Given a sequence of words, print all anagrams together | Set 1 - GeeksforGeeks

12 Apr, 2022
Given an array of words, print all anagrams together. For example, if the given array is {“cat”, “dog”, “tac”, “god”, “act”}, then output may be “cat tac act dog god”.
A simple method is to create a Hash Table. Calculate the hash value of each word in such a way that all anagrams have the same hash value. Populate the Hash Table with these hash values. Finally, print those words together with the same hash values. A simple hashing mechanism can be modulo sum of all characters. With modulo sum, two non-anagram words may have the same hash value. This can be handled by matching individual characters.
Following is another method to print all anagrams together. Take two auxiliary arrays, index array, and word array. Populate the word array with the given sequence of words. Sort each individual word of the word array. Finally, sort the word array and keep track of the corresponding indices. After sorting, all the anagrams cluster together. Use the index array to print the strings from the original array of strings.
Let us understand the steps with the following input Sequence of Words: 
"cat", "dog", "tac", "god", "act"
1) Create two auxiliary arrays index[] and words[]. Copy all given words to words[] and store the original indexes in index[] 
index[]:  0   1   2   3   4
words[]: cat dog tac god act
2) Sort individual words in words[]. Index array doesn’t change.
index[]:   0    1    2    3    4
words[]:  act  dgo  act  dgo  act
3) Sort the words array. Compare individual words using strcmp() to sort
index:     0    2    4    1    3
words[]:  act  act  act  dgo  dgo
4) All anagrams come together. But words are changed in the words array. To print the original words, take the index from the index array and use it in the original array. We get 
"cat tac act dog god"
Following are the implementations of the above algorithm. In the following program, an array of structure “Word” is used to store both index and word arrays. Dupray is another structure that stores an array of structure “Word”. 
C++
C
Java
Python
// A C++ program to print all anagrams together#include <bits/stdc++.h>#include <string.h>using namespace std; // structure for each word of duplicate arrayclass Word {public:    char* str; // to store word itself    int index; // index of the word in the original array}; // structure to represent duplicate array.class DupArray {public:    Word* array; // Array of words    int size; // Size of array}; // Create a DupArray object that contains an array of WordsDupArray* createDupArray(char* str[], int size){    // Allocate memory for dupArray and all members of it    DupArray* dupArray = new DupArray();    dupArray->size = size;    dupArray->array        = new Word[(dupArray->size * sizeof(Word))];     // One by one copy words from the given wordArray to    // dupArray    int i;    for (i = 0; i < size; ++i) {        dupArray->array[i].index = i;        dupArray->array[i].str            = new char[(strlen(str[i]) + 1)];        strcpy(dupArray->array[i].str, str[i]);    }     return dupArray;} // Compare two characters. Used in qsort() for// sorting an array of characters (Word)int compChar(const void* a, const void* b){    return *(char*)a - *(char*)b;} // Compare two words. Used in qsort()// for sorting an array of wordsint compStr(const void* a, const void* b){    Word* a1 = (Word*)a;    Word* b1 = (Word*)b;    return strcmp(a1->str, b1->str);} // Given a list of words in wordArr[],void printAnagramsTogether(char* wordArr[], int size){    // Step 1: Create a copy of all words present in given    // wordArr. The copy will also have original indexes of    // words    DupArray* dupArray = createDupArray(wordArr, size);     // Step 2: Iterate through all words in dupArray and    // sort individual words.    int i;    for (i = 0; i < size; ++i)        qsort(dupArray->array[i].str,              strlen(dupArray->array[i].str), sizeof(char),              compChar);     // Step 3: Now sort the array of words in dupArray    qsort(dupArray->array, size, sizeof(dupArray->array[0]),          compStr);     // Step 4: Now all words in dupArray are together, but    // these words are changed. Use the index member of word    // struct to get the corresponding original word    for (i = 0; i < size; ++i)        cout << wordArr[dupArray->array[i].index] << " ";} // Driver program to test above functionsint main(){    char* wordArr[] = { "cat", "dog", "tac", "god", "act" };    int size = sizeof(wordArr) / sizeof(wordArr[0]);    printAnagramsTogether(wordArr, size);    return 0;} // This is code is contributed by rathbhupendra
// A C program to print all anagrams together#include <stdio.h>#include <stdlib.h>#include <string.h> // structure for each word of duplicate arraystruct Word {    char* str; // to store word itself    int index; // index of the word in the original array}; // structure to represent duplicate array.struct DupArray {    struct Word* array; // Array of words    int size; // Size of array}; // Create a DupArray object that contains an array of Wordsstruct DupArray* createDupArray(char* str[], int size){    // Allocate memory for dupArray and all members of it    struct DupArray* dupArray        = (struct DupArray*)malloc(sizeof(struct DupArray));    dupArray->size = size;    dupArray->array = (struct Word*)malloc(        dupArray->size * sizeof(struct Word));     // One by one copy words from the given wordArray to    // dupArray    int i;    for (i = 0; i < size; ++i) {        dupArray->array[i].index = i;        dupArray->array[i].str            = (char*)malloc(strlen(str[i]) + 1);        strcpy(dupArray->array[i].str, str[i]);    }     return dupArray;} // Compare two characters. Used in qsort() for sorting an// array of characters (Word)int compChar(const void* a, const void* b){    return *(char*)a - *(char*)b;} // Compare two words. Used in qsort() for sorting an array// of wordsint compStr(const void* a, const void* b){    struct Word* a1 = (struct Word*)a;    struct Word* b1 = (struct Word*)b;    return strcmp(a1->str, b1->str);} // Given a list of words in wordArr[],void printAnagramsTogether(char* wordArr[], int size){    // Step 1: Create a copy of all words present in given    // wordArr. The copy will also have original indexes of    // words    struct DupArray* dupArray        = createDupArray(wordArr, size);     // Step 2: Iterate through all words in dupArray and    // sort individual words.    