Classify strings from an array using Custom Hash Function
Last Updated : 24 Jan, 2023
Given an array of strings arr[] consisting of N strings, the task is to categorize the strings according to the hash value obtained by adding ASCII values % 26 of the characters in the string.
Examples:
Input: arr[][] = {"geeks", "for", "geeks"}
Output:
geeks geeks
for
Explanation:
The hash value of string "for" is (102 + 111 + 114) % 26 = 14
The hash value of string "geeks" is (103 + 101 + 101 + 107 + 115) % 26 = 7
Therefore, two "geeks" are grouped together and "for" is kept in another group.
Input: arr[][] = {"adf", "aahe", "bce", "bgdb"}
Output:
aahe bgdb
bce
adf
Explanation:
The hash value of string "adf" is (97 + 100 + 102)%26 = 13
The hash value of string "aahe" is (97 + 97 + 104 + 101)%26 = 9
The hash value of string "bce" is (98 + 99 + 101)%26 = 12
The hash value of string "bgdb" is (98 + 103 + 100 + 98)%26 = 9
Therefore, strings "aahe" and "bgdb" have same hashed value, so they are grouped together.
Approach: The idea is to use Map Data Structure for grouping together the strings with the same hash values. Follow the steps below to solve the problem:
- Initialize a Map, say mp, to map hash values with respective strings in a vector.
- Traverse the given array of strings and perform the following steps:
- Calculate the hash value of the current string according to the given function.
- Push the string into the vector with the calculated hash values of the string as key in the map mp.
- Finally, traverse the map mp and print all the strings of respective keys.
Below is the implementation of the above approach:
C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the hash // value of the string s int stringPower(string s) { // Stores hash value of string int power = 0; int n = s.length(); // Iterate over characters of the string for (int i = 0; i < n; i++) { // Calculate hash value power = (power + (s[i])) % 26; } // Return the hash value return power; } // Function to classify the strings // according to the given condition void categorisation_Of_strings( vector<string> s, int N) { // Maps strings with their strings // respective hash values map<int, vector<string> > mp; // Traverse the array of strings for (int i = 0; i < N; i++) { // Find the hash value of the // of the current string int temp = stringPower(s[i]); // Push the current string in // value vector of temp key mp[temp].push_back(s[i]); } // Traverse over the map mp for (auto power : mp) { // Print the result for (auto str : power.second) { cout << str << " "; } cout << endl; } } // Driver Code int main() { vector<string> arr{ "adf", "aahe", "bce", "bgdb" }; int N = arr.size(); categorisation_Of_strings(arr, N); return 0; } // Java program for the above approach import java.util.Arrays; import java.util.HashMap; import java.util.Map; import java.util.Vector; class GFG{ // Function to find the hash // value of the string s static int stringPower(String s) { // Stores hash value of string int power = 0; int n = s.length(); char C[] = s.toCharArray(); // Iterate over characters of the string for(int i = 0; i < n; i++) { // Calculate hash value power = (power + (C[i])) % 26; } // Return the hash value return power; } // Function to classify the strings // according to the given condition static void categorisation_Of_strings(Vector<String> s, int N) { // Maps strings with their strings // respective hash values Map<Integer, Vector<String> > mp = new HashMap<>(); // Traverse the array of strings for(int i = 0; i < N; i++) { // Find the hash value of the // of the current string int temp = stringPower(s.get(i)); // Push the current string in // value vector of temp key if (mp.containsKey(temp)) { mp.get(temp).add(s.get(i)); } else { mp.put(temp, new Vector<String>()); mp.get(temp).add(s.get(i)); } } // Traverse over the map mp for(Map.Entry<Integer, Vector<String>> entry : mp.entrySet()) { // Print the result for(String str : entry.getValue()) { System.out.print(str + " "); } System.out.println(); } } // Driver code public static void main(String[] args) { String[] Sarr = { "adf", "aahe", "bce", "bgdb" }; Vector<String> arr = new Vector<String>( Arrays.asList(Sarr)); int N = arr.size(); categorisation_Of_strings(arr, N); } } // This code is contributed by abhinavjain194 # Python3 program for the above approach # Function to find the hash # value of the string s def stringPower(s): # Stores hash value of string power = 0 n = len(s) # Iterate over characters of the string for i in range(n): # Calculate hash value power = (power + ord(s[i])) % 26 # Return the hash value return power # Function to classify the strings # according to the given condition def categorisation_Of_strings(s, N): # Maps strings with their strings # respective hash values mp = {} # Traverse the array of strings for i in range(N): # Find the hash value of the # of the current string temp = stringPower(s[i]) # Push the current string in # value vector of temp key if temp in mp: mp[temp].append(s[i]) else: mp[temp] = [] mp[temp].append(s[i]) # Traverse over the map mp for i in sorted (mp) : # Print the result for str in mp[i]: print(str,end = " ") print("\n",end = "") # Driver Code if __name__ == '__main__': arr = ["adf", "aahe", "bce", "bgdb"] N = len(arr) categorisation_Of_strings(arr, N) # This code is contributed by ipg2016107. <script> // Javascript program for the above approach // Function to find the hash // value of the string s function stringPower(s) { // Stores hash value of string var power = 0; var n = s.length; // Iterate over characters of the string for(var i = 0; i < n; i++) { // Calculate hash value power = (power + (s[i].charCodeAt(0))) % 26; } // Return the hash value return power; } // Function to classify the strings // according to the given condition function categorisation_Of_strings(s, N) { // Maps strings with their strings // respective hash values var mp = new Map(); // Traverse the array of strings for(var i = 0; i < N; i++) { // Find the hash value of the // of the current string var temp = stringPower(s[i]); // Push the current string in // value vector of temp key if (!mp.has(temp)) { mp.set(temp, new Array()); } var tmp = mp.get(temp); tmp.push(s[i]); mp.set(temp, tmp); } var tmp = Array(); // Traverse over the map mp mp.forEach((value, key) => { tmp.push(key); }); tmp.sort((a, b) => a - b); // Traverse over the map mp tmp.forEach((value) => { // Print the result mp.get(value).forEach(element => { document.write(element + " "); }); document.write("<br>"); }); } // Driver Code var arr = [ "adf", "aahe", "bce", "bgdb" ]; var N = arr.length; categorisation_Of_strings(arr, N); // This code is contributed by rutvik_56 </script> using System; using System.Collections.Generic; using System.Linq; namespace ConsoleApp1 { class Program { // Function to find the hash // value of the string s static int StringPower(string s) { // Stores hash value of string int power = 0; int n = s.Length; char[] C = s.ToCharArray(); // Iterate over characters of the string for (int i = 0; i < n; i++) { // Calculate hash value power = (power + (C[i])) % 26; } // Return the hash value return power; } // Function to classify the strings // according to the given condition static void CategorisationOfStrings(List<string> s, int N) { // Maps strings with their strings // respective hash values SortedDictionary<int, List<string>> mp = new SortedDictionary<int, List<string>>(); // Traverse the array of strings for (int i = 0; i < N; i++) { // Find the hash value of the // of the current string int temp = StringPower(s[i]); // Push the current string in // value vector of temp key if (mp.ContainsKey(temp)) { mp[temp].Add(s[i]); } else { mp[temp] = new List<string>(); mp[temp].Add(s[i]); } } // Traverse over the map mp foreach (var entry in mp) { // Print the result foreach (string str in entry.Value) { Console.Write(str + " "); } Console.WriteLine(); } } static void Main(string[] args) { string[] Sarr = { "adf", "aahe", "bce", "bgdb" }; List<string> arr = new List<string>(Sarr); int N = arr.Count; CategorisationOfStrings(arr, N); } } } Time Complexity: O(N*M), where M is the length of the longest string.
Auxiliary Space: O(N*M)