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  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 6: Let's Review
  5. Discussions

Day 6: Let's Review

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3397 Discussions

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  • rohitshivdhaniy2
     + 0 comments

    const strsplit = input.split("\n").slice(1); Day 6 lets Review in JavaScript 

    for(let str of strsplit){
    let even = "";
    let odd = "";
   for(i=0; i<str.length; i++){
       if(i% 2 ==0){
           even+= str[i]
       }else{
           odd+= str[i]
       }
   }
   console.log(``${even} $`{odd}`)

}

  • adadasaivenkat01
     + 0 comments
    # Enter your code here. Read input from STDIN. Print output to STDOUT
n=int(input())
for i in range(n):
    l=list(input())
    for j in range(len(l)):
        if j%2==0:
            print(l[j],end="")
    print(end=" ")
    for k in range(len(l)):
        if k%2!=0:
            print(l[k],end="")
    print()

  • deekshithakandr1
     + 0 comments

    def string(s): even_char = s[::2] odd_char = s[1::2] return f"{even_char} {odd_char}" t = int(input()) for i in range(t): s = input() print(string(s))

  • lionheart_frozen
     + 0 comments

    My solution in C may not be the optimal one, but I enjoyed coding it

    #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

void procesar_cadena(char* str) {
    
    size_t lon = strlen(str);
    
    // Crear dos cadenas auxialiares para, impar
    char* str_p = malloc(sizeof(char)*lon);
    char* str_i = malloc(sizeof(char)*lon);

    if (!str_p || !str_i) exit(1);

    // Iterar sobre cadena original para copiar en las posiciones pares e impares
    int i=0, j=0;
    for(int k=0; k<strlen(str); k++){
        
        if (k%2 == 0) str_p[i++] = str[k]; 
        else str_i[j++] = str[k];

    }
    
    // Agregar 0 terminador cadena
    str_p[i] = 0; str_i[j] = 0;

    // Formatear salida original
    sprintf(str, "%s %s", str_p, str_i);
    
    // Liberar memoria;
    free(str_p); free(str_i);
        
}

int main() {

    // Leer numero de strings y limpiar buffer
    int num_cad = 0;
    scanf("%d", &num_cad);
    int c;
    while((c=getchar()) != '\n') {}
    
    char* cadenas[num_cad];    // Crear tantas cadenas como n num_cad
    char *linea = NULL;             // Buffer para almacenar la entrada
    size_t lon = 0;                     // Longitud cadena de entrada
        
    // Leer la entrada y almacenarlo en cada cadena
    for (int i=0; i<num_cad; i++) {
        
        // Leer la entrada con getline para manejar longitud de cadena desconocida
        if(getline(&linea, &lon, stdin) == -1) exit(1); 
            
        linea[strcspn(linea, "\n")] = 0;               // Limpiar salto de linea
        
        cadenas[i] = malloc(sizeof(char)*lon);  // Asignar memoria a cadenas
        strcpy(cadenas[i], linea);                      // Copiar el contenido de entrada a cadenas
        
    }
    
    // Procesar cadenas
    for (int i=0; i<num_cad; i++) {
        procesar_cadena(cadenas[i]);
    }
    
    // Printear cada cadena
    for (int i=0; i<num_cad; i++) {
        puts(cadenas[i]);
    }
    
    
    // Liberar memoria
    for (int i=0; i<num_cad; i++) {
        free(cadenas[i]);
    }
         
    return 0;
}

  • haqueasheerul86
     + 0 comments
    const lines = input.split('\n');
    const n = parseInt(lines[0], 10);

    for (let i = 1; i <= n; i++) {
        let evenStr = "";
        let oddStr = "";
        for (j = 0; j <= lines[i].length - 1; j++) {
            if (j % 2 === 0) {
                evenStr = evenStr + lines[i][j]
            } else {
                oddStr = oddStr + lines[i][j]
            }
        }
        console.log(evenStr,oddStr)
    }