Problem Statement A description of the problem can be found on Hackerrank. Solution Calculate count of chocolates: chocolates = money / price . Count of chocolates is count of wrappers, too. Free chocolates are calculated: divide chocolates with count of chocolates needed for discount add solution of the division to chocolates new wrappers count is: […]
Problem Statement A description of the problem can be found on Hackerrank. Solution Filter all matrix indexes that are at the periphery. For all the rest choose one index and compare its value with all neighbours (left, right, up, down). If it is greater than all neighbours, replace it with ‘X’. I created solution in […]
Problem Statement A description of the problem can be found on Hackerrank. Solution Find minimum in given array from index i to j. I created solution in 4 languages: Scala Java JavaScript Ruby All solutions are also available on my GitHub. Scala
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object ServiceLane extends App { val lines = io.Source.stdin.getLines().toList val lanes = lines(1).split(" ").map(_.toInt) val segments = lines.drop(2) val vehicles = segments.map(width(lanes, _)) println(vehicles.mkString("\n")) def width(lanes: Array[Int], inputLane: String): Int = { val arr = inputLane.split(" ").map(_.toInt) lanes.slice(arr(0), arr(1) + 1).min } } |
Java
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import java.util.Scanner; public class ServiceLane { private static final int BIKE = 1; private static final int TRUCK = 3; public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int laneLength = scanner.nextInt(); int testCases = scanner.nextInt(); int[] laneWidth = new int[laneLength]; for (int i = 0; i < laneLength; i++) { laneWidth[i] = scanner.nextInt(); } for (int j = 0; j < testCases; j++) { int laneEntry = scanner.nextInt(); int laneExit = scanner.nextInt(); System.out.println(largestVehicle(laneWidth, laneEntry, laneExit)); } scanner.close(); } private static int largestVehicle(int[] laneWidth, int laneEntry, int laneExit) { int car = TRUCK; while (car > BIKE && laneExit >= laneEntry) { if(laneWidth[laneEntry] < car) { car = laneWidth[laneEntry]; } laneEntry++; } return car; } } |
JavaScript
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function processData(input) { var lines = input.split('\n'); var lanes = lines[1].split(' ').map(function(item) {return parseInt(item)}); for(var i = 2; i < lines.length; i++) { var arr = lines[i].split(' ').map(function(item) {return parseInt(item)}); console.log(findMin(lanes, arr[0], arr[1])); } } function findMin(arr, i, j) { var min = 10000000; for(var k = i; k <= j; k++) { if(arr[k] < min) { min = arr[k]; } } return min; } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); }); |
Ruby
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(length, tests) = gets.split(" ").map{ |i| i.to_i } lanes = gets.split(" ").map{ |i| i.to_i } tests.times do arr = gets.split(" ").map{ |i| i.to_i } puts lanes.slice(arr[0], arr[1] - arr[0] + 1).min end |
Problem Statement A description of the problem can be found on Hackerrank. Solution Calculate a width and height of a required matrix: height = ceil(length(input_string) width = floor(length(input_string) Then create a height x width matrix containing a characters from input_string. Transpose the matrix and print all rows. Node: Instead we can print all columns of […]