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 list of string with given prefix using trie.


Input: str = [ “apk”, “app”, “apple”, “arp”, “array” ], pre = “ap”
Output: 3
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 3

Input: str = [ “gee”, “geek”, “geezer”, “geeksforgeeks”, “geekiness”, “geekgod” ], pre = “geek”
Output: 4

To solve this problem Trie Data Structure is used and each node of this Trie contains the following three fields:

  1. children: This field is used for mapping from a character to the next level trie node
  2. isEndOfWord: This field is used to distinguish the node as end of word node
  3. num: This field is used to count the number of times a node is visited during insertion in trie


Below is the implementation of the above approach:





// 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 efiicient
    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
            // corressponding 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++) {
    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;
        // 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;


Time Complexity: O(n*l) where n = number of words inserted in Trie and l = length of longest word inserted in Trie.

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