# Contacts

### Problem Statement :

```We're going to make our own Contacts application! The application must perform two types of operations:

1 . add name, where name  is a string denoting a contact name. This must store name as a new contact in the application.
find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with  and print the count on a new line.

Given n sequential add and find operations, perform each operation in order.

Input Format

The first line contains a single integer, n , denoting the number of operations to perform.
Each line i of the n subsequent lines contains an operation in one of the two forms defined above.

Constraints

1  <=  n  <=  10 ^5
1  <=  | name |  <= 21
1  <=  | partial |  <= 21

It is guaranteed that name and partial contain lowercase English letters only.
The input doesn't have any duplicate name for the add operation.
Output Format

For each find partial operation, print the number of contact names starting with partial on a new line.

Sample Input

4
find hac
find hak
Sample Output

2
0```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <bits/stdc++.h>
using namespace std;
struct node{
int h[26],n,f;
};
node null;
vector<node>trie;
int i,j,p,q,act=0;
for(i=0;i<A.size();i++){
p=A[i]-'a';
if(!trie[act].h[p]){
trie[act].h[p]=trie.size();
trie.push_back(null);
}
act=trie[act].h[p];
trie[act].n++;
}
}
int findi(string &A){
int i,j,p,q,act=0;
for(i=0;i<A.size();i++){
p=A[i]-'a';
if(!trie[act].h[p])return 0;
act=trie[act].h[p];
}
return trie[act].n;
}
int main(){
int i,j,p,q,N;
string A,B;
null.n=null.f=0;
for(i=0;i<26;i++)null.h[i]=0;
trie.push_back(null);
cin>>N;
for(i=0;i<N;i++){
cin>>A>>B;
else cout<<findi(B)<<"\n";
}
}

In Java :

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

class TrieNode{
int count = 0;
TrieNode[] trieNode = new TrieNode[26];
}

class Trie{
TrieNode contacts = new TrieNode();

TrieNode temp = contacts;
temp.count++;

for(char c : contact.toCharArray()){

int index = c - 'a';
if(temp.trieNode[index] != null){
temp = temp.trieNode[index];
}
else{
temp.trieNode[index] = new TrieNode();
temp = temp.trieNode[index];
}
temp.count++;
}

}

public int find(String contact){

TrieNode temp = contacts;

for(char c : contact.toCharArray()){

int index = c -'a';
if(temp.trieNode[index]!=null){
temp = temp.trieNode[index];
}
else{
return 0;
}

}
return temp.count;
}
}

public class Solution {

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
int numContacts;
Scanner in = new Scanner(System.in);
int numOperations = in.nextInt();

Trie trie = new Trie();

for(int i = 0; i <= numOperations; i++){
String op = in.nextLine();

String spl[] = op.split(" ");

//System.out.println("Input:"+op);
}
if(spl[0].equals("find")){
//System.out.println("finding"+spl[1]);
numContacts = trie.find(spl[1]);
System.out.println(numContacts);
}
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>
typedef struct node
{
bool isEOW;
int count;
struct node *letters[26];
} Trie;
Trie *createNode()
{
int i;
Trie *temp=malloc(sizeof(Trie));
temp->isEOW=false;
temp->count=0;
for(i=0; i<26; i++)
{
temp->letters[i]=NULL;
}
return temp;
}
Trie *insert(Trie *root,char *name)
{
int i;
Trie *temp=root;
for(i=0; name[i]!='\0'; i++)
{
if(root->letters[name[i]-'a']==NULL)
root->letters[name[i]-'a']=createNode();
root=root->letters[name[i]-'a'];
root->count++;
}
root->isEOW=true;
return temp;
}
/*int countNames(Trie *root)
{
int i,count=0;
if(root->isEOW)
count++;
for(i=0; i<26; i++)
{
if(root->letters[i])
count=count+countNames(root->letters[i]);
}
return count;
}*/
int main()
{

int i;
long n;
Trie* root=createNode();
scanf("%ld",&n);
char a[5],name[22];
while(n--)
{
scanf("%s",a);
scanf(" %s",name);
root= insert(root,name);
else if(strcmp(a,"find")==0)
{
Trie *temp=root;
for(i=0; i<strlen(name); i++)
{
temp=temp->letters[name[i]-'a'];
if(!temp)
{
printf("0\n");
break;
}
}
if(i==strlen(name))
printf("%d\n",temp->count);
}
}
return 0;
}

In Python3 :

N = int(input())

trie = {}

# Base case
if word == "": return

# TODO: default dict
if word[0] in trie:
trie[word[0]]['count'] += 1
else:
trie[word[0]] = { 'count': 1 }

def find_word(trie, word):
if word[0] not in trie:
return 0
else:
# Base case
if len(word) == 1:
return trie[word[0]]['count']
else:
return find_word(trie[word[0]], word[1:])

for i in range(N):
line = input()
command, word = line.split()
elif command == "find":
print(find_word(trie, word))
else:
print("WTF")```
```

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