int i;    for (i = 0; i < size; ++i)        qsort(dupArray->array[i].str,              strlen(dupArray->array[i].str), sizeof(char),              compChar);     // Step 3: Now sort the array of words in dupArray    qsort(dupArray->array, size, sizeof(dupArray->array[0]),          compStr);     // Step 4: Now all words in dupArray are together, but    // these words are changed. Use the index member of word    // struct to get the corresponding original word    for (i = 0; i < size; ++i)        printf("%s ", wordArr[dupArray->array[i].index]);} // Driver program to test above functionsint main(){    char* wordArr[] = { "cat", "dog", "tac", "god", "act" };    int size = sizeof(wordArr) / sizeof(wordArr[0]);    printAnagramsTogether(wordArr, size);    return 0;}
// A Java program to print all anagrams togetherimport java.util.Arrays;import java.util.Comparator;public class GFG {    // class for each word of duplicate array    static class Word {        String str; // to store word itself        int index; // index of the word in the        // original array         // constructor        Word(String str, int index)        {            this.str = str;            this.index = index;        }    }     // class to represent duplicate array.    static class DupArray {        Word[] array; // Array of words        int size; // Size of array         // constructor        public DupArray(String str[], int size)        {            this.size = size;            array = new Word[size];             // One by one copy words from the            // given wordArray to dupArray            int i;            for (i = 0; i < size; ++i) {                // create a word Object with the                // str[i] as str and index as i                array[i] = new Word(str[i], i);            }        }    }     // Compare two words. Used in Arrays.sort() for    // sorting an array of words    static class compStr implements Comparator<Word> {        public int compare(Word a, Word b)        {            return a.str.compareTo(b.str);        }    }     // Given a list of words in wordArr[],    static void printAnagramsTogether(String wordArr[],                                      int size)    {        // Step 1: Create a copy of all words present        // in given wordArr. The copy will also have        // original indexes of words        DupArray dupArray = new DupArray(wordArr, size);         // Step 2: Iterate through all words in        // dupArray and sort individual words.        int i;        for (i = 0; i < size; ++i) {            char[] char_arr                = dupArray.array[i].str.toCharArray();            Arrays.sort(char_arr);            dupArray.array[i].str = new String(char_arr);        }         // Step 3: Now sort the array of words in        // dupArray        Arrays.sort(dupArray.array, new compStr());         // Step 4: Now all words in dupArray are together,        // but these words are changed. Use the index        // member of word struct to get the corresponding        // original word        for (i = 0; i < size; ++i)            System.out.print(                wordArr[dupArray.array[i].index] + " ");    }     // Driver program to test above functions    public static void main(String args[])    {        String wordArr[]            = { "cat", "dog", "tac", "god", "act" };        int size = wordArr.length;        printAnagramsTogether(wordArr, size);    }}// This code is contributed by Sumit Ghosh
# A Python program to print all anagrams together # structure for each word of duplicate array  class Word(object):    def __init__(self, string, index):        self.string = string        self.index = index # Create a DupArray object that contains an array# of Words  def createDupArray(string, size):    dupArray = []     # One by one copy words from the given wordArray    # to dupArray    for i in xrange(size):        dupArray.append(Word(string[i], i))     return dupArray # Given a list of words in wordArr[]  def printAnagramsTogether(wordArr, size):    # Step 1: Create a copy of all words present in    # given wordArr.    # The copy will also have original indexes of words    dupArray = createDupArray(wordArr, size)     # Step 2: Iterate through all words in dupArray and sort    # individual words.    for i in xrange(size):        dupArray[i].string = ''.join(sorted(dupArray[i].string))     # Step 3: Now sort the array of words in dupArray    dupArray = sorted(dupArray, key=lambda k: k.string)     # Step 4: Now all words in dupArray are together, but    # these words are changed. Use the index member of word    # struct to get the corresponding original word    for word in dupArray:        print wordArr[word.index],  # Driver programwordArr = ["cat", "dog", "tac", "god", "act"]size = len(wordArr)printAnagramsTogether(wordArr, size) # This code is contributed by BHAVYA JAIN
Output: 
cat tac act dog god 
Time Complexity: Let there be N-words and each word has a maximum of M characters. The upper bound is O(NMLogM + MNLogN). Step 2 takes O(NMLogM) time. Sorting a word takes maximum O(MLogM) time. So sorting N-words takes O(NMLogM) time. step 3 takes O(MNLogN) Sorting array of words takes NLogN comparisons. A comparison may take maximum O(M) time. So time to sort an array of words will be O(MNLogN). 
