PROBLEM DESCRIPTION
You are given a Big String and an array of small strings. You need to check whether each small string is present in the big string. Do not use any in-built method.
SOLUTION
We can make use of Trie to solve this problem. We loop through the bigstring and keep adding the subarrays [0,n-1] [1,n-1], [2,n-1] and so on. Then, all we need is to check whether the smallstring is present in the Trie as a prefix.
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import java.util.*;
class Program {
public static List<Boolean> multiStringSearch(String bigString, String[] smallStrings) {
Trie trie = new Trie();
for(int i=0; i<bigString.length(); i++){
trie.insert(bigString.substring(i));
}
List<Boolean> list = new ArrayList<>();
for(int i=0; i<smallStrings.length; i++){
list.add(i, trie.startsWith(smallStrings[i]));
}
return list;
}
}
class Trie {
Node root;
class Node{
boolean isEnd;
Map<Character, Node> hm;
Node(Map<Character, Node> hm){
this.hm = hm;
}
}
public Trie() {
root = new Node(new HashMap<>());
}
public void insert(String word) {
int n = word.length();
Node temp = root;
for(int i=0; i<n; i++){
Character c = Character.valueOf(word.charAt(i));
if(temp.hm.containsKey(c)){
temp = temp.hm.get(c);
}else{
temp.hm.put(c, new Node(new HashMap<>()));
temp = temp.hm.get(c);
}
}
temp.isEnd = true;
}
//Return true if complete match is found
public boolean search(String word) {
int n = word.length();
Node temp = root;
for(int i=0; i<n; i++){
Character c = Character.valueOf(word.charAt(i));
if(!temp.hm.containsKey(c)){
return false;
}else{
temp = temp.hm.get(c);
}
}
return temp.isEnd;
}
//Returns true is the input string is found (can be a prefix)
public boolean startsWith(String prefix) {
int n = prefix.length();
Node temp = root;
for(int i=0; i<n; i++){
Character c = Character.valueOf(prefix.charAt(i));
if(!temp.hm.containsKey(c)){
return false;
}else{
temp = temp.hm.get(c);
}
}
return true;
}
}