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:
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.
- 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.
- Then iterate through the given string of array, sort the current string and check that string in the hashmap.
- 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.
- 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 occured 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; } |
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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 occured 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 |
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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 occured 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 |
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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 occured 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 |
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Output:
code geeks
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