Generate Anagram Substrings - Amazon Top Interview Questions


Problem Statement :


You are given a lowercase alphabet string s. Return all substrings in s where there is another substring in s at a different location that is an anagram. Return the list sorted in lexicographic order.

Constraints

1 ≤ n ≤ 100 where n is the length of s

Example 1

Input

s = "aba"

Output

["a", "a", "ab", "ba"]

Explanation

We have "a" and "a" which are anagrams. Also, "ab" and "ba" are anagrams.


Solution :



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                        Solution in C++ :

vector<string> solve(string s) {
    map<string, vector<int>> mp;
    int i, j, len, n = s.size();
    for (i = 0; i < n; i++) {
        for (len = 0; i + len <= n; len++) {
            string x = s.substr(i, len);
            sort(x.begin(), x.end());
            if (mp.count(x))
                mp[x].emplace_back(i);
            else {
                mp[x] = {i};
            }
        }
    }
    vector<string> ret;
    for (auto &rec : mp) {
        if (rec.second.size() > 1) {
            auto &v = rec.second;
            for (i = 0; i < v.size(); i++) {
                string tmp = (s.substr(v[i], rec.first.size()));
                if (tmp.size()) ret.emplace_back(tmp);
            }
        }
    }
    sort(ret.begin(), ret.end());
    return ret;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public String[] solve(String s) {
        Map<String, List<String>> map = new HashMap<>();

        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                char[] arr = s.substring(i, j + 1).toCharArray();
                Arrays.sort(arr);

                String key = String.valueOf(arr);

                List<String> list = map.getOrDefault(key, new ArrayList<>());
                list.add(s.substring(i, j + 1));

                map.putIfAbsent(key, list);
            }
        }
        List<String> res = new ArrayList<>();

        for (List<String> list : map.values())
            if (list.size() > 1)
                res.addAll(list);

        String[] ans = new String[res.size()];
        for (int i = 0; i < ans.length; i++) ans[i] = res.get(i);
        Arrays.sort(ans);

        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s):
        found = defaultdict(list)
        n = len(s)

        for i in range(n):
            for j in range(i, n):
                curr = s[i : j + 1]

                key = "".join(sorted(curr))

                found[key].append(curr)

        ans = []
        for vals in found.values():
            if len(vals) > 1:
                for v in vals:
                    ans.append(v)

        return sorted(ans)


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