Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Create groups containing same numbers from defined array. The result is length of the array without size of the biggest groups. Explanation: The only way to make a minimum deletion of other numbers from array. If I would delete any […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I created groups of socks with same color. Then for each of the group I calculated sock pairs – number_of_socks / 2. The result is the count of all pairs. I created solution in: Scala Java JavaScript Ruby All solutions […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Find a max height of given characters. Square is max height multiplied with length of the string. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java
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import java.util.*; public class Solution { public static void main(String[] args) { Scanner stdin = new Scanner(System.in); Object[] heights = Arrays.stream(stdin.nextLine().split(" ")) .map(Integer::parseInt).toArray(); String word = stdin.nextLine(); int maxHeight = 0; for(int i = 0; i < word.length(); i++) { int charIndex = word.charAt(i) - 'a'; int height = (int) heights[charIndex]; if(height > maxHeight) { maxHeight = height; } } long square = word.length() * maxHeight; System.out.println(square); stdin.close(); } } |
JavaScript
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'use strict'; const processData = input => { const lines = input.split('\n'); const heights = lines[0].split(' ').map(i => parseInt(i)); const word = lines[1]; let maxHeight = 0; for(let i = 0; i < word.length; i++) { const charIndex = word[i].codePointAt() - 97; const height = heights[charIndex]; if(height > maxHeight) { maxHeight = height; } } const square = word.length * maxHeight; console.log(square); }; process.stdin.resume(); process.stdin.setEncoding("ascii"); let _input = ""; process.stdin.on("data", input => _input += input); process.stdin.on("end", () => processData(_input)); |
Scala
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import scala.io.Source object DesignerPdfViewer extends App { val lines = Source.stdin.getLines().toList val heights = lines.head.split(" ").map(_.toInt) val word = lines(1) val characterIndexes = word.map(_ - 'a') val maxHeight = characterIndexes.map(heights(_)).max val square = word.length * maxHeight println(square) } } |
Ruby […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution A mathematical trick. just count total numbers of zero present in binary number of given n, and answer will be the 2 to the power of (total num of zero) Look at the truth table of XOR:
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0 ^ 0 = 0 0 ^ 1 = 1 1 ^ 0 = 1 1 ^ 1 = 0 |
Now look […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Fist I sorted the Space station, because they are not always ordered. Then I found out how many cities are between each pair of space stations i and i+1: Then I calculated the maximum distance for each city […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution For each number in the given interval between i and j count the numbers with valid given condition. I created solution in: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Java
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import java.util.*; public class Solution { public static void main(String[] args) { Scanner stdin = new Scanner(System.in); int start = stdin.nextInt(); int end = stdin.nextInt(); int k = stdin.nextInt(); int count = 0; for(int i = start; i <= end; i++) { String s = String.valueOf(i); String reversed = new StringBuilder(s).reverse().toString(); int diff = Math.abs(Integer.parseInt(s) - Integer.parseInt(reversed)); if(diff % k == 0) { count++; } } System.out.println(count); stdin.close(); } } |
JavaScript
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'use strict'; const reverse = (s) => s.split("").reverse().join(""); const processData = input => { const [start, end, d] = input.split(' ').map(i => parseInt(i)); let count = 0; for(let i = start; i <= end; i++) { const s = i.toString(); const reversed = reverse(s); const diff = Math.abs(parseInt(s) - parseInt(reversed)); if(diff % d == 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
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import scala.io.Source object BeautifulDaysAtTheMovies extends App { val nums = Source.stdin.bufferedReader().readLine().split(" ").map(_.toInt) val start = nums(0) val end = nums(1) val d = nums(2) val beautifulNum = (start to end).count(num => { val reversed = num.toString.reverse.toInt val diff = (num - reversed).abs diff % d == 0 }) println(beautifulNum) } |
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
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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
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'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
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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
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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 […]