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Hackerrank – Ice Cream Parlor

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Pre každú i-tú cenovku zisti, či súčet i-tej a (i+1)-ej ceny je rovný požadovanej sume. Vytvoril som riešenie v týchto programovacích jazykoch: Java JavaScript Scala Ruby Všetky riešenia sú dostupné aj na mojom GitHub profile. Java

import java.util.*;

public class IceCreamParlor {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int tests = Integer.parseInt(stdin.nextLine());
        for(int i = 0; i < tests; i++) {
            int amount = Integer.parseInt(stdin.nextLine());
            int length = Integer.parseInt(stdin.nextLine());
            int[] costs = new int[length];
            String[] strings = stdin.nextLine().split(" ");
            for(int j = 0; j < length; j++) {
                costs[j] = Integer.parseInt(strings[j]);
            }
            for(int j = 0; j < length - 1; j++) {
                for(int k = j + 1; k < length; k++) {
                    if(costs[j] + costs[k] == amount) {
                        System.out.println((j + 1) + " " + (k + 1));
                    }
                }
            }
        }
        stdin.close();
    }
}

import java.util.*;

public class IceCreamParlor {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int tests = Integer.parseInt(stdin.nextLine());

for(int i = 0; i < tests; i++) {

int amount = Integer.parseInt(stdin.nextLine());

int length = Integer.parseInt(stdin.nextLine());

int[] costs = new int[length];

String[] strings = stdin.nextLine().split(" ");

for(int j = 0; j < length; j++) {

costs[j] = Integer.parseInt(strings[j]);

}

for(int j = 0; j < length - 1; j++) {

for(int k = j + 1; k < length; k++) {

if(costs[j] + costs[k] == amount) {

System.out.println((j + 1) + " " + (k + 1));

}

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    let lines = input.split('\n');
    let tests = parseInt(lines[0]);
    let index = 1;
    for(let i = 0; i < tests; i++) {
        let amount = parseInt(lines[index++]);
        let lenght = parseInt(lines[index++]);
        let costs = lines[index++].split(" ").map(i => parseInt(i));
        let indexes = [];
        for(let j = 0; j < lenght - 1; j++) {
            for(let k = j + 1; k < lenght; k++) {
                if(costs[j] + costs[k] === amount) {
                    indexes.push(j + 1);
                    indexes.push(k + 1);
                }
            }
        }
        console.log(indexes.join(" "));
    }
};

process.stdin.resume();
process.stdin.setEncoding("ascii");

var _input = "";
process.stdin.on("data", input => _input += input);
process.stdin.on("end", () => processData(_input));

'use strict';

const processData = input => {

let lines = input.split('\n');

let tests = parseInt(lines[0]);

let index = 1;

for(let i = 0; i < tests; i++) {

let amount = parseInt(lines[index++]);

let lenght = parseInt(lines[index++]);

let costs = lines[index++].split(" ").map(i => parseInt(i));

let indexes = [];

for(let j = 0; j < lenght - 1; j++) {

for(let k = j + 1; k < lenght; k++) {

if(costs[j] + costs[k] === amount) {

indexes.push(j + 1);

indexes.push(k + 1);

}

console.log(indexes.join(" "));

}

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

var _input = "";

process.stdin.on("data", input => _input += input);

process.stdin.on("end", () => processData(_input));

Scala

import scala.collection.mutable.ListBuffer
import scala.io.Source

object IceCreamParlor extends App {
  val console = Source.stdin.bufferedReader()
  val tests = console.readLine().toInt
  (0 until tests).foreach(_ => {
    val amount = console.readLine().toInt
    val lenght = console.readLine().toInt
    val costs = console.readLine().split(" ").map(_.toInt)
    println(indexes(costs, amount).mkString(" "))
  })

  def indexes(costs: Array[Int], amount: Int): List[Int] = {
    val result: ListBuffer[Int] = ListBuffer.empty
    (0 until costs.length - 1).foreach(i => {
      (i + 1 until costs.length).foreach(j => {
        if(costs(i) + costs(j) == amount) {
          result += i + 1
          result += j + 1
        }
      })
    })
    result.toList
  }
}

import scala.collection.mutable.ListBuffer

import scala.io.Source

object IceCreamParlor extends App {

val console = Source.stdin.bufferedReader()

val tests = console.readLine().toInt

(0 until tests).foreach(_ => {

val amount = console.readLine().toInt

val lenght = console.readLine().toInt

val costs = console.readLine().split(" ").map(_.toInt)

println(indexes(costs, amount).mkString(" "))

