Algorithm
Problem Name: 30 days of code -
In this HackerRank in 30 Days of Code - 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!
Task
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!
Task
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
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
Not found
harry=12299933
Explanation
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 {
cout << "Not found" << endl;
}
}
return 0;
}
#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
printf("Not found\n");
}
return 0;
}
#2 Code Example with C# Programming
Code -
C# Programming
using System;
using System.Collections.Generic;
class Solution
{
static void Main(String[] args)
{
var n = int.Parse(Console.ReadLine());
var phoneBook = new Dictionary < string, int>();
for (var i = 0; i < n; i++)
{
var entry = Console.ReadLine().Split(' ');
var name = entry[0];
var phone = int.Parse(entry[1]);
phoneBook.Add(name, phone);
}
for (var i = 0; i < n; i++)
{
var name = Console.ReadLine();
if (phoneBook.ContainsKey(name))
{
var phone = phoneBook[name];
Console.WriteLine($"{name}={phone}");
}
else Console.WriteLine("Not found");
}
}
}
#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
System.out.println("Not found");
}
in.close();
}
}
#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{
console.log('Not found');
}
}
}
#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:
print("Not found")
except:
break