Find all substrings that are anagrams of another substring of the string S

Given a string S, the task is to find all the substrings in the string S which is an anagram of another different substring in the string S. The different substrings mean the substring starts at a different index.

Examples:

Input: S = “aba”
Output: a a ab ba
Explanation:
Following substrings are anagrams of another substring of the string S:

1. “a”: Substring “a” is anagram of the substring “a”(= {S[0]}).
2. “a”: Substring “a” is anagram of the substring “a”(= {S[2]}).
3. “ab”: Substring “ab” is anagram of the substring “ba”(= {S[1], S[2]}).
4. “ba”: Substring “ba” is anagram of the substring “ab”(= {S[0], S[2]}).

Input: S = “abcd”
Output: []

Approach: The given problem can be solved by using Hashing by storing the anagrams of each substring of the string S and printing the resultant substring. Follow the below steps to solve the given problem:

• Initialize a HashMap that stores all the anagrams of each substring of the string S.
• Generate all the possible substrings of S and for each substring, say str store the substring in the HashMap mapped with the key as the sorted string str.
• Traverse the HashMap and print all the strings associated with each key whose number of strings associated with each string is at least 1.

Below is the implementation of the above approach:

C++

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

// Function to find all the substrings
// whose anagram exist as a different
// substring in S
void findAnagrams(string S)
{

  // Stores the length of S

  int N = S.length();

  // Stores the lists of anagrams of
  // each substring of S
  map<string, vector<string>> mp;

  // Generate all substrings of S

  for (int i = 0; i < N; i++) {
    for (int j = i; j < N; j++) {
      string curr = "";

      // Store the current substring
      // of the string S
      for (int k = i; k < j + 1; k++) curr.push_back(S[k]);

      string key = curr;
      // Key is the sorted substring

      sort(key.begin(), key.end());

      // Add the sorted substring
      // to the dictionary
      mp[key].push_back(curr);
    }
  }

  // Iterate over values of map

  for (auto itr : mp) {

    // If length of list > 1
    if (itr.second.size() > 1) {

      // Print all the strings
      for (string str : itr.second) cout << str << " ";
    }
  }
}

// Driver Code
signed main() {
  string S = "aba";
  findAnagrams(S);
  return 0;
}

// This code is contributed by sdeadityasharma

Java

import java.util.*;

class Main
{

    // Function to find the anagrams of the substring
    public static void findAnagrams(String s)
    {
        // stores the length of s
        int n = s.length();

        // stores the lists of anagrams of
        // each substring of s
        HashMap<String, ArrayList<String> > mp
            = new HashMap<String, ArrayList<String> >();

        // Generate all the substrings of s
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                String curr = "";

                // Store the current substring
                // of the string s
                curr = s.substring(i, j + 1);

                String key = curr;

                // key is the sorted substring
                char[] chars = key.toCharArray();
                Arrays.sort(chars);
                key = new String(chars);

                // Add the sorted substring
                // to the dictionary
                if (mp.containsKey(key)) {
                    mp.get(key).add(curr);
                }
                else {
                    ArrayList<String> list
                        = new ArrayList<String>();
                    list.add(curr);
                    mp.put(key, list);
                }
            }
        }
        // Iterate over the values of map
        ArrayList<String> result = new ArrayList<String>();
        for (String itr : mp.keySet()) {
            if (mp.get(itr).size() > 1) {
                for (String str : mp.get(itr)) {
                    result.add(str);
                }
            }
        }
        String res = String.join(" ", result);
        System.out.println(res);
    }
    public static void main(String[] args)
    {
        String s = "aba";
        findAnagrams(s);
    }
} // this code is contributed by devendra

Python3

# Python program for the above approach

import collections

# Function to find all the substrings
# whose anagram exist as a different
# substring in S
def findAnagrams(S):

    # Stores the lists of anagrams of
    # each substring of S
    Map = collections.defaultdict(list)

    # Stores the length of S
    N = len(S)

    # Generate all substrings of S
    for i in range(N):
        for j in range(i, N):

            # Store the current substring
            # of the string S
            curr = S[i: j + 1]

            # Key is the sorted substring
            key = "".join(sorted(curr))

            # Add the sorted substring
            # to the dictionary
            Map[key].append(curr)

    # Store the final result
    result = []

    # Iterate over values of dictionary
    for vals in Map.values():

        # If length of list > 1
        if len(vals) > 1:

            # Print all the strings
            for v in vals:
                  print(v, end =" ")

# Driver Code

S = "aba"
findAnagrams(S)

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        string S = "aba";
        FindAnagrams(S);
    }

    // Function to find all the substrings
    // whose anagram exist as a different
    // substring in S
    static void FindAnagrams(string S)
    {
        // Stores the length of S
        int N = S.Length;

        // Stores the lists of anagrams of
        // each substring of S
        Dictionary<string, List<string>> mp = new Dictionary<string, List<string>>();

        // Generate all substrings of S
        for (int i = 0; i < N; i++)
        {
            for (int j = i; j < N; j++)
            {
                string curr = "";

                // Store the current substring
                // of the string S
                for (int k = i; k < j + 1; k++)
                {
                    curr += S[k];
                }

                string key = new string(curr.ToCharArray().OrderBy(c => c).ToArray());

                // Add the sorted substring
                // to the dictionary
                if (!mp.ContainsKey(key))
                {
                    mp.Add(key, new List<string>());
                }
                mp[key].Add(curr);
            }
        }

        // Iterate over values of map
        foreach (var itr in mp)
        {
            // If length of list > 1
            if (itr.Value.Count > 1)
            {
                // Print all the strings
                foreach (string str in itr.Value)
                {
                    Console.Write(str + " ");
                }
            }
        }
    }
}

Javascript

function findAnagrams(s)
{

    // stores the length of s
    let n = s.length;

    // stores the lists of anagrams of
    // each substring of s
    mp = {};

    // Generate all the substrings of s
    for(let i = 0; i < n; i++){
        for(let j = i; j < n; j++){
            let curr = "";

            // Store the current substring
            // of the string s
            curr = s.slice(i,j + 1);

            let key = curr;

            // console.log(curr);
            // key is the sorted substring
            key = key.split('').sort().join('');

            // Add the sorted substring
            // to the dictionary
            if (key in mp){
                mp[key].push(curr);
            }
            else{
                mp[key] = [curr];
            }
        }
    }

    // Iterate over the values of map
    result = [];
    for(let itr in mp){
        if(mp[itr].length > 1){
            for(let str of mp[itr]){
                result.push(str);
            }
        }
    }
    result = result.join(" ");
    console.log(result);
}

let s = "aba";
findAnagrams(s)

// This code is contributed by sdeadityasharma.

Output:

ab ba a a

Time Complexity: O(N³ log N)
Auxiliary Space: O(N²)