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

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  • Difficulty Level : Easy
  • Last Updated : 23 May, 2022

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:
“abba”, “abbb”, “abbc” and “abbd” are the matching strings.
Input: arr[] = {“geeks”, “geeksforgeeks”, “forgeeks”}, str = “geeks”, k = 2 
Output:
 

 

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: 
 

C++




// 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;
}

Java




// Java implementation of the approach
class GFG
{
 
    // Trie node (considering only lowercase alphabets)
    static class Node
    {
        Node[] arr = new Node[26];
        int freq;
    };
 
    // Function to insert a node in the trie
    static Node insert(String s, Node root)
    {
        int in;
        Node cur = root;
        for (int i = 0; i < s.length(); i++)
        {
            in = s.charAt(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
    static 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.charAt(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
    public static void main(String[] args)
    {
        String arr[] = { "abba", "abbb", "abbc",
                        "abbd", "abaa", "abca" };
        int n = arr.length;
 
        Node root = new Node();
 
        // Insert the Strings in the trie
        for (int i = 0; i < n; i++)
            root = insert(arr[i], root);
 
        // Query 1
        System.out.print(find("abbg", 3, root) + "\n");
 
        // Query 2
        System.out.print(find("abg", 2, root) + "\n");
 
        // Query 3
        System.out.print(find("xyz", 2, root) + "\n");
 
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the approach
 
# Trie node (considering only lowercase alphabets)
class Node :
    def __init__(self):
        self.arr = [None]*26
        self.freq = 0
     
class Trie:
     
    # Trie data structure class
    def __init__(self):
        self.root = self.getNode()
 
    def getNode(self):
     
        # Returns new trie node (initialized to NULLs)
        return Node()
 
    # Function to insert a node in the trie
    def insert(self, s):
         
        _in = 0
        cur = self.root
        for i in range(len(s)):
            _in = ord(s[i]) - ord('a')
 
            # If there is no node created then create one
            if not cur.arr[_in]:
                cur.arr[_in] = self.getNode()
 
            # Increase the frequency of the node
            cur.arr[_in].freq += 1
 
            # Move to the next node
            cur = cur.arr[_in]
 
    # Function to return the count of strings
    # whose prefix of length k matches with the
    # k length prefix of the given string
    def find(self, s, k):
         
        _in = 0
        count = 0
        cur = self.root
 
        # Traverse the string
        for i in range(len(s)):
            _in = ord(s[i]) - ord('a')
 
            # If there is no node then return 0
            if cur.arr[_in] == None:
                return 0
 
            # Else traverse to the required node
            cur = cur.arr[_in]
 
            count += 1
 
            # Return the required count
            if count == k:
                return cur.freq
        return 0
 
# Driver code
def main():
     
    arr = [ "abba", "abbb", "abbc", "abbd", "abaa", "abca" ]
    n = len(arr)
 
    root = Trie();
 
    # Insert the strings in the trie
    for i in range(n):
        root.insert(arr[i])
 
    # Query 1
    print(root.find("abbg", 3))
 
    # Query 2
    print(root.find("abg", 2))
 
    # Query 3
    print(root.find("xyz", 2))
 
if __name__ == '__main__':
    main()
     
# This code is contributed by divyamohan123
    

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Trie node (considering only lowercase alphabets)
    public class Node
    {
        public Node[] arr = new Node[26];
        public int freq;
    };
 
    // Function to insert a node in the trie
    static 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
    static 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
    public static void Main(String[] args)
    {
        String []arr = { "abba", "abbb", "abbc",
                        "abbd", "abaa", "abca" };
        int n = arr.Length;
 
        Node root = new Node();
 
        // Insert the Strings in the trie
        for (int i = 0; i < n; i++)
            root = insert(arr[i], root);
 
        // Query 1
        Console.Write(find("abbg", 3, root) + "\n");
 
        // Query 2
        Console.Write(find("abg", 2, root) + "\n");
 
        // Query 3
        Console.Write(find("xyz", 2, root) + "\n");
 
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Trie node (considering only lowercase alphabets)
class Node
{
    constructor()
    {
        this.arr=new Array(26);
        this.freq=0;
    }
}
 
// Function to insert a node in the trie
function insert(s,root)
{
    let In;
        let cur = root;
        for (let i = 0; i < s.length; i++)
        {
            In = s[i].charCodeAt(0) - 'a'.charCodeAt(0);
   
