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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++




#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 strings
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]))
                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




// Java program to Print all pairs of
// anagrams in a given array of strings
public 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




# Python3 program to find
# best meeting point in 2D array
NO_OF_CHARS = 256
 
# function to check whether two strings
# are anagram of each other
def 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 strings
def 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 Code
if __name__ == "__main__":
 
    arr = ["geeksquiz", "geeksforgeeks",
           "abcd", "forgeeksgeeks", "zuiqkeegs"]
    n = len(arr)
    findAllAnagrams(arr, n)
 
# This code is contributed by
# sanjeev2552


C#




// C# program to Print all pairs of
// anagrams in a given array of strings
using 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


Javascript




<script>
// JavaScript program to find
// best meeting point in 2D array
let NO_OF_CHARS = 256
 
// function to check whether two strings
// are anagram of each other
function 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 strings
function 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)

Below is the implementation of this approach : 

C++




#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 strings
void findAllAnagrams(string arr[], int n)
{
    for (int i = 0; i < n-1; i++) {
        for (int j = i+1; j < n; j++) {
            // If the original strings are anagrams, print them
            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




import java.util.Arrays;
 
public class Anagram {
 
    final static 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];
        Arrays.fill(count, 0);
 
        // For each character in input strings, increment count in
        // the corresponding count array
        for (int 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 (int 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 - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                // If the original strings are anagrams, print them
                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);
    }
}


Python3




NO_OF_CHARS = 256
 
 
def areAnagram(str1, str2):
    NO_OF_CHARS = 256
    count = [0] * NO_OF_CHARS
 
    # For each character in input strings, increment count in
    # the corresponding count array
    for i in range(max(len(str1), len(str2))):
        if i < len(str1):
            count[ord(str1[i])] += 1
        if i < len(str2):
            count[ord(str2[i])] -= 1
 
    # 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
 
 
def findAllAnagrams(arr, n):
    for i in range(n-1):
        for j in range(i+1, n):
            # If the original strings are anagrams, print them
            if areAnagram(arr[i], arr[j]):
                print(arr[i], "is anagram of", arr[j])
 
 
# Driver code
if __name__ == "__main__":
    arr = ["geeksquiz", "geeksforgeeks", "abcd", "forgeeksgeeks", "zuiqkeegs"]
    n = len(arr)
    findAllAnagrams(arr, n)


C#




using System;
public class Anagram
{
  const 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];
    Array.Fill(count, 0);
 
    // For each character in input strings, increment count in
    // the corresponding count array
    for (int 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 (int 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 - 1; i++)
    {
      for (int j = i + 1; j < n; j++)
      {
        // If the original strings are anagrams, print them
        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[] args)
  {
    string[] arr = { "geeksquiz", "geeksforgeeks", "abcd", "forgeeksgeeks", "zuiqkeegs" };
    int n = arr.Length;
    FindAllAnagrams(arr, n);
  }
}


Javascript




const NO_OF_CHARS = 256;
 
/* function to check whether two strings are anagram of each other */
function areAnagram(str1, str2) {
  // Create two count arrays and initialize all values as 0
  const count = Array(NO_OF_CHARS).fill(0);
 
  // For each character in input strings, increment count in
  // the corresponding count array
  for (let i = 0; i < str1.length && i < str2.length; i++) {
    count[str1.charCodeAt(i)]++;
    count[str2.charCodeAt(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 (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 strings
function findAllAnagrams(arr) {
  const n = arr.length;
  for (let i = 0; i < n-1; i++) {
    for (let j = i+1; j < n; j++) {
      // If the original strings are anagrams, print them
      if (areAnagram(arr[i], arr[j]))
        console.log(`${arr[i]} is anagram of ${arr[j]}`);
    }
  }
}
 
/* Driver program to test to print printDups*/
const arr = ["geeksquiz", "geeksforgeeks", "abcd", "forgeeksgeeks", "zuiqkeegs"];
findAllAnagrams(arr);


Output

geeksquiz is anagram of zuiqkeegs
geeksforgeeks is anagram of forgeeksgeeks

Space complexity :- O(1)

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)



Last Updated : 17 Apr, 2023
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