Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created an accumulator with initial value as empty string. I iterated through each character and updated the accumulator if an actual character is equals to last accumulator character, delete this character from the accumulator else append the character to […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created a pointer for an actual position which starts at index 0. I always tried to jump at position i + 2. If it is not possible because of thundercloud I step at i + 1 position. I do […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I replaced all occurences of “010” with an arbitrary placeholder. Then I counted the placeholders. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
import java.util.*; public class BeautifulBinaryString { public static void main(String[] args) { Scanner stdin = new Scanner(System.in); int l = Integer.parseInt(stdin.nextLine()); String s = stdin.nextLine(); String beautiful = s.replaceAll("010", "b"); int changes = beautiful.replaceAll("[01]", "").length(); System.out.println(changes); stdin.close(); } } |
JavaScript
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
'use strict'; const processData = input => { const lines = input.split('\n'); const s = lines[1]; const beautiful = s.replace(/010/g, "b"); const changes = beautiful.replace(/[01]/g, "").length; console.log(changes); }; process.stdin.resume(); process.stdin.setEncoding("ascii"); let _input = ""; process.stdin.on("data", input => _input += input); process.stdin.on("end", () => processData(_input)); |
Scala
|
1 2 3 4 5 6 7 8 9 10 11 12 |
object BeautifulBinaryString { def main(args: Array[String]) { val sc = new java.util.Scanner (System.in) val n = sc.nextInt() val B = sc.next() val beautiful = B.replaceAll("010", "b") val changes = beautiful.count(_ == 'b') println(changes) } } |
Ruby
|
1 2 3 4 5 6 |
l = gets.to_i s = gets.to_s.strip s.gsub!(/010/, "b") s.gsub!(/[01]/, "") puts s.length() |
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution We rotate the array in circles. That why it is enough to calculate a position after all rotations using modulo. We split the array into two halves at position array_length – calculated_position. We change halves and concat them – the […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I jumped for k positions and lowered the energy. If I reached position 0 I printed the remaining enenergy. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
import java.util.*; public class JumpingOnTheCloudsRevisited { public static void main(String[] args) { Scanner stdin = new Scanner(System.in); int n = stdin.nextInt(); int k = stdin.nextInt(); int[] clouds = new int[n]; for(int i = 0; i < n; i++) { clouds[i] = stdin.nextInt(); } int energy = 100; int pos = k % n; energy -= ((clouds[pos] == 1) ? 3 : 1); while(pos != 0) { pos = (pos + k) % n; energy -= ((clouds[pos] == 1) ? 3 : 1); } System.out.println(energy); stdin.close(); } } |
JavaScript
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
'use strict'; const processData = input => { const lines = input.split('\n'); const [n, k] = lines[0].split(' ').map(i => parseInt(i)); const clouds = lines[1].split(' ').map(i => parseInt(i)); let energy = 100; let pos = k % n; energy -= ((clouds[pos] == 1) ? 3 : 1); while(pos != 0) { pos = (pos + k) % n; energy -= ((clouds[pos] == 1) ? 3 : 1); } console.log(energy); }; process.stdin.resume(); process.stdin.setEncoding("ascii"); let _input = ""; process.stdin.on("data", input => _input += input); process.stdin.on("end", () => processData(_input)); |
Scala […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I divided the whole string into to 3-character parts. Then I checked for each triplet if first character is equal to S, second to O and third to S. And counted the changed letters. I created solution in: Scala Java […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Create index pairs variations for all indexes that pair are unique – it means when (i, j) exists (j, i) is the same. Filter only this pairs of indexes, which have the same value. Make an absolute difference of them […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution The answer is the count of distinct characters. If the string does not contain a specific character, only then it costs 1 dollar. Because one character is a substring. When we are able to find it in a partially created […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution (((sweets % prisoners) + startId) % prisoners) – 1 Modulo finds the id of prisoner who gets the last poisoned candy after all circles distribution. Since the starting id S is not from first prisoner we make an offset. Do […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution There are two checks: I checked if is less then and immediatelly print “NO”. That means, kangaroo 1 would be never able to reach kangaroo 2. I calculated differences between kangaroo positions and their […]