Given a string S in the form of a sentence, the task is to find the word from the text with the 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”.
Method 1:
- Initialise a map say unmap, to keep track of the frequency of each sorted word.
- Initialise a variable maxx, this will keep track of the previous anagram count
- Initialise a variable result for answer.
- Iterate over each word of given string s
- Keep the original word
- Sort the word
- Check if this sorted word exists in the map
- If not exist then store unmap[word] = {current occurrence, original word}
- Otherwise, increment the count of the frequency
- Check if the anagram count is greater than the previous anagram count
- Update the previous anagram count
- Update the result
- Return the result
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;string largestAnagramGrp(string s){ // Initialise a map for keep track of frequency of // each sorted word. unordered_map<string, pair<int, string> > unmap; stringstream ss(s); string word; // Initialise a variable maxx, this will keep track of // previous anagram count int maxx = 0; // Initialise a variable result for answer. string result; // Iterate over each word of given string s while (ss >> word) { // Keep the original word string temp = word; // Sort the word sort(word.begin(), word.end()); // Check if this sorted word exist in map if (unmap.count(word) == 0) { // If not exist then store unmap[word] = // {current occurrence, original word} unmap[word] = { 1, temp }; } else { // Otherwise, increment the count of frequency unmap[word].first++; } // Check if the anagram count are greater than // previous anagram count if (unmap[word].first > maxx) { // Update the previous anagram count maxx = unmap[word].first; // Update the result result = unmap[word].second; } } // Return the result return result;}// Driver Codeint main(){ string S = "cat is not a dog and sword has no words " "when government creates act so what is tac"; cout << largestAnagramGrp(S) << endl; return 0;}//This code is contributed by hkdass001 |
Java
import java.util.Arrays;import java.util.HashMap;import java.util.Map;import java.util.StringTokenizer;class Pair<K, V> { private K key; private V value; public Pair(K key, V value) { this.key = key; this.value = value; } public void setKey(K key) { this.key = key; } public void setValue(V value) { this.value = value; } public K getKey() { return key; } public V getValue() { return value; }}public class Gfg { public static String largestAnagramGrp(String s) { // Initialise a map for keep track of frequency of // each sorted word. Map<String, Pair<Integer, String> > unmap = new HashMap<>(); StringTokenizer st = new StringTokenizer(s); String word; // Initialise a variable maxx, this will keep track // of previous anagram count int maxx = 0; // Initialise a variable result for answer. String result = ""; // Iterate over each word of given string s while (st.hasMoreTokens()) { // Keep the original word String temp = st.nextToken(); // Sort the word char[] wordArray = temp.toCharArray(); Arrays.sort(wordArray); word = new String(wordArray); // Check if this sorted word exist in map if (!unmap.containsKey(word)) { // If not exist then store unmap[word] = // {current occurrence, original word} Pair<Integer, String> pair = new Pair<>(1, temp); unmap.put(word, pair); } else { // Otherwise, increment the count of // frequency Pair<Integer, String> pair = unmap.get(word); pair.setKey(pair.getKey() + 1); } // Check if the anagram count are greater than // previous anagram count if (unmap.get(word).getKey() > maxx) { // Update the previous anagram count maxx = unmap.get(word).getKey(); // Update the result result = unmap.get(word).getValue(); } } // Return the result return result; } // Driver Code public static void main(String[] args) { String S = "cat is not a dog and sword has no words when government creates act so what is tac"; System.out.println(largestAnagramGrp(S)); }} |
Python3
# python Program to implement# the above approachdef stringstream(s): res = [] temp = [] for i in range(len(s)): if s[i] != ' ': temp.append(s[i]); else: res.append(''.join(temp)) temp.clear() res.append(''.join(temp)) return resdef largestAnagramGrp(s): # Initialise a map for keep track of frequency of # each sorted word. unmap = {} ss = stringstream(s) # Initialise a variable maxx, this will keep track of # previous anagram count maxx = 0 # Initialise a variable result for answer. result = [] i = 0 # Iterate over each word of given string s while (i < len(ss)): # Keep the original word word = ss[i] temp = ss[i] # Sort the word word = ''.join(sorted(word)) # Check if this sorted word exist in map if word in unmap: # Otherwise, increment the count of frequency unmap[word][0] = unmap[word][0] + 1 else: # If not exist then store unmap[word] = # {current occurrence, original word} unmap[word] = [1, temp] # Check if the anagram count are greater than # previous anagram count if unmap[word][0] > maxx: # Update the previous anagram count maxx = unmap[word][0] # Update the result result = list(unmap[word][1]) i = i + 1 # Return the result return ''.join(result)# Driver CodeS = "cat is not a dog and sword has no words when government creates act so what is tac"print(largestAnagramGrp(S))# This code is contributed by Nidhi goel. |
C#
