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 Create a recursive function with condition described – see implementation. I created solution in: Scala All solutions are also available on my GitHub. Scala
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import scala.io.Source object SuperDigit extends App { val line = Source.stdin.bufferedReader().readLine() val nk = line.split(" ") val n = nk(0) val k = nk(1).toInt println(superDigit(n, k)) def superDigit(num: String, k: Int): String = { if(num.length == 1) { return num } else { val sum: BigInt = num.foldLeft(BigInt(0))(_ + Character.digit(_, 10)) * k return superDigit(sum.toString, 1) } } } |
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) } |