In this HackerRank Day 8 Dictionaries and maps 30 days of code problem you have 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 the name is not found, print Not found instead.


Day 8 Dictionaries and maps 30 days of code solution hackerrank


Problem solution in Python 2 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT

N=int(raw_input())
d={}
for i in range(N):
    tmp=raw_input()
    d[tmp]=raw_input()
#print d
for i in range(N):
    tmp=raw_input()
    if tmp in d:
        print str(tmp)+'='+str(d[tmp])
    else:
        print 'Not found'


Problem solution in Python 3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT

x = int(input())

dictt = {}

for i in range(x):
    text = input().split()
    dictt[text[0]] = text[1]

while True:
    try:
        inpt = input()
        if inpt in dictt:
            print(inpt+"="+dictt[inpt])
        else:
            print("Not found")
    except EOFError:
        break



Problem solution in java programming.

//Complete this code or write your own from scratch
import java.util.*;
import java.io.*;

class Solution{
    public static void main(String []argh){
        Map<String,Integer> phoneBook = new HashMap<String,Integer>();
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        in.nextLine();
        for(int i=0;i<n;i++){
            String name = in.nextLine();
            int phone = Integer.parseInt(in.nextLine());
            phoneBook.put(name, phone);
        }
        while(in.hasNext()){
            String s = in.nextLine();
            Integer phoneNumber = phoneBook.get(s);
            System.out.println(
                (phoneNumber != null) 
                ? s + "=" + phoneNumber 
                : "Not found"
            );
        }
        in.close();
    }
}


Problem solution in c++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    string temp;
    getline(cin, temp);
    int N = stoi(temp);
            
    map<string, string> phoneList;
    for(int n = 0; n < N; n++){
        string name;
        string number;
        getline(cin, name);
        getline(cin, number);
        phoneList.insert(std::pair<string, string>(name, number));
    }
    
    string name;
    while(getline(cin, name))
    {
        std::map<string, string>::iterator it;
        it = phoneList.find(name);
        if (it == phoneList.end()){
            cout << "Not found" << endl;
        } else {
            cout << name << "=" << it->second << endl;
        }
    }
    
    return 0;
}


Problem solution in 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;
}


Problem solution in Javascript programming.

function processData(input) {
    //Enter your code here
    var input = input.split('\n');
    var numLines = input[0];
    var phoneBook = {};
    
    for (var i = 1; i < numLines*2; i=i+2){
        // 1,2; 3,4; 5,6
        phoneBook[input[i]] = input[i+1];
    }
    
    for (var j = numLines*2 + 1; j < input.length; j++){
        if (input[j] in phoneBook) console.log(input[j] + '=' + phoneBook[input[j]]);
        else console.log('Not found');
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});