Hackerrank - Equalize the Array

Hackerrank – Popis problému

Celý popis zadania sa nacháza – Hackerrank.

Riešenie

Vytvoríme skupiny čísel, ktoré budú obsahovať rovnaké čísla. Výsledok je veľkosť poľa bez veľkosti najväčšej skupiny.

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 EqualizeTheArray {

    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int[] array = new int[n];
        for(int i = 0; i < n; i++) {
            array[i] = stdin.nextInt();
        }
        Map<Integer, Integer> numCount = new HashMap<>();
        for(int i = 0; i < n; i++) {
            int count = numCount.getOrDefault(array[i], 0);
            numCount.put(array[i], count + 1);
        }
        int maxCount = numCount.values().stream().max(
                (Integer o1, Integer o2) -> o1 - o2
        ).orElse(0);

        System.out.println(n - maxCount);
        stdin.close();
    }
}

import java.util.*;

public class EqualizeTheArray {

public static void main(String[] args) {

Scanner stdin = new Scanner(System.in);

int n = stdin.nextInt();

int[] array = new int[n];

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

array[i] = stdin.nextInt();

}

Map<Integer, Integer> numCount = new HashMap<>();

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

int count = numCount.getOrDefault(array[i], 0);

numCount.put(array[i], count + 1);

}

int maxCount = numCount.values().stream().max(

(Integer o1, Integer o2) -> o1 - o2

).orElse(0);

System.out.println(n - maxCount);

stdin.close();

}

JavaScript

'use strict';

const processData = input => {
    const lines = input.split('\n');
    const array = lines[1].split(' ');
    const numCounts = {};
    for(let i = 0; i < array.length; i++) {
        let count = numCounts[array[i]] || 0;
        numCounts[array[i]] = count + 1;
    }
    let maxCount = 0;
    for(let p in numCounts) {
        if(numCounts.hasOwnProperty(p) && numCounts[p] > maxCount) {
            maxCount = numCounts[p];
        }
    }
    console.log(array.length - maxCount);
};

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 array = lines[1].split(' ');

const numCounts = {};

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

let count = numCounts[array[i]] || 0;

numCounts[array[i]] = count + 1;

}

let maxCount = 0;

for(let p in numCounts) {

if(numCounts.hasOwnProperty(p) && numCounts[p] > maxCount) {

maxCount = numCounts[p];

}

console.log(array.length - maxCount);

};

process.stdin.resume();

process.stdin.setEncoding("ascii");

let _input = "";

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

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

Scala

import scala.io.Source

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

  val maxOccuredNumber = array.groupBy(n => n).map(_._2.length).max
  val toRemove = array.length - maxOccuredNumber
  println(toRemove)
}

import scala.io.Source

object EqualizeTheArray extends App {

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

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

val maxOccuredNumber = array.groupBy(n => n).map(_._2.length).max

val toRemove = array.length - maxOccuredNumber

println(toRemove)

}

Ruby

n = gets.strip.to_i
arr = gets.strip.split

num_counts = Hash.new
arr.each do |num|
  count = 0
    count = num_counts[num] if num_counts.has_key?(num)
  num_counts[num] = count + 1
end

max_count = 0
num_counts.each_key do |key|
  count = num_counts[key]
  max_count = count if count > max_count
end

puts n - max_count

n = gets.strip.to_i

arr = gets.strip.split

num_counts = Hash.new

arr.each do |num|

count = 0

count = num_counts[num] if num_counts.has_key?(num)

num_counts[num] = count + 1

end

max_count = 0

num_counts.each_key do |key|

count = num_counts[key]

max_count = count if count > max_count

end

puts n - max_count

Pavol Pidanič

Dokážem na 10 prstoch napočítať do 1023

Hackerrank – Equalize the Array