Removing string that is an anagram of an earlier string

Given an array arr of strings, the task is to remove the strings that are an anagram of an earlier string, then print the remaining array in sorted order.

Examples:

Input: arr[] = { “geeks”, “keegs”, “code”, “doce” }, N = 4 
Output: [“code”, “geeks”] 
Explanation:
“geeks” and “keegs” are anagrams, so we remove “keegs”. 
Similarly, “code” and “doce” are anagrams, so we remove “doce”.

Input : arr[] = {“tea”, “ate”, “anagram”, “eat”, “gramaan”}, N = 5 
Output : [“anagram”, “tea”] 
Explanation: “ate” and “eat” are anagram of “tea”. 
“gramaan” is an anagram of “anagram” hence, array becomes [“anagram”, “tea”]. 

Approach: In order to check whether the given two strings are anagrams are not, we can simply sort both the strings and compare them. Also, to check if a string has occurred or not, we can use a hashmap. 

1. Create an auxiliary array to keep the resultant strings, and a hashmap to keep a mark of the string that we have found so far.
2. Then iterate through the given string of array, sort the current string and check that string in the hashmap.
3. If the current string is not found in the hashmap, then push arr[i] in the resultant array, and insert the sorted string in hashmap.
4. Finally, sort the resultant array and print each string.

Below is the implementation of above approach.

C++

// C++ implementation to remove
// all the anagram strings
#include <bits/stdc++.h>
using namespace std;

// Function to remove the anagram string
void removeAnagrams(string arr[], int N)
{
    // vector to store the final result
    vector<string> ans;

    // data structure to keep a mark
    // of the previously occurred string
    unordered_set<string> found;

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

        string word = arr[i];

        // Sort the characters
        // of the current string
        sort(begin(word), end(word));

        // Check if current string is not
        // present inside the hashmap
        // Then push it in the resultant vector
        // and insert it in the hashmap
        if (found.find(word) == found.end()) {

            ans.push_back(arr[i]);
            found.insert(word);
        }
    }

    // Sort the resultant vector of strings
    sort(begin(ans), end(ans));

    // Print the required array
    for (int i = 0; i < ans.size(); ++i) {
        cout << ans[i] << " ";
    }
}

// Driver code
int main()
{
    string arr[]
        = { "geeks", "keegs",
            "code", "doce" };
    int N = 4;

    removeAnagrams(arr, N);

    return 0;
}

Java

// Java implementation to remove
// all the anagram Strings
import java.util.*;

class GFG{

// Function to remove the anagram String
static void removeAnagrams(String arr[], int N)
{
    // vector to store the final result
    Vector<String> ans = new Vector<String>();

    // data structure to keep a mark
    // of the previously occurred String
    HashSet<String> found = new HashSet<String> ();

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

        String word = arr[i];

        // Sort the characters
        // of the current String
        word = sort(word);

        // Check if current String is not
        // present inside the hashmap
        // Then push it in the resultant vector
        // and insert it in the hashmap
        if (!found.contains(word)) {

            ans.add(arr[i]);
            found.add(word);
        }
    }

    // Sort the resultant vector of Strings
    Collections.sort(ans);

    // Print the required array
    for (int i = 0; i < ans.size(); ++i) {
        System.out.print(ans.get(i)+ " ");
    }
}
static String sort(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);
}

// Driver code
public static void main(String[] args)
{
    String arr[]
        = { "geeks", "keegs",
            "code", "doce" };
    int N = 4;

    removeAnagrams(arr, N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to remove
# all the anagram strings

# Function to remove the anagram string
def removeAnagrams(arr, N):

    # vector to store the final result
    ans = []

    # data structure to keep a mark
    # of the previously occurred string
    found = dict()

    for i in range(N):

        word = arr[i]

        # Sort the characters
        # of the current string
        word = " ".join(sorted(word))

        # Check if current is not
        # present inside the hashmap
        # Then push it in the resultant vector
        # and insert it in the hashmap
        if (word not in found):

            ans.append(arr[i])
            found[word] = 1

    # Sort the resultant vector of strings
    ans = sorted(ans)

    # Print the required array
    for i in range(len(ans)):
        print(ans[i], end=" ")

# Driver code
if __name__ == '__main__':
    arr=["geeks", "keegs","code", "doce"]
    N = 4

    removeAnagrams(arr, N)

# This code is contributed by mohit kumar 29

C#

// C# implementation to remove
// all the anagram Strings
using System;
using System.Collections.Generic;

class GFG{

// Function to remove the anagram String
static void removeAnagrams(String []arr, int N)
{
    // vector to store the readonly result
    List<String> ans = new List<String>();

    // data structure to keep a mark
    // of the previously occurred String
    HashSet<String> found = new HashSet<String> ();

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

        String word = arr[i];

        // Sort the characters
        // of the current String
        word = sort(word);

        // Check if current String is not
        // present inside the hashmap
        // Then push it in the resultant vector
        // and insert it in the hashmap
        if (!found.Contains(word)) {

            ans.Add(arr[i]);
            found.Add(word);
        }
    }

    // Sort the resultant vector of Strings
    ans.Sort();

    // Print the required array
    for (int i = 0; i < ans.Count; ++i) {
        Console.Write(ans[i]+ " ");
    }
}
static String sort(String inputString)
{
    // convert input string to char array
    char []tempArray = inputString.ToCharArray();

    // sort tempArray
    Array.Sort(tempArray);

    // return new sorted string
    return String.Join("",tempArray);
}

// Driver code
public static void Main(String[] args)
{
    String []arr
        = { "geeks", "keegs",
            "code", "doce" };
    int N = 4;

    removeAnagrams(arr, N);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation to remove
// all the anagram Strings

// Function to remove the anagram String
function removeAnagrams(arr, N)
{

    // Vector to store the final result
    let ans = [];

    // Data structure to keep a mark
    // of the previously occurred String
    let found = new Set();

    for(let i = 0; i < N; i++)
    {
        let word = arr[i];

        // Sort the characters
        // of the current String
        word = sort(word);

        // Check if current String is not
        // present inside the hashmap
        // Then push it in the resultant vector
        // and insert it in the hashmap
        if (!found.has(word))
        {
            ans.push(arr[i]);
            found.add(word);
        }
    }

    // Sort the resultant vector of Strings
    (ans).sort();

    // Print the required array
    for(let i = 0; i < ans.length; ++i)
    {
        document.write(ans[i] + " ");
    }
}

function sort(inputString)
{

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

    // Sort tempArray
    (tempArray).sort();

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

// Driver code
let arr = [ "geeks", "keegs",
            "code", "doce" ];
let N = 4;

removeAnagrams(arr, N);

// This code is contributed by unknown2108

</script>

Output:

code geeks

Time Complexity: O(n*mlogm) where n is the size of the array and m is the length of the word.

Auxiliary space: O(n). 
Please suggest if someone has a better solution which 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