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 :
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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