containsKey(s)) { // Get new node TrieNode p = getNewTrieNode(); // Insert s in character // field of current node // with reference to node p current.children.put(s, p); // Insert s in num field // of current node with // value 1 current.num.put(s, 1 ); } else { // Increment the count // corresponding to the // character s current.num.put(s, current.num.get(s) + 1 ); } // Go to next node current = current.children.get(s); } current.isEndOfWord = true ; } // Function to count the number of // words in trie with given prefix static int countWords(Vector<String> words, String prefix) { root = getNewTrieNode(); // Size of list of string int n = words.size(); // Construct trie containing // all the words for ( int i = 0 ; i < n; i++) { insertWord(words.get(i)); } TrieNode current = root; char s; // Initialize the wordCount = 0 int wordCount = 0 ; for ( int i = 0 ; i < prefix.length(); i++) { s = prefix.charAt(i); // If the complete prefix is // not present in the trie if (!current.children.containsKey(s)) { // Make wordCount 0 and // break out of loop wordCount = 0 ; break ; } // Update the wordCount wordCount = current.num.get(s); // Go to next node current = current.children.get(s); } return wordCount; } // Driver Code public static void main(String[] args) { // input list of words Vector<String> words = new Vector<>(); words.add( "apk" ); words.add( "app" ); words.add( "apple" ); words.add( "arp" ); words.add( "array" ); // Given prefix to find String prefix = "ap" ; // Print the number of words with // given prefix System.out.println(countWords(words, prefix)); } } // This code is contributed by Aman Kumar. |
Python3
# Python3 implementation of counting the # number of words in a trie with a # given prefix # Trie Node class TrieNode: def __init__( self ): # Using map to store the pointers # of children nodes for dynamic # implementation, for making the # program space efficient self .children = dict () # If isEndOfWord is true, then # node represents end of word self .isEndOfWord = False # num represents number of times # a character has appeared during # insertion of the words in the # trie self .num = dict () # Declare root node root = None # Function to create New Trie Node def getNewTrieNode(): pNode = TrieNode() return pNode # Function to insert a string in trie def insertWord(word): global root # To hold the value of root current = root # To hold letters of the word s = '' # Traverse through strings in list for i in range ( len (word)): s = word[i] # If s is not present in the # character field of current node if (s not in current.children): # Get new node p = getNewTrieNode() # Insert s in character # field of current node # with reference to node p current.children[s] = p # Insert s in num field # of current node with # value 1 current.num[s] = 1 else : # Increment the count # corresponding to the # character s current.num[s] = current.num[s] + 1 # Go to next node current = current.children[s] current.isEndOfWord = True # Function to count the number of # words in trie with given prefix def countWords(words, prefix): global root root = getNewTrieNode() # Size of list of string n = len (words) # Construct trie containing # all the words for i in range (n): insertWord(words[i]) current = root s = '' # Initialize the wordCount = 0 wordCount = 0 for i in range ( len (prefix)): s = prefix[i] # If the complete prefix is # not present in the trie if (s not in current.children): # Make wordCount 0 and # break out of loop wordCount = 0 break # Update the wordCount wordCount = (current.num)[s] # Go to next node current = (current.children)[s] return wordCount # Driver Code if __name__ = = '__main__' : # input list of words words = [ "apk" , "app" , "apple" , "arp" , "array" ] # Given prefix to find prefix = "ap" # Print the number of words with # given prefix print (countWords(words, prefix)) # This code is contributed by rutvik_56 |
C#
using System; using System.Collections.Generic; // Define the TrieNode class to represent a node in the Trie data structure. public class TrieNode { public Dictionary< char , TrieNode> Children; // Dictionary to store child nodes. public bool IsEndOfWord; // Indicates whether this node represents the end of a word. public Dictionary< char , int > Num; // Dictionary to store the count of words that start with a particular character. // Constructor to initialize the TrieNode object. public TrieNode() { Children = new Dictionary< char , TrieNode>(); IsEndOfWord = false ; Num = new Dictionary< char , int >(); } } // Define the Trie class to represent the Trie data structure. public class