Using vector of pair :
The problem can easily be solved with the use of a vector of pairs. The pair will be of string and int. The string will require to store the input string and int will require to store their respective indexes. 
here is the implementation of the above approach:
C++
#include <bits/stdc++.h>#include <strings.h>using namespace std; void createDuplicateArray(    vector<pair<string, int> >& dupArray,    vector<string>& wordAr, int size){    for (int i = 0; i < size; i++) {        dupArray.push_back(make_pair(wordAr[i], i));        // pair.first contains the input words and        // pair.second contains its index    }} void printAnagramsTogether(vector<string>& wordArr,                           int size){     vector<pair<string, int> >        dupArray; // dupArray to store the word-index pair    createDuplicateArray(        dupArray, wordArr,        size); // making copy of all the words and their               // respective index     // Iterate through all words in dupArray and sort    // characters in each word.    int i;    for (i = 0; i < size; ++i) {        sort(dupArray[i].first.begin(),             dupArray[i].first.end());    }     // now sort the whole vector to get the identical words    // together    sort(dupArray.begin(), dupArray.end());     // now all the indentical words are together but we have    // lost the original form of string    // so through index stored in the word-index pair fetch    // the original word from main input    for (i = 0; i < size; ++i)        cout << wordArr[dupArray[i].second] << " ";} int main(){    vector<string> wordArr        = { "cat", "dog", "tac", "god", "act" };    printAnagramsTogether(wordArr, wordArr.size());    return 0;}
cat tac act dog god 
Time complexity: 
Let there be N-words and each word has a maximum of M characters.
 O(NMLogM + MNLogN). 
Using hashmapHere, we first sort each word, use the sorted word as a key and then put an original word on a map. The value of the map will be a list containing all the words which have the same word after sorting. Lastly, we will print all values from the hashmap where the size of values will be greater than 1.
C++
Java
Python3
C#
// C++ program to print anagrams// together using dictionary#include <bits/stdc++.h>using namespace std; void printAnagrams(string arr[], int size){    unordered_map<string, vector<string> > map;     // Loop over all words    for (int i = 0; i < size; i++) {         // Convert to char array, sort and        // then re-convert to string        string word = arr[i];        char letters[word.size() + 1];        strcpy(letters, word.c_str());        sort(letters, letters + word.size() + 1);        string newWord = "";        for (int i = 0; i < word.size() + 1; i++) {            newWord += letters[i];        }         // Calculate hashcode of string        // after sorting        if (map.find(newWord) != map.end()) {            map[newWord].push_back(word);        }        else {             // This is the first time we are            // adding a word for a specific            // hashcode            vector<string> words;            words.push_back(word);            map[newWord] = words;        }    }     // Print all the values where size is > 1    // If you want to print non-anagrams,    // just print the values having size = 1    unordered_map<string, vector<string> >::iterator it;    for (it = map.begin(); it != map.end(); it++) {        vector<string> values = map[it->first];        if (values.size() > 1) {            cout << "[";            for (int i = 0; i < values.size() - 1; i++) {                cout << values[i] << ", ";            }            cout << values[values.size() - 1];            cout << "]";        }    }} // Driver codeint main(){    string arr[] = { "cat", "dog", "tac", "god", "act" };    int size = sizeof(arr) / sizeof(arr[0]);     printAnagrams(arr, size);     return 0;} // This code is contributed by Ankit Garg
