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Add and Search Word - Data structure design in C++
Suppose we have to design a data structure that supports the following two operations −
addWord(word)
search(word)
The search(word) method can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter. So for example, if we add some words like “bad”, “dad”, “mad”, then search for search(“pad”) → false, search(“bad”) → true, search(“.ad”) → true and search(“b..”) → true
To solve this, we will follow these steps −
There are some methods, initially define insertNode(), this will take the head reference and the string s, this will work as follows −
curr := head, n := size of s
for i in range 0 to n – 1
x := s[i]
if child[x] of curr is not null, then child[x] of curr := new node
curr := child[x] of curr
set isEnd of curr as true
From the addWord(), call this insertNode() method
Define another method called check(), this will take curr, string s, and index. Initially index is 0. This will work as follows −
if index = size of s, then return isEnd of curr
set ok := true
if s[index] is dot, then
for i in range 0 to 25
x := ASCII of ‘a’ + i
if child[x] of curr AND check(child[x] of curr, s, index + 1) is true, then return true.
otherwise
x := s[index]
if child[x] of curr AND check(child[x] of curr, s, index + 1) is true, then return true.
return false
From the search method, set curr := head and return check(curr, word, 0)
Example (C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
struct Node{
bool isEnd;
map <char, Node*> child;
Node(){
isEnd = false;
}
};
class WordDictionary {
public:
Node* head;
WordDictionary() {
head = new Node();
}
void insertNode(Node* head, string s){
Node* curr = head;
int n = s.size();
for(int i = 0; i < n; i++){
char x = s[i];
if(!curr->child[x]){
curr->child[x] = new Node();
}
curr = curr->child[x];
}
curr->isEnd = true;
}
void addWord(string word) {
insertNode(head, word);
}
bool check(Node* curr, string s, int idx = 0){
if(idx == s.size()) return curr->isEnd;
bool ok = false;
if(s[idx] == '.'){
for(int i = 0; i < 26; i++){
char x = 'a' + i;
if(curr->child[x] && check(curr->child[x], s, idx + 1))return true;
}
} else {
char x = s[idx];
if(curr->child[x] && check(curr->child[x], s, idx + 1))return true;
}
return false;
}
bool search(string word) {
Node* curr = head;
return check(curr, word);
}
};
main(){
WordDictionary ob;
ob.addWord("bad");
ob.addWord("dad");
ob.addWord("mad");
cout << (ob.search("pad")) << endl;
cout << (ob.search("bad")) << endl;
cout << (ob.search(".ad")) << endl;
cout << (ob.search("b..")) << endl;
}Input
Initialize the WordDictionary, then call the addWord() methods ans search methods. See in the main() function
Output
0 1 1 1
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