Digital Dictionary

#1

Bhaiya isme trie bnali mene , ab aage kese karu?

#2

hello @CODER_JATIN

ab search functinality add karo , jisme pehle jo string type ki hai wo match karwao trie me aur, uske baad remaining part me dfs run karo

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#3

dfs kya hota hai bhaiya?

#4

graph nahi padha hao tumne abhi?

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#5

nahi ye to hashing or trie wale module m hai ye question

#6

ok , tum ye samjho ki tumhe traverse karna hai trie ko.

recursion ki help se

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#7

backtracking bhi use hoegi bhaiya?

#8

tumhari implementation pe depend karega.

agar string ko value se pass karoge to backtracking ki jarurat nahi hogi otherwise jarurat padegi

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#9

ab fir kese karu bhaiya ?
pehle graphs padhu?

#10

nahi , recusion to padha hi hua hai tumne . … . . . .
utna kaafi hai (dfs kuch alag chiz nahi hai , graph ya tree/trie ko depth wise order me traverse karne ko dfs traversal bolte hai)

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#11

thik hai bhaiya koshish krta hu krne ki

#12

https://ide.codingblocks.com/s/444953 , bhaiya dekhna , ab smjh nhi aa rha

#13

ok chalo main ek ek kar ke samjha ta hun.

	bool find(string w) {
		node * temp = root;
		for(int i = 0; w[i] != '\0'; i++) {
			char ch = w[i];
			if(temp->children.count(ch) == 0) {
				return false;
			}
			else 
				temp = temp->children[ch];
		}

		return temp->terminal;
	}

is ko dekho aur batao isme hum kya kar rahe hai

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#14

isme to ham bta rhe hai ki agar string hai to true return kar rhe ni to false

#15

ha hum dekh rahe hai ki given string hamare trie me hai ya nahi . . . .

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#16

Hanji bhaiya . . … .

#17 

	void insert(string w) {
		node * temp = root;
		for(int i = 0; w[i] != '\0'; i++) {
			char ch = w[i];
			if(temp->children.count(ch) != 0) {
				temp = temp->children[ch];
			}
			else {
				node * n = new node(ch);
				temp->children[ch] = n;
				temp = n;
                
			}

		}
		temp->terminal = true;
	}

```
iska batao
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#18

ye bhi aa gya smjh , insert kra rhe hai trie mai given string ko

#19

hanji bhaiya,ye done hai . .

#20 
    void PrintAll(node*temp,string out,vector<string>&ans){
		if(temp->terminal){
			ans.push_back(out);
		}

		for(auto it:temp->children){
			out.push_back(it.first);
			PrintAll(it.second,out,ans);
			out.pop_back();
		}
	}

ye batao yaha kya kar rahe hai

is diagram ko dekh ke batao (maan lo tum abhi a) pe ho

ss

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