Count of strings whose prefix match with the given string to a given length k

Given an array of strings arr[] and given some queries where each query consists of a string str and an integer k. The task is to find the count of strings in arr[] whose prefix of length k matches with the k length prefix of str.

Examples:

Input: arr[] = {“abba”, “abbb”, “abbc”, “abbd”, “abaa”, “abca”}, str = “abbg”, k = 3
Output: 4
“abba”, “abbb”, “abbc” and “abbd” are the matching strings.

Input: arr[] = {“geeks”, “geeksforgeeks”, “forgeeks”}, str = “geeks”, k = 2
Output: 2



Prerequisite: Trie | (Insert and Search)

Approach: We will form a trie and insert all the strings in the trie and we will create another variable (frequency) for each node which will store the frequency of prefix of the given strings. Now to get the count of strings whose prefix matches with the given string to a given length k we will have to traverse the trie to the length k from the root, the frequency of the Node will give the count of such strings.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Trie node (considering only lowercase alphabets)
struct Node {
    Node* arr[26];
    int freq;
};
  
// Function to insert a node in the trie
Node* insert(string s, Node* root)
{
    int in;
    Node* cur = root;
    for (int i = 0; i < s.length(); i++) {
        in = s[i] - 'a';
  
        // If there is no node created then create one
        if (cur->arr[in] == NULL)
            cur->arr[in] = new Node();
  
        // Increase the frequency of the node
        cur->arr[in]->freq++;
  
        // Move to the next node
        cur = cur->arr[in];
    }
  
    // Return the updated root
    return root;
}
  
// Function to return the count of strings
// whose prefix of length k matches with the
// k length prefix of the given string
int find(string s, int k, Node* root)
{
    int in, count = 0;
    Node* cur = root;
  
    // Traverse the string
    for (int i = 0; i < s.length(); i++) {
        in = s[i] - 'a';
  
        // If there is no node then return 0
        if (cur->arr[in] == NULL)
            return 0;
  
        // Else traverse to the required node
        cur = cur->arr[in];
  
        count++;
  
        // Return the required count
        if (count == k)
            return cur->freq;
    }
    return 0;
}
  
// Driver code
int main()
{
    string arr[] = { "abba", "abbb", "abbc", "abbd", "abaa", "abca" };
    int n = sizeof(arr) / sizeof(string);
  
    Node* root = new Node();
  
    // Insert the strings in the trie
    for (int i = 0; i < n; i++)
        root = insert(arr[i], root);
  
    // Query 1
    cout << find("abbg", 3, root) << endl;
  
    // Query 2
    cout << find("abg", 2, root) << endl;
  
    // Query 3
    cout << find("xyz", 2, root) << endl;
  
    return 0;
}

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Output:

4
6
0


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Second year Department of Information Technology Jadavpur University

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