Count of anagrams of each string in an array present in another array

Given two arrays arr1[] and arr2[] consisting of strings, the task is to print the count of anagrams of every string in arr2[] that are present in arr1[].

Examples:

Input: arr1[] = [“geeks”, “learn”, “for”, “egeks”, “ealrn”], arr2[] = [“kgees”, “rof”, “nrael”] 
Output: 2 1 2 
Explanation:
Anagrams of arr2[0] (“kgees”) in arr1 : “geeks” and “egeks”. 
Anagrams of arr2[1] (“rof”) in arr1 : “for”. 
Anagrams of arr2[2] (“nrael”) in arr1 : “learn” and “ealrn”.

Input: arr1[] = [“code”, “to”, “grow”, “odce”], arr2[] = [“edoc”, “wgor”, “ot”] 
Output: 2 1 1 
Explanation:
Anagrams of arr2[0] (“edoc”) in arr1 “code” and “odce”. 
Anagrams of arr2[1] (“wgor”) in arr1 “grow”. 
Anagrams of arr2[2] (“ot”) in arr1 “to” 

Approach:
To solve the problem, the idea is to use frequency-counting with the help of HashMap. Store the frequencies of every string in arr1[] in hashmap in their sorted form. Traverse arr2[], sort strings in arr2[], and print their respective frequencies in HashMap.

Below is the implementation of the above approach:  

C++

// C++ Program to count the
// number of anagrams of
// each string in a given
// array present in
// another array

#include <bits/stdc++.h>
using namespace std;

// Function to return the
// count of anagrams
void count(string arr1[],
           string arr2[],
           int n, int m)
{
    // Store the frequencies
    // of strings in arr1[]
    unordered_map<string, int> freq;

    for (int i = 0; i < n; i++) {

        // Sort the string
        sort(arr1[i].begin(),
             arr1[i].end());

        // Increase its frequency
        // in the map
        freq[arr1[i]]++;
    }

    for (int i = 0; i < m; i++) {

        // Sort the string
        sort(arr2[i].begin(),
             arr2[i].end());

        // Display its anagrams
        // in arr1[]
        cout << freq[arr2[i]]
             << " ";
    }
}

// Driver Code
int main()
{

    string arr1[] = { "geeks", "learn",
                      "for", "egeks",
                      "ealrn" };
    int n = sizeof(arr1)
            / sizeof(string);

    string arr2[] = { "kgees", "rof",
                      "nrael" };
    int m = sizeof(arr2)
            / sizeof(string);

    count(arr1, arr2, n, m);
}

Java

// Java program to count the number
// of anagrams of each String in a
// given array present in
// another array
import java.util.*;

class GFG{

static String sortString(String inputString)
{

    // Convert input string to char array
    char tempArray[] = inputString.toCharArray();

    // Sort tempArray
    Arrays.sort(tempArray);

    // Return new sorted string
    return new String(tempArray);
}

// Function to return the
// count of anagrams
static void count(String arr1[],
                  String arr2[],
                  int n, int m)
{

    // Store the frequencies
    // of Strings in arr1[]
    HashMap<String, Integer> freq = new HashMap<>();

    for(int i = 0; i < n; i++)
    {

        // Sort the String
        arr1[i] = sortString(arr1[i]);

        // Increase its frequency
        // in the map
        if (freq.containsKey(arr1[i]))
        {
            freq.put(arr1[i],
            freq.get(arr1[i]) + 1);
        }
        else
        {
            freq.put(arr1[i], 1);
        }
    }

    for(int i = 0; i < m; i++)
    {

        // Sort the String
        arr2[i] = sortString(arr2[i]);

        // Display its anagrams
        // in arr1[]
        System.out.print(freq.get(arr2[i]) + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    String arr1[] = { "geeks", "learn",
                      "for", "egeks",
                      "ealrn" };
    int n = arr1.length;

    String arr2[] = { "kgees", "rof",
                      "nrael" };
    int m = arr2.length;

    count(arr1, arr2, n, m);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to count the number
# of anagrams of each string in a
# given array present in another array

