No Prefix Set

Problem Statement :

There is a given list of strings where each string contains only lowercase letters from a - j, 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.

Function Description

Complete the noPrefix function in the editor below.

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

- 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 n, the size of words[].
Then next n lines each contain a string, words[ i ] .


1  <=   n  <=  10^5
1  <= the length of words[i]   <= 60
All letters in words[ i ] are in the range 'a' through 'j', inclusive.

Solution :


                            Solution in C :

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

struct node {
    char end;
    struct node * next[10];
} head;

int check_prefix(string str){
    struct node *nptr = &head; 
    int i;
    int index;
    int new_node = 0;

        index = ((int)str[i])%'a';
        if(nptr->next[index]==0) {
            nptr->next[index] = new node;
            nptr = nptr->next[index];
            memset (nptr,0,sizeof(struct node));
            new_node = 1;
        else {
            nptr = nptr->next[index];
        if(nptr->end == 1) {cout <<"BAD SET"<<endl; return 1;}
    if(new_node == 0) {cout <<"BAD SET"<<endl; return 1;}
    nptr->end = 1; 
    return 0;

int main() {
    string str;
    int n;
    int i;
    cin >>n;
    for (i=0;i<n;i++){
        cin >> str;
        if (check_prefix(str))  break;
    if(i!=n) cout << str <<endl;
    else cout << "GOOD SET" <<endl;
    return 0;

In Java :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

class Trie{
    char c;
    boolean isEndOfString;
    Trie []children;

public class Solution {

    Trie root;
    void input(){
        Scanner sin = new Scanner(;
        int N = sin.nextInt();
        root = new Trie();
        root.children = new Trie[26];                        
        boolean flag = false;
        for(int i = 0; i < N; i++){
            String s =;
                flag = insert(s);
           System.out.println("GOOD SET"); 
    Trie newNode(char c){
        Trie newNode = new Trie();
        newNode.c = c;
        newNode.isEndOfString = false;
        newNode.children = new Trie[26];
        return newNode;
    boolean insert(String s){
        Trie ptr = root;
        boolean flag = false;
        int i = 0;
        int ctr = 0;
        for(; i < s.length(); i++){
            char c = s.charAt(i);
            int index = (int)c - 97;
            if(ptr.children[index] == null){
                Trie newNode = newNode(c);
                ptr.children[index] = newNode;                
                ptr = ptr.children[index];
                ptr = ptr.children[index];
                    flag = true;
        if(i == s.length()){
            ptr.isEndOfString = true;    
            System.out.println("BAD SET");
        else if(ctr == s.length()){
            System.out.println("BAD SET");
            flag = true;
        return flag;
    public static void main(String[] args) {
        Solution s = new Solution();

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

typedef struct node
    int is_word;
    struct node* children[10];


NODE GetNewNode()
    NODE temp = (NODE)malloc(sizeof(struct node));
    temp->is_word = 0;
    int i;
    return temp;

void Insert(char word[])
    int digit;
    NODE temp;
        head = GetNewNode();
    int i=0;
        digit = word[i];
        temp = temp->children[digit];

int Search(char word[])
    int digit,found=0;

    NODE temp = head;
    if(temp==NULL)return 0;
    int i=0;
            return 0;
        if(temp->is_word==1)return 1;
    if(word[i]=='\0')//Check if we have scanned all characters....if we have, then prefix present       
   return found;

int main(void)
  char words[61];
  int number;
    int bad_set=0;
          printf("BAD SET\n%s",words);
        printf("GOOD SET");

In Python3 :

class solution:
    def checkSET(self, strs):
        trie = dict()
        for cstr in strs:
            curr = trie
            for ch in cstr:
                if ch not in curr:
                    curr[ch] = dict()
                curr = curr[ch]
                if '_end' in curr:
                    return cstr
            if len(curr) > 0:
                return cstr
            curr['_end'] = ''
        return ''
n = int(input())
strs = []
for i in range(n):
sol = solution()
ret = sol.checkSET(strs)
if ret == '':
    print('GOOD SET')
    print('BAD SET')

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