Print all pairs of anagrams in a given array of strings

22 Jun, 2022
Given an array of strings, find all anagram pairs in the given array. Example:  
Input: arr[] =  {"geeksquiz", "geeksforgeeks", "abcd",
                 "forgeeksgeeks", "zuiqkeegs"};
Output: (geeksforgeeks, forgeeksgeeks), (geeksquiz, zuiqkeegs)
We can find whether two strings are anagram or not in linear time using count array (see method 2 of this). One simple idea to find whether all anagram pairs is to run two nested loops. The outer loop picks all strings one by one. The inner loop checks whether remaining strings are anagram of the string picked by outer loop. Below is the implementation of this approach : 
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256 /* function to check whether two strings are anagram of each other */bool areAnagram(string str1, string str2){    // Create two count arrays and initialize all values as 0    int count[NO_OF_CHARS] = {0};    int i;     // For each character in input strings, increment count in    // the corresponding count array    for (i = 0; str1[i] && str2[i];  i++)    {        count[str1[i]]++;        count[str2[i]]--;    }     // If both strings are of different length. Removing this condition    // will make the program fail for strings like "aaca" and "aca"    if (str1[i] || str2[i])      return false;     // See if there is any non-zero value in count array    for (i = 0; i < NO_OF_CHARS; i++)        if (count[i])            return false;     return true;} // This function prints all anagram pairs in a given array of stringsvoid findAllAnagrams(string arr[], int n){    for (int i = 0; i < n; i++)        for (int j = i+1; j < n; j++)            if (areAnagram(arr[i], arr[j]))                cout << arr[i] << " is anagram of " << arr[j] << endl;}  /* Driver program to test to print printDups*/int main(){    string arr[] = {"geeksquiz", "geeksforgeeks", "abcd",                    "forgeeksgeeks", "zuiqkeegs"};    int n = sizeof(arr)/sizeof(arr[0]);    findAllAnagrams(arr, n);    return 0;}
// Java program to Print all pairs of// anagrams in a given array of stringspublic class GFG{    static final int NO_OF_CHARS = 256;         /* function to check whether two    strings are anagram of each other */    static boolean areAnagram(String str1, String str2)    {        // Create two count arrays and initialize        // all values as 0        int[] count = new int[NO_OF_CHARS];        int i;         // For each character in input strings,        // increment count in the corresponding        // count array        for (i = 0; i < str1.length() && i < str2.length();                                                   i++)        {            count[str1.charAt(i)]++;            count[str2.charAt(i)]--;        }         // If both strings are of different length.        // Removing this condition will make the program        // fail for strings like "aaca" and "aca"        if (str1.length() != str2.length())          return false;         // See if there is any non-zero value in        // count array        for (i = 0; i < NO_OF_CHARS; i++)            if (count[i] != 0)                return false;         return true;    }     // This function prints all anagram pairs in a    // given array of strings    static void findAllAnagrams(String arr[], int n)    {        for (int i = 0; i < n; i++)            for (int j = i+1; j < n; j++)                if (areAnagram(arr[i], arr[j]))                    System.out.println(arr[i] +                       " is anagram of " + arr[j]);    }     /* Driver program to test to print printDups*/    public static void main(String args[])    {        String arr[] = {"geeksquiz", "geeksforgeeks",                        "abcd", "forgeeksgeeks",                        "zuiqkeegs"};        int n = arr.length;        findAllAnagrams(arr, n);    }}// This code is contributed by Sumit Ghosh
# Python3 program to find# best meeting point in 2D arrayNO_OF_CHARS = 256 # function to check whether two strings# are anagram of each otherdef areAnagram(str1: str, str2: str) -> bool:     # Create two count arrays and    # initialize all values as 0    count = [0] * NO_OF_CHARS    i = 0     # For each character in input strings,    # increment count in the corresponding    # count array    while i < len(str1) and i < len(str2):        count[ord(str1[i])] += 1        count[ord(str2[i])] -= 1        i += 1     # If both strings are of different length.    # Removing this condition will make the program    # fail for strings like "aaca" and "aca"    if len(str1) != len(str2):        return False     # See if there is any non-zero value    # in count array    for i in range(NO_OF_CHARS):        if count[i]:            return False        return True # This function prints all anagram pairs# in a given array of stringsdef findAllAnagrams(arr: list, n: int):    for i in range(n):        for j in range(i + 1, n):            if areAnagram(arr[i], arr[j]):                print(arr[i], "is anagram of", arr[j]) # Driver Codeif __name__ == "__main__":     arr = ["geeksquiz", "geeksforgeeks",           "abcd", "forgeeksgeeks", "zuiqkeegs"]    n = len(arr)    findAllAnagrams(arr, n) # This code is contributed by# sanjeev2552
// C# program to Print all pairs of// anagrams in a given array of stringsusing System; class GFG{    static int NO_OF_CHARS = 256;         /* function to check whether two    strings are anagram of each other */    static bool areAnagram(String str1, String str2)    {        // Create two count arrays and initialize        // all values as 0        int[] count = new int[NO_OF_CHARS];        int i;         // For each character in input strings,        // increment count in the corresponding        // count array        for (i = 0; i < str1.Length &&                    i < str2.Length; i++)        {            count[str1[i]]++;            count[str2[i]]--;        }         // If both strings are of different length.        // Removing this condition will make the program        // fail for strings like "aaca" and "aca"        if (str1.Length != str2.Length)        return false;         // See if there is any non-zero value in        // count array        for (i = 0; i < NO_OF_CHARS; i++)            if (count[i] != 0)                return false;        return true;    }     // This function prints all anagram pairs in a    // given array of strings    static void findAllAnagrams(String []arr, int n)    {        for (int i = 0; i < n; i++)            for (int j = i+1; j < n; j++)                if (areAnagram(arr[i], arr[j]))                    Console.WriteLine(arr[i] +                    " is anagram of " + arr[j]);    }     /* Driver program to test to print printDups*/    public static void Main()    {        String []arr = {"geeksquiz", "geeksforgeeks",                        "abcd", "forgeeksgeeks",                        "zuiqkeegs"};        int n = arr.Length;        findAllAnagrams(arr, n);    }} // This code is contributed by nitin mittal
<script>// JavaScript program to find// best meeting point in 2D arraylet NO_OF_CHARS = 256 // function to check whether two strings// are anagram of each otherfunction areAnagram(str1, str2){     // Create two count arrays and    // initialize all values as 0    let count = [];    for(let i = 0;i< NO_OF_CHARS;i++)        count[i] = 0;    let i = 0;     // For each character in input strings,    // increment count in the corresponding    // count array    while(i < (str1).length && i < (str2).length){        count[ord(str1[i])] += 1;        count[ord(str2[i])] -= 1;        i += 1;    }     // If both strings are of different length.    // Removing this condition will make the program    // fail for strings like "aaca" and "aca"    if ( (str1).length !=  (str2).length)        return false;     // See if there is any non-zero value    // in count array    for(let i = 0; i < NO_OF_CHARS; i++){        if (count[i])            return false;        return True;     }} // This function prints all anagram pairs// in a given array of stringsfunction findAllAnagrams(arr, n){    for(let i = 0; i < n; i++){        for(let j = i + 1; j < n; j++){            if areAnagram(arr[i], arr[j])                document.write(arr[i]+"is anagram of"+arr[j])        }    }}    // Driver Code    let arr = ["geeksquiz", "geeksforgeeks",           "abcd", "forgeeksgeeks", "zuiqkeegs"];    let n = (arr).length;    findAllAnagrams(arr, n); // This code is contributed by Rohitsingh07052.</script>
Output:  
geeksquiz is anagram of zuiqkeegs
geeksforgeeks is anagram of forgeeksgeeks
The time complexity of the above solution is O(n2*m) where n is number of strings and m is maximum length of a string.
Auxiliary Space: O(1) or O(256).Optimizations: We can optimize the above solution using following approaches. 1) Using sorting: We can sort array of strings so that all anagrams come together. Then print all anagrams by linearly traversing the sorted array. The time complexity of this solution is O(mnLogn) (We would be doing O(nLogn) comparisons in sorting and a comparison would take O(m) time)2) Using Hashing: We can build a hash function like XOR or sum of ASCII values of all characters for a string. Using such a hash function, we can build a hash table. While building the hash table, we can check if a value is already hashed. If yes, we can call areAnagrams() to check if two strings are actually anagrams (Note that xor or sum of ASCII values is not sufficient, see Kaushik Lele’s comment here)This article is contributed by Abhishek. Please write comments if you find anything incorrect or want to share more information about above topic.
nitin mittal
sanjeev2552
rohitsingh07052
kk773572498
prachisoda1234
sagartomar9927
anandkumarshivam2266
anagram
Strings
Strings
anagram
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.