// Java program to print anagrams// together using dictionaryimport java.util.*; public class FindAnagrams {     private static void printAnagrams(String arr[])    {        HashMap<String, List<String> > map            = new HashMap<>();         // loop over all words        for (int i = 0; i < arr.length; i++) {             // convert to char array, sort and            // then re-convert to string            String word = arr[i];            char[] letters = word.toCharArray();            Arrays.sort(letters);            String newWord = new String(letters);             // calculate hashcode of string            // after sorting            if (map.containsKey(newWord)) {                 map.get(newWord).add(word);            }            else {                 // This is the first time we are                // adding a word for a specific                // hashcode                List<String> words = new ArrayList<>();                words.add(word);                map.put(newWord, words);            }        }         // print all the values where size is > 1        // If you want to print non-anagrams,        // just print the values having size = 1        for (String s : map.keySet()) {            List<String> values = map.get(s);            if (values.size() > 1) {                System.out.print(values);            }        }    }     public static void main(String[] args)    {         // Driver program        String arr[]            = { "cat", "dog", "tac", "god", "act" };        printAnagrams(arr);    }}
from collections import defaultdict  def printAnagramsTogether(words):    groupedWords = defaultdict(list)     # Put all anagram words together in a dictionary    # where key is sorted word    for word in words:        groupedWords["".join(sorted(word))].append(word)     # Print all anagrams together    for group in groupedWords.values():        print(" ".join(group))  if __name__ == "__main__":    arr = ["cat", "dog", "tac", "god", "act"]    printAnagramsTogether(arr)
// C# program to print anagrams// together using dictionaryusing System;using System.Collections.Generic; class GFG {    private static void printAnagrams(String[] arr)    {        Dictionary<String, List<String> > map            = new Dictionary<String, List<String> >();         // loop over all words        for (int i = 0; i < arr.Length; i++) {             // convert to char array, sort and            // then re-convert to string            String word = arr[i];            char[] letters = word.ToCharArray();            Array.Sort(letters);            String newWord = new String(letters);             // calculate hashcode of string            // after sorting            if (map.ContainsKey(newWord)) {                map[newWord].Add(word);            }            else {                 // This is the first time we are                // adding a word for a specific                // hashcode                List<String> words = new List<String>();                words.Add(word);                map.Add(newWord, words);            }        }         // print all the values where size is > 1        // If you want to print non-anagrams,        // just print the values having size = 1        List<String> value = new List<String>();        foreach(KeyValuePair<String, List<String> > entry in                    map)        {            value.Add(entry.Key);        }        int k = 0;        foreach(KeyValuePair<String, List<String> > entry in                    map)        {            List<String> values = map[value[k++]];            if (values.Count > 1) {                Console.Write("[");                int len = 1;                foreach(String s in values)                {                    Console.Write(s);                    if (len++ < values.Count)                        Console.Write(", ");                }                Console.Write("]");            }        }    }     // Driver Code    public static void Main(String[] args)    {        String[] arr            = { "cat", "dog", "tac", "god", "act" };        printAnagrams(arr);    }} // This code is contributed by Princi Singh
[dog, god][cat, tac, act]
Output : 
[cat, tac, act][dog, god]
HashMap with O(NM) SolutionIn the previous approach, we were sorting every string in order to maintain a similar key, but that cost extra time in this approach will take the advantage of another hashmap to maintain the frequency of the characters which will generate the same hash function for different string having same frequency of characters.Here, we will take HashMap<HashMap, ArrayList>, the inner hashmap will count the frequency of the characters of each string and the outer HashMap will check whether that hashmap is present or not if present then it will add that string to the corresponding list. 