            // 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
function find(s,k,root)
{
    let In, count = 0;
        let cur = root;
   
        // Traverse the String
        for (let i = 0; i < s.length; i++)
        {
            In = s[i].charCodeAt(0) - 'a'.charCodeAt(0);
   
            // 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
let arr=["abba", "abbb", "abbc",
                        "abbd", "abaa", "abca"];
let n = arr.length;
 
let root = new Node();
 
// Insert the Strings in the trie
for (let i = 0; i < n; i++)
    root = insert(arr[i], root);
 
// Query 1
document.write(find("abbg", 3, root) + "<br>");
 
// Query 2
document.write(find("abg", 2, root) + "<br>");
 
// Query 3
document.write(find("xyz", 2, root) + "<br>");
 
 
// This code is contributed by rag2127
 
</script>

Output

4
6
0

Approach without using Trie:

The idea is to use substring of length k. Since we are dealing with prefixes we need to consider the substring of length k starting from index 0. This substring will always be the prefix of the string. Store the k-length substring of str from index 0 in a string a. Run a loop for each word in the string array and for each word which is of length greater than k take the substring of length k starting from index 0. Now check for equality of the two substrings a and b. If the two are equal increase the count. Finally after coming out of the loop return the count.

C++




#include <bits/stdc++.h>
using namespace std;
 
class Solution {
public:
    int klengthpref(string arr[], int n, int k, string str)
    {
 
        string a = str.substr(
            0, k); // storing k-length substring of str
        int count
            = 0; // counter to count the matching condition
 
        // looping through each word of arr
        for (int i = 0; i < n; i++) {
            if (arr[i].length()
                < k) // if the length of string from arr is
                     // less than k
                continue; // then there is no point in
                          // finding the substring so we
                          // skip
            string b = arr[i].substr(
                0, k); // storing k-length substring of each
                       // string from arr
            if (a == b) // checking equality of the two
                        // substring a and b
                count++; // if condition matches increase
                         // the counter
        }
 
        return count; // finally return the count
    }
};
 
int main()
{
    string arr[] = { "abba", "abbb", "abbc",
                     "abbd", "abaa", "abca" };
    string str = "abbg";
    int n = sizeof(arr) / sizeof(string), k = 3;
 
    Solution ob;
    cout << ob.klengthpref(arr, n, k, str);
 
    return 0;
}

Java




class GFG {
    public static void main(String args[])
    {
        String[] arr = { "abba", "abbb", "abbc",
                         "abbd", "abaa", "abca" };
        String str = "abbg";
        int n = arr.length, k = 3;
        Solution obj = new Solution();
        int ans = obj.klengthpref(arr, n, k, str);
        System.out.println(ans);
    }
}
 
class Solution {
    public int klengthpref(String[] arr, int n, int k,
                           String str)
    {
        String a = str.substring(
            0, k); // storing k-length substring of str
        int count
            = 0; // counter to count the matching condition
 
        // looping through each word of arr
        for (int i = 0; i < n; i++) {
            if (arr[i].length()
                < k) // if the length of string from arr is
                     // less than k
                continue; // then there is no point in
                          // finding the substring so we
                          // skip
 
            String b = arr[i].substring(
                0, k); // storing k-length substring of each
                       // string from arr
            if (a.equals(b)) // checking equality of the two
                             // substring a and b
                count++; // if condition matches increase
                         // the counter
        }
 
        return count; // finally return the count
    }
}

Python3




class Solution:
    def klengthpref(self, arr, n, k, s):
        a = s[:k]  # storing k-length substring of str
        count = 0  # counter to count the matching condition
 
        # looping through each word of arr
        for i in range(n):
            if(len(arr[i]) < k):  # if the length of string from arr is less than k
                continue  # then there is no point in finding the substring so we skip
            t = arr[i]
            b = t[:k]  # storing k-length substring of each string from arr
            if(a == b):  # checking equality of the two substring a and b
                count += 1  # if condition matches increase the counter by 1
        return count  # finally return the count
 
 
arr = ["abba", "abbb", "abbc", "abbd", "abaa", "abca"]
str = "abbg"
n = len(arr)
k = 3
 
obj = Solution()
print(obj.klengthpref(arr, n, k, str))

Output

4

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