Hackerrank - Divisible Sum Pairs

Hackerrank – Popis problému

Celý popis zadania sa nacháza – Hackerrank.

Riešenie

Vytvorím si kombinácie prvkov poľa, tak aby boli unikátne((i, j) je rovnaké ako (j, i). Pre každú dvojicu urobím ich súčet. Ak je deliteľná zadaným deliteľom, pridám ju do celkového počtu.

Vytvoril som riešenie v týchto programovacích jazykoch:

Scala
Java
JavaScript
Ruby

Všetky riešenia sú dostupné aj na mojom GitHub profile.

Java

import java.util.*;

public class DivisibleSumPairs {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int divisor = stdin.nextInt();
        int[] array = new int[n];
        for(int i = 0; i < n; i++) {
            array[i] = stdin.nextInt();
        }

        int count = 0;
        for(int i = 0; i < n - 1; i++) {
            for(int j = i + 1; j < n; j++) {
                if((array[i] + array[j]) % divisor == 0) {
                    count++;
                }
            }
        }
        System.out.println(count);
        stdin.close();
    }
}

import java.util.*;

public class DivisibleSumPairs {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int divisor = stdin.nextInt();

int[] array = new int[n];

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

array[i] = stdin.nextInt();

}

int count = 0;

for(int i = 0; i < n - 1; i++) {

for(int j = i + 1; j < n; j++) {

if((array[i] + array[j]) % divisor == 0) {

count++;

}

System.out.println(count);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const line1 = lines[0];
    const divisor = parseInt(line1.split(' ')[1]);
    const array = lines[1].split(' ').map(item => parseInt(item));
    let count = 0;
    for(let i = 0; i < array.length - 1; i++) {
        for(let j = i + 1; j < array.length; j++) {
            if((array[i] + array[j]) % divisor === 0) {
                count++;
            }
        }
    }
    console.log(count);
};

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

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

'use strict';

const processData = input => {

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

const line1 = lines[0];

const divisor = parseInt(line1.split(' ')[1]);

const array = lines[1].split(' ').map(item => parseInt(item));

let count = 0;

for(let i = 0; i < array.length - 1; i++) {

for(let j = i + 1; j < array.length; j++) {

if((array[i] + array[j]) % divisor === 0) {

count++;

}

console.log(count);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

object DivisibleSumPairs extends App {
  val lines = Source.stdin.getLines().toList
  val divisor = lines(0).split(" ").map(_.toInt).toList(1)
  val array = lines(1).split(" ").map(_.toInt)

  val indexPairs = array.indices.flatMap(i => array.indices.slice(i + 1, array.length).map(j => (i, j)))
  val divisible = indexPairs.count(pair => (array(pair._1) + array(pair._2)) % divisor == 0)
  println(divisible)
}

object DivisibleSumPairs extends App {

val lines = Source.stdin.getLines().toList

val divisor = lines(0).split(" ").map(_.toInt).toList(1)

val array = lines(1).split(" ").map(_.toInt)

val indexPairs = array.indices.flatMap(i => array.indices.slice(i + 1, array.length).map(j => (i, j)))

val divisible = indexPairs.count(pair => (array(pair._1) + array(pair._2)) % divisor == 0)

println(divisible)

}

Ruby

(n, divisor) = gets.split(" ").map {|i| i.to_i}
array = gets.split(" ").map {|i| i.to_i}
count = 0
for a in 0..(n-2) do
  for j in (a+1)..(n-1) do
    count += 1 if ((array[a] + array[j]) % divisor == 0)
  end
end

puts count

(n, divisor) = gets.split(" ").map {|i| i.to_i}

array = gets.split(" ").map {|i| i.to_i}

count = 0

for a in 0..(n-2) do

for j in (a+1)..(n-1) do

count += 1 if ((array[a] + array[j]) % divisor == 0)

end

puts count

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Hackerrank – Divisible Sum Pairs