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There is a given list of strings where each string contains only lowercase letters from , inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked.

Note If two strings are identical, they are prefixes of each other.

Example

Here ‘abcd’ is a prefix of ‘abcde’ and ‘bcd’ is a prefix of ‘bcde’. Since ‘abcde’ is tested first, print 

BAD SET  
abcde

.

No string is a prefix of another so print 

GOOD SET

Function Description
Complete the noPrefix function in the editor below.

noPrefix has the following parameter(s):
– string words[n]: an array of strings

Prints
– string(s): either GOOD SET or BAD SET on one line followed by the word on the next line. No return value is expected.

Input Format
First line contains , the size of .
Then next lines each contain a string, .

Constraints

the length of words[i]
All letters in are in the range ‘a’ through ‘j’, inclusive.

Sample Input00 

STDIN       Function
-----       --------
7            words[] size n = 7
aab          words = ['aab', 'defgab', 'abcde', 'aabcde', 'bbbbbbbbbb', 'jabjjjad']
defgab  
abcde
aabcde
cedaaa
bbbbbbbbbb
jabjjjad

Sample Output00 

BAD SET
aabcde

Explanation
‘aab’ is prefix of ‘aabcde’ so it is a BAD SET and fails at string ‘aabcde’.

Sample Input01

4
aab
aac
aacghgh
aabghgh

Sample Output01 

BAD SET
aacghgh

Explanation
‘aab’ is a prefix of ‘aabghgh’, and aac’ is prefix of ‘aacghgh’. The set is a BAD SET. ‘aacghgh’ is tested before ‘aabghgh’, so and it fails at ‘aacghgh’.

Limbajul de programare folosit: cpp14

Cod:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <string>
#include <memory>
#include <unordered_map>

using namespace std;

class Trie
{
    public:
        bool insert(const string& s)
        {
            bool prefixFound = false;
            Node* node = &root;
            for (char c : s)
            {
                auto found = node->children.find(c);
                if (found == node->children.end())
                {
                    if (node->isWord) {
                        prefixFound = true;
                    }
                    Node* newNode = new Node();

                    node->children[c] = unique_ptr<Node>(newNode);
                    node = newNode;
                }
                else
                {
                    node = found->second.get();
                }
            }
            if (node->isWord) {
                prefixFound = true;
            }
            node->isWord = true;
            return prefixFound || node->children.size() > 0;
        }

    private:
        struct Node
        {
            bool isWord = false;
            unordered_map<char, unique_ptr<Node> > children;
        };

        Node root;
};



int main() {
    int n;
    cin >> n;
    
    Trie trie;
    string s;
    while (n--) {
        cin >> s;
        if (trie.insert(s)) {
            cout << "BAD SET" << endl;
            cout << s << endl;
            return 0;
        }
    }
    cout << "GOOD SET" << endl;
    return 0;
}

Scor obtinut: 1.0

Submission ID: 464653295

Link challenge: https://www.hackerrank.com/challenges/no-prefix-set/problem

No Prefix Set