C++
Java
Python3
// C++ code to print all anagrams together#include <bits/stdc++.h>using namespace std; void solver(vector<string> my_list){         // Inner hashmap counts frequency    // of characters in a string.    // Outer hashmap for if same    // frequency characters are present in    // in a string then it will add it to    // the vector.    map<map<char, int>, vector<string>> my_map;         // Loop over all words    for(string str : my_list)    {                 // Counting the frequency of the        // characters present in a string        map<char, int> temp_map;        vector<string> temp_my_list;        for(int i = 0; i < str.length(); ++i)        {            ++temp_map[str[i]];        }                 // If the same frequency of characters        // are already present then add that        // string into that arraylist otherwise        // created a new arraylist and add that        // string        auto it = my_map.find(temp_map);        if (it != my_map.end())        {            it->second.push_back(str);        }        else        {            temp_my_list.push_back(str);            my_map.insert({ temp_map, temp_my_list });        }    }         // Stores the result in a vector    vector<vector<string>> result;     for(auto it = my_map.begin();             it != my_map.end(); ++it)    {        result.push_back(it->second);    }     for(int i = 0; i < result.size(); ++i)    {          cout << "[";        for(int j = 0; j < result[i].size(); ++j)        {            cout << result[i][j] << ", ";        }          cout << "]";    }} // Driver codeint main(){    vector<string> my_list = { "cat", "dog", "ogd",                               "god", "atc" };    solver(my_list);    return 0;} // This code is contributed by// Apurba Kumar Gorai(coolapurba05)
// Java code tp print all anagrams togetherimport java.util.ArrayList;import java.util.HashMap; public class FindAnagrams {     private static ArrayList<ArrayList<String> >    solver(        ArrayList<String> list)    {         // Inner hashmap counts frequency        // of characters in a string.        // Outer hashmap for if same        // frequency characters are present in        // in a string then it will add it to        // the arraylist.        HashMap<HashMap<Character, Integer>,                ArrayList<String> >            map = new HashMap<HashMap<Character, Integer>,                              ArrayList<String> >();        for (String str : list) {            HashMap<Character, Integer>                tempMap = new HashMap<Character, Integer>();             // Counting the frequency of the            // characters present in a string            for (int i = 0; i < str.length(); i++) {                if (tempMap.containsKey(str.charAt(i))) {                    int x = tempMap.get(str.charAt(i));                    tempMap.put(str.charAt(i), ++x);                }                else {                    tempMap.put(str.charAt(i), 1);                }            }             // If the same frequency of characters            // are already present then add that            // string into that arraylist otherwise            // created a new arraylist and add that string            if (map.containsKey(tempMap))                map.get(tempMap).add(str);            else {                ArrayList<String>                    tempList = new ArrayList<String>();                tempList.add(str);                map.put(tempMap, tempList);            }        }         // Stores the result in a arraylist        ArrayList<ArrayList<String> >            result = new ArrayList<>();        for (HashMap<Character, Integer>                 temp : map.keySet())            result.add(map.get(temp));        return result;    }     // Drivers Method    public static void main(String[] args)    {        ArrayList<String> list = new ArrayList<>();        list.add("cat");        list.add("dog");        list.add("ogd");        list.add("god");        list.add("atc");         System.out.println(solver(list));    }} // This code is contributed by Arijit Basu(ArijitXfx)
# Python code to print all anagrams togetherfrom collections import Counter, defaultdictuser_input = ["cat", "dog", "tac", "edoc", "god", "tacact",              "act", "code", "deno", "node", "ocde", "done", "catcat"]  def solve(words: list) -> list:    # defaultdict will create a new list if the key is not found in the dictionary    m = defaultdict(list)     # loop over all the words    for word in words:        # Counter('cat') :        #    counts the frequency of the characters present in a string        #    >>> Counter({'c': 1, 'a': 1, 't': 1})         # frozenset(dict(Counter('cat')).items()) :        #    frozenset takes an iterable object as input and makes them immutable.        #    So that hash(frozenset(Counter('cat'))) is equal to        #   hash of other 'cat' anagrams        #    >>> frozenset({('c', 1), ('a', 1), ('t', 1)})        m[frozenset(dict(Counter(word)).items())].append(word)    return [v for k, v in m.items()]  print(solve(user_input)) # This code is contributed by# Rohan Kumar(@r0hnx)
[cat, atc, ][dog, ogd, god, ]
Output:
[[cat, atc], [dog, ogd, god]]
Time Complexity: Let there be N-words and each word has a maximum of M characters. The upper bound is O(NM). Space Complexity: Let there be N-words and each word has maximum M characters, therefore max. storage space for each word with at max. M characters will be O(M), therefore for max N-words, it will be O(N*M). Therefore, the upper bound is O(NM).
Given a sequence of words, print all anagrams together | Set 2 Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
subharaj01
rituraj_jain
rathbhupendra
Akanksha_Rai
princi singh
Arijit Basu 1
coolapurba05
ankit19146
gabaa406
r0hnx
sooda367
simmytarika5
arorakashish0911
RishMeh19
surinderdawra388
rupeshsk30
sumitgumber28
Amazon
anagram
Morgan Stanley
Snapdeal
Hash
Strings
Morgan Stanley
Amazon
Snapdeal
Hash
Strings
anagram
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Hashing | Set 2 (Separate Chaining)
Sort string of characters
Counting frequencies of array elements
Most frequent element in an array
Double Hashing
Write a program to reverse an array or string
Reverse a string in Java
Write a program to print all permutations of a given string
C++ Data Types
Longest Common Subsequence | DP-4