Trie { public TrieNode Root; // The root node of the Trie. // Constructor to initialize the Trie object. public Trie() { Root = new TrieNode(); } // Method to insert a word into the Trie. public void InsertWord( string word) { TrieNode current = Root; foreach ( char c in word) { if (!current.Children.ContainsKey(c)) // If the character is not present in the child nodes, add it. { current.Children = new TrieNode(); current.Num = 1; // Since this is the first occurrence of the character, set its count to 1. } else // If the character is already present in the child nodes, increment its count. { current.Num++; } current = current.Children; // Move to the next node. } current.IsEndOfWord = true ; // Set the IsEndOfWord flag to true for the last node in the word. } // Method to count the number of words in the Trie that start with a given prefix. public int CountWords(List< string > words, string prefix) { Trie trie = new Trie(); // Create a new Trie object. foreach ( string word in words) // Insert all the words in the input list into the Trie. { trie.InsertWord(word); } TrieNode current = trie.Root; // Start at the root node of the Trie. int wordCount = 0; foreach ( char c in prefix) // Traverse the Trie using the characters in the prefix. { if (!current.Children.ContainsKey(c)) // If the character is not present in the child nodes, there are no words in the Trie that start with the given prefix. { wordCount = 0; break ; } wordCount = current.Num; // Get the count of words that start with the given prefix. current = current.Children; // Move to the next node. } return wordCount; // Return the count of words that start with the given prefix. } } // Define the Program class to test the Trie data structure. public class Program { public static void Main() { List< string > words = new List< string > { "apk" , "app" , "apple" , "arp" , "array" }; string prefix = "ap" ; Trie trie = new Trie(); int count = trie.CountWords(words, prefix); // Count the number of words that start with the given prefix. Console.WriteLine(count); // Print the count. } } // This code is contributed by rudra1807raj |
Javascript
// JavaScript implementation of counting the // number of words in a trie with a // given prefix // Trie Node class TrieNode { constructor() { // Using map to store the pointers // of children nodes for dynamic // implementation, for making the // program space efficient this .children = new Map(); // If isEndOfWord is true, then // node represents end of word this .isEndOfWord = false ; // num represents number of times // a character has appeared during // insertion of the words in the // trie this .num = new Map(); } } // Declare root node let root = null ; // Function to create New Trie Node function getNewTrieNode() { let pNode = new TrieNode(); return pNode; } // Function to insert a string in trie function insertWord(word) { // To hold the value of root let current = root; // To hold letters of the word let s = '' ; // Traverse through strings in list for (let i = 0; i < word.length; i++) { s = word[i]; // If s is not present in the // character field of current node if (!current.children.has(s)) { // Get new node let p = getNewTrieNode(); // Insert s in character // field of current node // with reference to node p current.children.set(s, p); // Insert s in num field // of current node with // value 1 current.num.set(s, 1); } else { // Increment the count // corresponding to the // character s current.num.set(s, current.num.get(s) + 1); } // Go to next node current = current.children.get(s); } current.isEndOfWord = true ; } // Function to count the number of // words in trie with given prefix function countWords(words, prefix) { root = getNewTrieNode(); // Size of list of string let n = words.length; // Construct trie containing // all the words for (let i = 0; i < n; i++) { insertWord(words[i]); } let current = root; let s = '' ; // Initialize the wordCount = 0 let wordCount = 0; for (let i = 0; i < prefix.length; i++) { s = prefix[i]; // If the complete prefix is // not present in the trie if (!current.children.has(s)) { // Make wordCount 0 and // break out of loop wordCount = 0; break ; } // Update the wordCount wordCount = current.num.get(s); // Go to next node current = current.children.get(s); } return wordCount; } // Driver Code let words = [ "apk" , "app" , "apple" , "arp" , "array" ]; let prefix = "ap" ; console.log(countWords(words, prefix)); // Output: 3 |
3
Time Complexity: O(n*l) where n = the number of words inserted in Trie and l = the length of the longest word inserted in Trie.
Auxiliary Space: O(n*l).