# Function to return the count of anagrams
def count(arr1, arr2, n, m):

    # Store the frequencies of
    # strings in arr1
    freq = {}

    for word in arr1:

        # Sort the string
        word = ' '.join(sorted(word))

        # Increase its frequency
        if word in freq.keys():
            freq[word] = freq[word] + 1
        else:
            freq[word] = 1

    for word in arr2:

        # Sort the string
        word = ' '.join(sorted(word))

        # Display its anagrams
        # in arr1
        if word in freq.keys():
            print(freq[word], end = " ")
        else:
            print(0, end = " ")

    print()

# Driver Code
if __name__ == '__main__':

    arr1 = [ "geeks", "learn", "for",
             "egeks", "ealrn" ]
    n = len(arr1)

    arr2 = [ "kgees", "rof", "nrael" ]
    m = len(arr2)

    count(arr1, arr2, n, m)

# This code is contributed by Pawan_29

C#

// C# program to count the number
// of anagrams of each String in a
// given array present in
// another array
using System;
using System.Collections.Generic;

class GFG{

static String sortString(String inputString)
{

    // Convert input string to char array
    char []tempArray = inputString.ToCharArray();

    // Sort tempArray
    Array.Sort(tempArray);

    // Return new sorted string
    return new String(tempArray);
}

// Function to return the
// count of anagrams
static void count(String []arr1,
                  String []arr2,
                  int n, int m)
{

    // Store the frequencies
    // of Strings in arr1[]
    Dictionary<String,
               int> freq = new Dictionary<String,
                                          int>();

    for(int i = 0; i < n; i++)
    {

        // Sort the String
        arr1[i] = sortString(arr1[i]);

        // Increase its frequency
        // in the map
        if (freq.ContainsKey(arr1[i]))
        {
            freq[arr1[i]] =
            freq[arr1[i]] + 1;
        }
        else
        {
            freq.Add(arr1[i], 1);
        }
    }

    for(int i = 0; i < m; i++)
    {

        // Sort the String
        arr2[i] = sortString(arr2[i]);

        // Display its anagrams
        // in arr1[]
        Console.Write(freq[arr2[i]] + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    String []arr1 = { "geeks", "learn",
                      "for", "egeks",
                      "ealrn" };
    int n = arr1.Length;

    String []arr2 = { "kgees", "rof",
                      "nrael" };
    int m = arr2.Length;

    count(arr1, arr2, n, m);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program to count the number
// of anagrams of each String in a
// given array present in
// another array

function sortString(inputString)
{

    // Convert input string to char array
    let tempArray = inputString.split('');

    // Sort tempArray
    tempArray.sort();

    // Return new sorted string
    return tempArray.join("");
}

// Function to return the
// count of anagrams
function count(arr1, arr2, n, m)
{

    // Store the frequencies
    // of Strings in arr1[]
    let freq = new Map();

    for(let i = 0; i < n; i++)
    {

        // Sort the String
        arr1[i] = sortString(arr1[i]);

        // Increase its frequency
        // in the map
        if (freq.has(arr1[i]))
        {
            freq.set(arr1[i],
            freq.get(arr1[i]) + 1);
        }
        else
        {
            freq.set(arr1[i], 1);
        }
    }

    for(let i = 0; i < m; i++)
    {

        // Sort the String
        arr2[i] = sortString(arr2[i]);

        // Display its anagrams
        // in arr1[]
        document.write(freq.get(arr2[i]) + " ");
    }
}

// Driver code

    let arr1 = [ "geeks", "learn",
                      "for", "egeks",
                      "ealrn" ];
    let n = arr1.length;

    let arr2 = [ "kgees", "rof",
                      "nrael" ];
    let m = arr2.length;

    count(arr1, arr2, n, m);

 // This code is contributed by souravghosh0416.
</script>

Output:

2 1 2

Time Complexity: O(n*plog(p)+m*qlog(q)) where n and m are the sizes of both the array and p and q are the maximum size of string present in arr1 and arr2 respectively.
Auxiliary space: O(n+m). 

Please suggest if someone has a better solution that is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above