# HackerRank – 30 Days of Code – Day 8: Dictionaries and Maps – Ruby Program

## 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.

3
sam 99912222
tom 11122222
harry 12299933
sam
edward
harry

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.

Ruby Implementation:

```n = gets.to_i
phoneBook = Hash.new

## getting & storing data in phoneBook

n.times do
input = gets.split(" ")
phoneBook[input[0]] = input[1]
end

## find the phonenumber based on name

n.times do
name = gets.strip

if phoneBook.key? name
puts "#{name}=#{phoneBook[name]}"
else
end
end```

# HackerRank – 30 Days of Code – Day 7: Arrays – Ruby Program

## Objective

Today, we’re learning about the Array data structure. Check out the Tutorial tab for learning materials and an instructional video!

Given an array, A, of N integers, print A’s elements in reverse order as a single line of space-separated numbers.

### Input Format

The first line contains an integer, N(the size of our array).
The second line contains N space-separated integers describing array A’s elements.

### Constraints

• 1 <= N <= 1000
• 1 <- Ai <= 10000, where Ai is the ith integer in the array.
Output Format

Print the elements of array A in reverse order as a single line of space-separated numbers.

4
1 4 3 2

2 3 4 1

### Ruby Implementation

```#!/bin/ruby

n = gets.strip.to_i
arr = gets.strip
arr = arr.split(' ').map(&:to_i)

count = arr.size
op = ""
until (count == 0)
count -= 1
op += arr[count].to_s + " "
end

puts op
```

# HackerRank – 30 Days of Code – Day 6: Let’s Review – Ruby Program

## Objective

Today we’re expanding our knowledge of Strings and combining it with what we’ve already learned about loops. Check out the Tutorial tab for learning materials and an instructional video!

Given a string, S, of length N that is indexed from 0 to N-1, print its even-indexed and odd-indexed characters as 2 space-separated strings on a single line (see the Sample below for more detail).

Note: 0 is considered to be an even index.

### Input Format

The first line contains an integer, T (the number of test cases).
Each line of the T subsequent lines contain a String, S.

### Constraints

• 1 <= T <= 10
• 2 <= length of S <= 10000

### Output Format

For each String Sj(where 0<=j<=T-1), print Sj’s even-indexed characters, followed by a space, followed by Sj’s odd-indexed characters.

2
Hacker
Rank

Hce akr
Rn ak
Explanation

### Ruby Implementation

```T = gets.to_i
S = Array.new
Output = Array.new

T.times do | array_index |
S.push(gets.chomp)
end

T.times do | array_index |
temp_string = S[array_index]
string_count = S[array_index].to_s.length
odd = ""
even = ""

string_count.times do | string_index |
if(string_index %2 == 0)
even = even + temp_string[string_index]
else
odd = odd + temp_string[string_index]
end
end

Output.push(even + " " + odd)
end

puts Output```