Print n numbers using recursion
The following is the java method to print n numbers using recursion :- public static void print(int n) { if (n == 0) return; print(n - 1); System.out.print(n + " "); }
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Jewel Game (Hack The Interview V : Asia-Pacific)
Contest Link import java.util.*; class Solution{ public static void main(String[] args){ Scanner sc = new Scanner(System.in); int t = sc.nextInt(); while(t-->0){ String jewels = sc.next(); StringBuilder s = new StringBuilder(); s.append(jewels); int count = 0; while(true){ int temp = 0; for(int i=0;i<s.length()-1;i++){ if(s.charAt(i)==s.charAt(i+1)){ count++; ...
The XOR Problem (Hack The Interview V : Asia-Pacific)
Contest Link import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.function.*; import java.util.regex.*; import java.util.stream.*; import static java.util.stream.Collectors.joining; import static java.util.stream.Collectors.toList; class Result { /* * Complete the 'maxXorValue' function below. * * The function is expected to return a STRING. * The function accepts following parameters: * 1. STRING x * 2. INTEGER k */ public static String maxXorValue(String x, int k) { StringBuilder s = new StringBuilder(); int n = 0; int i = 0; for(i=0;i<x.length();i++){ if(n<k){ if(x.charAt(i)=='0'){ s.append('1'); n += 1; } else{ s.append('0'); } ...
New KeyBoard (Hack The Interview V : Asia-Pacific)
Contest Link import java.util.*; class Solution{ public static void main(String[] args){ Scanner sc = new Scanner(System.in); String s = sc.next(); StringBuilder res = new StringBuilder(""); boolean isPressed = false; int count = 0; boolean cursorStart = false; ArrayList<Character> c = new ArrayList<>(); for(int i=0;i<10;i++){ c.add((char)i); } for(int i=0;i<s.length();i++){ if(s.charAt(i)=='<'){ res = res.reverse(); cursorStart = true; continue; } else if(s.charAt(i)=='>'){ if(cursorStart == true){ res = res.reverse(); cursorStart = false; } continue; } else if(s.charAt(i)=='*'){ res.delete(res.length()-1, res.length()); ...
Hackerrank Solution :- print-the-elements-of-a-linked-list
Question Link import java.io.*; import java.math.*; import java.security.*; import java.text.*; import java.util.*; import java.util.concurrent.*; import java.util.regex.*; public class Solution { static class SinglyLinkedListNode { public int data; public SinglyLinkedListNode next; public SinglyLinkedListNode(int nodeData) { this.data = nodeData; this.next = null; } } static class SinglyLinkedList { public SinglyLinkedListNode head; public SinglyLinkedListNode tail; public SinglyLinkedList() { ...
Seive of Eratosthenis algorithm to generate prime numbers.
Sieve of Eratosthenis Algorithm:- 1. Create a list of consecutive integers from 2 to n: (2, 3, 4, …, n). 2. Initially, let p equal 2, the first prime number. 3. Starting from p2, count up in increments of p and mark each of these numbers greater than or equal to p2 itself in the list. These numbers will be p(p+1), p(p+2), p(p+3), etc.. 4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this number (which is the next prime), and repeat from step 3. <--> When the algorithm terminates, all the numbers in the list that are not marked false are prime. If you want the code for it then comment down.