// C# Program to implement// the above approachusing System;using System.Collections;using System.Collections.Generic;using System.Linq;class HelloWorld { public static string largestAnagramGrp(string s) { // Initialise a map for keep track of frequency of // each sorted word. Dictionary<string, KeyValuePair<int, string>> unmap = new Dictionary<string, KeyValuePair<int, string>>(); var ss = s.Split(' '); // Initialise a variable maxx, this will keep track of // previous anagram count int maxx = 0; // Initialise a variable result for answer. string result = ""; // Iterate over each word of given string s for(int i = 0;i < ss.Length; i++) { // Keep the original word string word = ss[i]; string temp = ss[i]; // Sort the word word = String.Concat(word.OrderBy(c => c)); // Check if this sorted word exist in map if (unmap.ContainsKey(word) == false) { // If not exist then store unmap[word] = // {current occurrence, original word} unmap[word] = new KeyValuePair<int, string>(1, temp); } else { // Otherwise, increment the count of frequency unmap[word] = new KeyValuePair<int, string>(unmap[word].Key + 1, unmap[word].Value); } // Check if the anagram count are greater than // previous anagram count if (unmap[word].Key > maxx) { // Update the previous anagram count maxx = unmap[word].Key; // Update the result result = unmap[word].Value; } } // Return the result return result; } // Driver code. static void Main() { string S = "cat is not a dog and sword has no words when government creates act so what is tac"; Console.WriteLine(largestAnagramGrp(S)); }}// The code is contributed by Nidhi goel. |
Javascript
// javascript Program to implement// the above approachfunction stringstream(s){ let res = []; let temp = []; for(let i = 0; i < s.length; i++){ if(s[i] != ' '){ temp.push(s[i]); } else{ res.push(temp.join("")); temp.splice(0, temp.length) } } res.push(temp.join("")); return res;}function largestAnagramGrp(s){ // Initialise a map for keep track of frequency of // each sorted word. unmap = {}; let ss = stringstream(s); // Initialise a variable maxx, this will keep track of // previous anagram count let maxx = 0; // Initialise a variable result for answer. let result = []; let i = 0 // Iterate over each word of given string s while (i < ss.length) { // Keep the original word let word = ss[i]; let temp = ss[i]; // Sort the word word = word.split('').sort().join(""); // Check if this sorted word exist in map if (word in unmap) { // Otherwise, increment the count of frequency unmap[word] = unmap[word] + 1; } else { // If not exist then store unmap[word] = // {current occurrence, original word} unmap[word] = [1, temp]; } // Check if the anagram count are greater than // previous anagram count if (unmap[word][0] > maxx) { // Update the previous anagram count maxx = unmap[word][0]; // Update the result result = unmap[word][1].split(''); } i = i + 1; } // Return the result return result.join("");}// Driver Codelet S = "cat is not a dog and sword has no words when government creates act so what is tac";console.log(largestAnagramGrp(S));// This code is contributed by Nidhi goel. |
Output
cat
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Approach:
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 anagramsstring 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 Codeint 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 sentenceimport 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 anagramsdef 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 Codeif __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 sentenceusing System;using System.Collections.Generic;class GFG{// Function to find the word with// maximum number of anagramsprivate 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 Codepublic 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 |
Javascript
<script>// JavaScript program to find the word// with most anagrams in a sentence// Function to find the word with// maximum number of anagramsfunction largestAnagramGrp(arr){ // Primes assigned to 26 alphabets var 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, ]; var max = -1; var maxpdt = -1; // Stores the product and // word mappings var W = {}; // Stores the frequencies // of products var P = {}; for (const temp of arr) { var c = temp.split(""); var pdt = 1; // Calculate the product of // primes assigned for(const t of c) { pdt *= prime[t.charCodeAt(0) - "a".charCodeAt(0)]; } // If product already exists if (P.hasOwnProperty(pdt)) { P[pdt] = P[pdt] + 1; } // Otherwise else { W[pdt] = temp; P[pdt] = 1; } } // Fetch the most frequent product for (const [key, value] of Object.entries(P)) { if (max < value) { max = value; maxpdt = key; } } // Return a string // with that product return W[maxpdt];}// Driver Codevar S = "please be silent and listen" + " to what the professor says ";var arr = S.split(" ");document.write(largestAnagramGrp(arr));// This code is contributed by rdtank</script> |
Output
silent
Time Complexity: O(N), N is the length of the string excluding the blankspaces.
Auxiliary Space: O(N)
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