## Algorithm

Problem Name: 30 days of code - Day 8: Dictionaries and Maps

In this HackerRank in 30 Days of Code - Day 8: Dictionaries and Maps problem solution,

Objective
Today, we're learning about Key-Value pair mappings using a Map or Dictionary data structure. Check out the Tutorial tab for learning materials and an instructional video!

Given

Objective
Today, we're learning about Key-Value pair mappings using a Map or Dictionary data structure. Check out the Tutorial tab for learning materials and an instructional video!

Given n names and phone numbers, assemble a phone book that maps friends' names to their respective phone numbers. You will then be given an unknown number of names to query your phone book for. For each name queried, print the associated entry from your phone book on a new line in the form `name=phoneNumber`; if an entry for name is not found, print `Not found` instead.

Note: Your phone book should be a Dictionary/Map/HashMap data structure.

Input Format

The first line contains an integer, n, denoting the number of entries in the phone book.
Each of the n subsequent lines describes an entry in the form of 2 space-separated values on a single line. The first value is a friend's name, and the second value is an 8-digit phone number.

After the n lines of phone book entries, there are an unknown number of lines of queries. Each line (query) contains a name to look up, and you must continue reading lines until there is no more input.

Note: Names consist of lowercase English alphabetic letters and are first names only.

Constraints

• 1 <= n <= 10**5
• 1 <= queries <= 10**5

Output Format

On a new line for each query, print `Not found` if the name has no corresponding entry in the phone book; otherwise, print the full name and PhoneNumber in the format `name=phoneNumber`.

Sample Input

``````3
sam 99912222
tom 11122222
harry 12299933
sam
edward
harry
``````

Sample Output

``````sam=99912222
harry=12299933
``````

Explanation

We add the following n = 3 (Key,Value) pairs to our map so it looks like this:

phoneBook = {(sam,99912222),(tom,11122222),(harry,12299933)}
We then process each query and print `key=value` if the queried key is found in the map; otherwise, we print `Not found`.

Query 0: `sam`
Sam is one of the keys in our dictionary, so we print `sam=99912222`.

Query 1: `edward`
Edward is not one of the keys in our dictionary, so we print `Not found`.

Query 2: `harry`
Harry is one of the keys in our dictionary, so we print `harry=12299933`.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <map>
#include <iostream>

using namespace std;

int main() {
int N;
string name;
cin >> N;

map < string, int> phone_book;

for (int i = 0; i  <  N; i++) {
cin >> name;

if (!phone_book[name]) {
cin >> phone_book[name];
}
}

while (cin>>name) {

if (phone_book[name]) {
cout << name << "=" << phone_book[name] << endl;
} else {
}
}

return 0;
}
``````
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### #6 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <strings.h>

typedef struct pair {
char* first;
char* second;
struct pair* next;
} pair;

typedef struct dict {
int size;
pair** table;
} dict;

unsigned int hash(char* s) {
unsigned int hashval = 1337;
for (int i=0; i < (int)strlen(s); i++) {
hashval = hashval * s[i] + 0xdeadbeef;
hashval %= 0x3f3f3f3f;
}
return hashval;
}

void setsize(dict* d, int s) {
d->table = malloc(s * sizeof(pair*));
d->size = s;
bzero(d->table, s * sizeof(pair*));
}

void insert(dict* d, char* k, char* v) {
pair* p = malloc(sizeof(pair));

char* s = malloc(strlen(k) * sizeof(char) + 1);
char* t = malloc(strlen(v) * sizeof(char) + 1);
strcpy(s, k);
strcpy(t, v);

p->first = s;
p->second = t;
p->next = NULL;

unsigned int hashval = hash(k);
if (d->table[hashval % d->size] == NULL)
d->table[hashval % d->size] = p;
else {
pair* q = d->table[hashval % d->size];
while (q->next)
q = q->next;
q->next = p;
}
}

char* retreive(dict* d, char* k) {
unsigned int hashval = hash(k);
pair* p = d->table[hashval % d->size];

if (!p) return NULL;

while (strcmp(p->first, k) != 0) {
if (p->next) p = p->next;
else return NULL;
}

return p->second;
}

int main(void) {
dict* d = malloc(sizeof(dict));
setsize(d, 10000007);

int T;
scanf("%d", &T);

char s[37];
char t[37];
while(T--) {
scanf("%s %s", s, t);
insert(d, s, t);
}

while(scanf("%s", s) != EOF) {
if (retreive(d, s) != NULL)
printf("%s=%s\n", s, retreive(d, s));
else
}

return 0;
}
``````
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### #2 Code Example with C# Programming

```Code - C# Programming```

``````
using System;
using System.Collections.Generic;

class Solution
{
static void Main(String[] args)
{
var phoneBook = new Dictionary < string, int>();

for (var i = 0; i  <  n; i++)
{
var name = entry[0];
var phone = int.Parse(entry[1]);
}

for (var i = 0; i  <  n; i++)
{
if (phoneBook.ContainsKey(name))
{
var phone = phoneBook[name];
Console.WriteLine(\$"{name}={phone}");
}
}
}
}
``````
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### #3 Code Example with Java Programming

```Code - Java Programming```

``````
import java.util.*;
import java.io.*;

class Solution{
public static void main(String []argh){
Scanner in = new Scanner(System.in);
int n = in.nextInt();

Map < String,Integer> myMap = new HashMap();

for(int i = 0; i  <  n; i++){
String name = in.next();
int phone = in.nextInt();
// Write code here
myMap.put(name, phone);
}
while(in.hasNext()){
String s = in.next();
// Write code here
if (myMap.get(s)!=null)
System.out.println(s + "=" + myMap.get(s) );
else
}
in.close();
}
}
``````
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### #4 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
function processData(input) {
input = input.split('\n');
var phoneBook =  [];

for(i = 1; i  < = parseInt(input[0]); i++) {
var contactArray = input[i].split(' ');
phoneBook[contactArray[0]] = contactArray[1];
}

for(i = (parseInt(input[0])+1); i  <  input.length; i++){
if(phoneBook[input[i]]) {
console.log(input[i] + '=' + phoneBook[input[i]]);
}else{
}
}
}
``````
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### #5 Code Example with Python Programming

```Code - Python Programming```

``````
n = int(input())
phonebook = dict()
for i in range(n):
line = input()
line = line.split()
phonebook[line[0]] = phonebook.get(line[0],line[1])

while 1:
try:
q = input()
if q in phonebook:
print(str(q) + "=" + str(phonebook[q]))
else: