Algorithm
Problem Name: 30 days of code -
In this HackerRank in 30 Days of Code - problem solution,
Objective
Today we will expand our knowledge of strings, combining it with what we have already learned about loops. Check out the Tutorial tab for learning materials and an instructional video.
Task
Given a string, S, of length N of length 0 to N - 1 , print its even-indexed and odd-indexed characters as 2 space-separated strings on a single line (see the Sample below for more detail).
Note: 0 is considered to be an even index.
Example
s = abdecf
Print abc def
Input Format
The first line contains an integer, T (the number of test cases).
Each line i of the T subsequent lines contain a string, S.
Constraints
- 1 <= T <= 10
- 2 <= length of S <= 10000
Output Format
For each String Sj (where 0 <= j <= T - 1 ), print Sj's even-indexed characters, followed by a space, followed by Sj's odd-indexed characters.
Sample Input
2
Hacker
Rank
Sample Output
Hce akr
Rn ak
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int tests;
scanf("%d", &tests);
for (int i = 0; i < tests; i++) {
char s[10000];
scanf("%s", s);
int j = 0;
while (s[j] != '\0') {
if (j % 2 == 0) {
printf("%c", s[j]);
}
j++;
}
printf(" ");
j = 0;
while (s[j] != '\0') {
if (j % 2 != 0) {
printf("%c", s[j]);
}
j++;
}
printf("\n");
}
return 0;
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
int main() {
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string str;
cin >> str;
for (int j = 0; j < str.length(); j++) {
if (j % 2 == 0) cout << str[j];
}
cout << " ";
for (int j = 0; j < str.length(); j++) {
if (j % 2 != 0) cout << str[j];
}
cout << endl;
}
return 0;
}
#3 Code Example with C# Programming
Code -
C# Programming
using System;
class Solution
{
static void Main(String[] args)
{
var N = int.Parse(Console.ReadLine());
for (var i = 0; i < N; i++)
{
var str = Console.ReadLine();
for (var j = 0; j < str.Length; j++)
{
if (j % 2 == 0) Console.Write(str[j]);
}
Console.Write(" ");
for (var j = 0; j < str.Length; j++)
{
if (j % 2 != 0) Console.Write(str[j]);
}
Console.Write(Environment.NewLine);
}
}
}
#4 Code Example with Golang Programming
Code -
Golang Programming
package main
import (
"bufio"
"fmt"
"os"
"strings"
)
const (
EVEN = iota
ODD
)
func print(oddOrEven int, words string) {
for i := oddOrEven; i < len(words); i += 2 {
fmt.Printf("%c", words[i])
}
fmt.Print(" ")
}
func main() {
var T int
fmt.Scan(&T)
words := make([]string, T)
reader := bufio.NewReader(os.Stdin)
for i := 0; i < T; i++ {
words[i], _ = reader.ReadString('\n')
words[i] = strings.TrimSuffix(words[i], "\n")
}
for k := range words {
print(EVEN, words[k])
print(ODD, words[k])
fmt.Printf("\n")
}
}
#5 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
in.nextLine();
for (int i = 0; i < N; i++) {
String string = in.nextLine();
char[] charArray = string.toCharArray();
for (int j = 0; j < charArray.length; j++) {
if (j % 2 == 0) {
System.out.print(charArray[j]);
}
}
System.out.print(" ");
for (int j = 0; j < charArray.length; j++) {
if (j % 2 != 0) {
System.out.print(charArray[j]);
}
}
System.out.println();
}
in.close();
}
}
#6 Code Example with Javascript Programming
Code -
Javascript Programming
function processData(input) {
//Enter your code here
input = input.split('\n')
for(let i = 1; i < input.length; i++){
var splitWord = input[i].split('');
var even = '';
var odd = '';
for(x = 0; x < splitWord.length; x++){
if(x%2 == 0){
even = even + splitWord[x];
}else{
odd = odd +splitWord[x];
}
}
console.log(even + ' '+ odd);
}
}
#7 Code Example with Python Programming
Code -
Python Programming
# First we have to take the input of the number of Strings
NumberOfStrings = int(input())
# for loop from 0 to the length of the String
for i in range(0, NumberOfStrings):
# Taking input from the User
string = input()
# The below line has two parts 1. string[::2] & 2. string[1::2].
# General format is [start:stop:step].
# 1. string[::2] basically means that start from 0 to the end of the String skipping 2 characters hence taking all even strings
# 2. string[1::2] same as the above but we start from 1 and not 0
print(string[::2],string[1::2])
#8 Code Example with PHP Programming
Code -
PHP Programming
0 && $j % 2 != 0 ) {
$oddString .= $s[$j];
}
}
echo $evenString . ' ' . $oddString . "\n";
}
fclose( $_fp );
?>