})

def indexes(costs: Array[Int], amount: Int): List[Int] = {

val result: ListBuffer[Int] = ListBuffer.empty

(0 until costs.length - 1).foreach(i => {

(i + 1 until costs.length).foreach(j => {

if(costs(i) + costs(j) == amount) {

result += i + 1

result += j + 1

}

})

result.toList

}

Ruby

tests = gets.chomp.to_i
tests.times do
  m = gets.chomp.to_i
  n = gets.chomp.to_i
  flavors = gets.chomp.split.map { |e| e.to_i }
  for i in 0...flavors.length - 1 do
    for j in i + 1...flavors.length do
      puts "#{i + 1} #{j + 1}" if flavors[i] + flavors[j] == m
    end
  end
end

tests = gets.chomp.to_i

tests.times do

m = gets.chomp.to_i

n = gets.chomp.to_i

flavors = gets.chomp.split.map { |e| e.to_i }

for i in 0...flavors.length - 1 do

for j in i + 1...flavors.length do

puts "#{i + 1} #{j + 1}" if flavors[i] + flavors[j] == m

end

Hackerrank – Weather Observation Station 5

Popis problému Celý popis zadania sa nacháza – Hackerrank. Riešenie Vytvoril som riešenie v týchto programovacích jazykoch: SQL Všetky riešenia sú dostupné aj na mojom GitHub profile. SQL

select * from
( 
    select city, length(city)
    from station
    order by length(city) asc, city asc
)
where rownum = 1
union 
select * from
( 
    select city, length(city)
    from station
    order by length(city) desc, city asc
)
where rownum = 1;

select * from

(

select city, length(city)

from station

order by length(city) asc, city asc

)

where rownum = 1

union

select * from

(

select city, length(city)

from station

order by length(city) desc, city asc

)

where rownum = 1;

Hackerrank – Weather Observation Station 6

select distinct city
from station
where lower(substr(city, 1, 1)) in ('a', 'e', 'i', 'o', 'u')
    order by city asc;

select distinct city

from station

where lower(substr(city, 1, 1)) in ('a', 'e', 'i', 'o', 'u')

order by city asc;

Hackerrank – Weather Observation Station 7

select distinct city
  from station
where 
  substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u')
  order by city;

select distinct city

from station

where

substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u')

order by city;

Hackerrank – Weather Observation Station 8

select distinct city
  from station
where 
  substr(lower(city), 1, 1) in ('a', 'e', 'i', 'o', 'u')
     AND
  substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u')
  order by city;

select distinct city

from station

where

substr(lower(city), 1, 1) in ('a', 'e', 'i', 'o', 'u')

AND

substr(lower(city), length(city), 1) in ('a', 'e', 'i', 'o', 'u')

order by city;

Hackerrank – Weather Observation Station 9

select distinct city
from station
where lower(substr(city, 1, 1)) not in ('a', 'e', 'i', 'o', 'u')
    order by city asc;

select distinct city

from station

where lower(substr(city, 1, 1)) not in ('a', 'e', 'i', 'o', 'u')

order by city asc;

Hackerrank – Weather Observation Station 10

select distinct city
  from station
where 
  substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')
  order by city;

select distinct city

from station

where

substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')

order by city;

Hackerrank – Weather Observation Station 11

select distinct city
  from station
where 
  substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u')
     or
  substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')
  order by city;

select distinct city

from station

where

substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u')

substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')

order by city;

Hackerrank – Weather Observation Station 12

select distinct city
  from station
where 
  substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u')
     and
  substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')
  order by city;

select distinct city

from station

where

substr(lower(city), 1, 1) not in ('a', 'e', 'i', 'o', 'u')

and

substr(lower(city), length(city), 1) not in ('a', 'e', 'i', 'o', 'u')

order by city;

Hackerrank – Higher Than 75 Marks

select name from 
(
    select name, substr(name, length(name)-2) as sub_name 
    from Students
    where marks > 75 order by sub_name, id asc
);

select name from

(

select name, substr(name, length(name)-2) as sub_name

from Students

where marks > 75 order by sub_name, id asc

);