Approach Name: Prefix Match Count
Steps:
- Initialize a variable count to 0 to keep track of the number of strings in arr[] that match the prefix of length k in str.
- Loop through all the strings in arr[].
- Check if the prefix of length k in the current string matches the prefix of length k in str.
- If it does, increment the count variable.
- Return the count variable as the result of the query.
C++
#include <iostream> #include <string> #include <vector> using namespace std; // Function to count the number of strings in the vector 'arr' that have a matching prefix of length 'k' with the input string 'str' int PrefixMatchCount(vector<string>& arr, string str, int k) { int count = 0; // Initialize a variable to store the count of matching strings for ( const string& s : arr) { // Iterate through each string 's' in the vector 'arr' using a range-based for loop if (s.substr(0, k) == str.substr(0, k)) { // Check if the first 'k' characters of 's' and 'str' are equal count++; // If there is a match, increment the count variable } } return count; // Return the final count of matching strings } int main() { vector<string> arr = { "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" }; // Initialize a vector 'arr' containing some strings string str = "abbg" ; // Initialize the input string 'str' int k = 3; // Initialize the prefix length 'k' cout << PrefixMatchCount(arr, str, k) << endl; // Call the PrefixMatchCount function and print the result return 0; // Exit the program with status 0 } |
Java
import java.io.*; class GFG { public static int prefixMatchCount(String[] arr, String str, int k) { int count = 0 ; for (String s : arr) { if (s.substring( 0 , k).equals( str.substring( 0 , k))) { count++; } } return count; } // Driver Code public static void main(String[] args) { String[] arr = { "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" }; String str = "abbg" ; int k = 3 ; System.out.println(prefixMatchCount(arr, str, k)); } } |
Python3
def prefixMatchCount(arr, str , k): count = 0 for s in arr: if s[:k] = = str [:k]: count + = 1 return count arr = [ "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" ] str = "abbg" k = 3 print (prefixMatchCount(arr, str , k)) |
C#
using System; class Program { static int PrefixMatchCount( string [] arr, string str, int k) { int count = 0; foreach ( string s in arr) { if (s.Substring(0, k) == str.Substring(0, k)) { count++; } } return count; } static void Main( string [] args) { string [] arr = { "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" }; string str = "abbg" ; int k = 3; Console.WriteLine(PrefixMatchCount(arr, str, k)); } } |
Javascript
// Function to count the number of strings in the 'arr' that have a matching prefix of length 'k' with the input string 'str' function prefixMatchCount(arr, str, k) { let count = 0; for (let s of arr) { if (s.substring(0, k) === str.substring(0, k)) { count++; } } return count; } // Driver code let arr = [ "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" ]; let str = "abbg" ; let k = 3; // Calling the function and printing the result console.log(prefixMatchCount(arr, str, k)); |
4
Time Complexity: O(N*L), where N is the number of strings in arr[] and L is the maximum length of a string in arr[] and str.
Auxiliary Space: O(1)
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”, “w3wiki”, “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 prefixes 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> |
4 6 0
Time Complexity: O(N*M), where N is the size of the array and M is the maximum length of string present in that array.
Auxiliary Space: O(N*M)
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 )) |
C#
using System; class GFG { static void Main( string [] args) { string [] arr = { "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" }; string str = "abbg" ; int n = arr.Length; int k = 3; Solution obj = new Solution(); int ans = obj.Klengthpref(arr, n, k, str); Console.WriteLine(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 } } // This code is contributed by Aman Kumar. |
Javascript
<script> class Solution { klengthpref(arr, n, k, str) { // storing k-length substring of str let a = str.substr(0, k); // counter to count the matching condition let count = 0; // looping through each word of arr for (let i = 0; i < n; i++) { // if the length of string from arr is // less than k // then there is no point in // finding the substring so we // skip if (arr[i].length < k) { continue ; } // storing k-length substring of each // string from arr let b = arr[i].substr(0, k); // checking equality of the two // substring a and b if (a == b) { // if condition matches increase // the counter count++; } } // finally return the count return count; } } let arr = [ "abba" , "abbb" , "abbc" , "abbd" , "abaa" , "abca" ]; let str = "abbg" ; let n = arr.length; let k = 3; let ob = new Solution(); document.write(ob.klengthpref(arr, n, k, str)); // This code is contributed by Utkarsh Kumar. </script> |
4
Time Complexity: O(N*M), where N is the size of the array and M is the maximum length of string present in that array.
Auxiliary Space: O(M)
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