Given a string S in the form of a sentence, the task is to find the word from the text with maximum number of its anagrams present in the given sentence.
Example:
Input: S = “please be silent and listen to what the professor says”
Output: silent
Explanation:
Only the word “silent” has an anagram(listen) present in the sentence.
Input: S = “cat is not a dog and sword has no words when government creates act so what is tac”
Output: cat
Explanation:
The word “cat” has two anagrams (“act”, “tac”) present in the sentence.
The word “words” has an anagram (“sword”) present in the sentence.
Hence, the word with the maximum anagrams is “cat”.
Approach:
Following observations need to be made to solve the program:
Property of Prime Numbers: The product of any combination of prime numbers generates an unique number which cannot be obtained by any other combination of prime numbers.
Follow the steps below to solve the problem:
- Assign each alphabet with a distinct prime number.
- For each word in the given string, calculate the product of prime numbers assigned to the characters of that word and store it in a HashMap.
- Calculate the product of prime numbers assigned to the characters of a word and store it in a HashMap.
- Find the product with maximum frequency in the HashMap and print one of the corresponding anagrams as the answer.
Below is the implementation of the above approach:
C++14
// C++ Program to find the word // with most anagrams in a sentence #include <bits/stdc++.h> using namespace std; // Function to find the word with // maximum number of anagrams string largestAnagramGrp(vector<string> arr) { // Primes assigned to 26 alphabets int prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101}; int max = -1; long maxpdt = -1; // Stores the product and // word mappings unordered_map< long long , string> W; // Stores the frequencies // of products unordered_map< long long , long > P; for (string temp : arr) { long pdt = 1; // Calculate the product of // primes assigned for ( char t : temp) { pdt *= prime[t - 'a' ]; } // If product already exists if (P.find(pdt) != P.end()) { P[pdt]++; } // Otherwise else { W[pdt] = temp; P[pdt] = 1; } } // Fetch the most frequent product for ( auto e : P) { if (max < e.second) { max = e.second; maxpdt = e.first; } } // Return a string // with that product return W[maxpdt]; } // Driver Code int main() { char S[] = "please be silent and listen to what the professor says " ; vector<string> arr; char *token = strtok (S, " " ); while (token != NULL) { arr.push_back(token); token = strtok (NULL, " " ); } cout << largestAnagramGrp(arr) << endl; return 0; } // This code is contributed by // sanjeev2552 |
Java
// Java Program to find the word // with most anagrams in a sentence import java.util.*; class GFG { // Function to find the word with // maximum number of anagrams private static String largestAnagramGrp( String arr[]) { // Primes assigned to 26 alphabets int prime[] = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 }; int max = - 1 ; long maxpdt = - 1 ; // Stores the product and // word mappings HashMap<Long, String> W = new HashMap<>(); // Stores the frequencies // of products HashMap<Long, Integer> P = new HashMap<>(); for (String temp : arr) { char c[] = temp.toCharArray(); long pdt = 1 ; // Calculate the product of // primes assigned for ( char t : c) { pdt *= prime[t - 'a' ]; } // If product already exists if (P.containsKey(pdt)) { P.put(pdt, P.get(pdt) + 1 ); } // Otherwise else { W.put(pdt, temp); P.put(pdt, 1 ); } } // Fetch the most frequent product for (Map.Entry<Long, Integer> e : P.entrySet()) { if (max < e.getValue()) { max = e.getValue(); maxpdt = e.getKey(); } } // Return a string // with that product return W.get(maxpdt); } // Driver Code public static void main(String args[]) { String S = "please be silent and listen" + " to what the professor says " ; String arr[] = S.split( "[ ]+" ); System.out.println(largestAnagramGrp(arr)); } } |
Python3
# Python3 Program to find the word # with most anagrams in a sentence # Function to find the word with # maximum number of anagrams def largestAnagramGrp(arr): # Primes assigned to 26 alphabets prime = [ 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 ] max = - 1 maxpdt = - 1 # Stores the product and # word mappings W = {} # Stores the frequencies # of products P = {} for temp in arr: c = [i for i in temp] pdt = 1 # Calculate the product of # primes assigned for t in c: pdt * = prime[ ord (t) - ord ( 'a' )] # If product already exists if (pdt in P): P[pdt] = P.get(pdt, 0 ) + 1 # Otherwise else : W[pdt] = temp P[pdt] = 1 # Fetch the most frequent product for e in P: if ( max < P[e]): max = P[e] maxpdt = e # Return a string # with that product return W[maxpdt] # Driver Code if __name__ = = '__main__' : S = "please be silent and listen to what the professor says" arr = S.split( " " ) print (largestAnagramGrp(arr)) # This code is contributed by mohit kumar 29 |
C#
// C# program to find the word // with most anagrams in a sentence using System; using System.Collections.Generic; class GFG{ // Function to find the word with // maximum number of anagrams private static String largestAnagramGrp(String []arr) { // Primes assigned to 26 alphabets int []prime = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 }; int max = -1; long maxpdt = -1; // Stores the product and // word mappings Dictionary< long , String> W = new Dictionary< long , String>(); // Stores the frequencies // of products Dictionary< long , int > P = new Dictionary< long , int >(); foreach (String temp in arr) { char []c = temp.ToCharArray(); long pdt = 1; // Calculate the product of // primes assigned foreach ( char t in c) { pdt *= prime[t - 'a' ]; } // If product already exists if (P.ContainsKey(pdt)) { P[pdt] = P[pdt] + 1; } // Otherwise else { W.Add(pdt, temp); P.Add(pdt, 1); } } // Fetch the most frequent product foreach (KeyValuePair< long , int > e in P) { if (max < e.Value) { max = e.Value; maxpdt = e.Key; } } // Return a string // with that product return W[maxpdt]; } // Driver Code public static void Main(String []args) { String S = "please be silent and listen" + " to what the professor says " ; String []arr = S.Split( ' ' ); Console.WriteLine(largestAnagramGrp(arr)); } } // This code is contributed by sapnasingh4991 |
silent
Time Complexity: O(N), N is the length of the string excluding the blankspaces.
Auxiliary Space: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.