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 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++

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// 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[]
    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);
}

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# 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 



















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Output:


2 1 2







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