Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution First, I have found common characters in both strings (intersection of characters) and calculate the count. Then I haves substracted string one length and intersection chars count. Similar way substraction of string 2 length and intersection chars count. The result […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Look at the implementation. I created solution in: Java All solutions are also available on my GitHub profile. Java
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import java.util.*; public class Solution { public static void insertionSortPart2(int[] ar) { for(int i = 1; i < ar.length; i++) { int num = ar[i]; int swapPos = i - 1; int l = i - 1; boolean isSwap = false; while(l >= 0) { if(num < ar[l]) { isSwap = true; swapPos = l; l--; } else { break; } } if(isSwap) { for(int j = i - 1; j >= swapPos ; j--) { ar[j + 1] = ar[j]; } ar[swapPos] = num; } printArray(ar); } } public static void main(String[] args) { Scanner in = new Scanner(System.in); int s = in.nextInt(); int[] ar = new int[s]; for(int i=0;i<s;i++){ ar[i]=in.nextInt(); } insertionSortPart2(ar); } private static void printArray(int[] ar) { for(int n: ar){ System.out.print(n+" "); } System.out.println(""); } } |
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution Look at the implementation. I created solution in: Scala Java Javascript All solutions are also available on my GitHub profile. Scala
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import scala.io.Source object Solution{ def countingSort(arr: Array[Int]): Array[Int] = { val countsOfNumbers = arr.sorted.groupBy(i => i).map(a => (a._1, a._2.length)) (0 until 100).map(countsOfNumbers.getOrElse(_, 0)).toArray } def main(args: Array[String]) { val lines = Source.stdin.getLines().toList val arr = lines.tail.head.split(" ").map(_.toInt) val result = countingSort(arr) println (result.mkString(" ")) } } |
Java
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import java.util.*; public class Solution { static int[] countingSort(int[] arr) { int[] counts = new int[100]; for(int i = 0; i < arr.length; i++) { counts[arr[i]]++; } return counts; } public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] arr = new int[n]; for(int arr_i = 0; arr_i < n; arr_i++){ arr[arr_i] = in.nextInt(); } int[] result = countingSort(arr); for (int i = 0; i < result.length; i++) { System.out.print(result[i] + (i != result.length - 1 ? " " : "")); } System.out.println(""); in.close(); } } |
Javascript
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'use strict'; const counting = (arr) => { const counts = [] for(let i = 0; i < 100; i++) { counts[i] = 0; } for(let i = 0; i < arr.length; i++) { counts[arr[i]]++; } console.log(counts.join(' ')); } const processData = input => { const lines = input.split('\n'); const arr = lines[1].split(' ').map(i => parseInt(i)); counting(arr); }; process.stdin.resume(); process.stdin.setEncoding("ascii"); let _input = ""; process.stdin.on("data", input => _input += input); process.stdin.on("end", () => processData(_input)); |
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I explain my solution with following examples: Example 1234 I start with a beginning of a string, which would be one digit number – 1 and try to examine a rest of the given string 234. Next, I check the […]
Hackerrank – Problem Statement A description of the problem can be found on Hackerrank. Solution I started with finding all unique characters. Then I created all possible combinations of pairs. I deleted other characters except the pair. I found if the examined pair contains alternating characters in deleted strings If yes I count the length […]