Dada una secuencia de ‘n’ números (sin ceros a la izquierda), la tarea es encontrar si es posible crear una string binaria concatenando estos números secuencialmente.
Si es posible, imprima la string binaria formada; de lo contrario, imprima «-1».
Ejemplos:
Entrada: arr[] = {10, 11, 1, 0, 10}
Salida: 10111010
Todos los números contienen los dígitos ‘1’ y ‘0’ solamente. Entonces, es posible formar una string binaria concatenando
estos números secuencialmente, que es 10111010.
Entrada: arr[] = {1, 2, 11, 10}
Salida: -1
Uno de los números contiene el dígito ‘2’ que no se puede una parte de cualquier string binaria.
Entonces, la salida es -1.
Enfoque: La observación principal es que solo podemos concatenar aquellos números que contienen los dígitos ‘1’ y ‘0’ solamente. De lo contrario, es imposible formar una string binaria.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
bool isBinary(int n)
{
while (n != 0) {
int temp = n % 10;
if (temp != 0 && temp != 1) {
return false;
}
n = n / 10;
}
return true;
}
//Function that checks whether the
//binary string can be formed or not
void formBinaryStr(int n, int a[])
{
bool flag = true;
// Empty string for storing
// the binary number
string s = "";
for (int i = 0; i < n; i++) {
// check if a[i] can be a
// part of the binary string
if (isBinary(a[i]))
// Conversion of int into string
s += to_string(a[i]);
else {
// if a[i] can't be a part
// then break the loop
flag = false;
break;
}
}
// possible to create binary string
if (flag)
cout << s << "\n";
// impossible to create binary string
else
cout << "-1\n";
}
// Driver code
int main()
{
int a[] = { 10, 1, 0, 11, 10 };
int N = sizeof(a) / sizeof(a[0]);
formBinaryStr(N, a);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class Solution
{
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
static boolean isBinary(int n)
{
while (n != 0) {
int temp = n % 10;
if (temp != 0 && temp != 1) {
return false;
}
n = n / 10;
}
return true;
}
//Function that checks whether the
//binary String can be formed or not
static void formBinaryStr(int n, int a[])
{
boolean flag = true;
// Empty String for storing
// the binary number
String s = "";
for (int i = 0; i < n; i++) {
// check if a[i] can be a
// part of the binary String
if (isBinary(a[i]))
// Conversion of int into String
s += ""+a[i];
else {
// if a[i] can't be a part
// then break the loop
flag = false;
break;
}
}
// possible to create binary String
if (flag)
System.out.print( s + "\n");
// impossible to create binary String
else
System.out.print( "-1\n");
}
// Driver code
public static void main(String args[])
{
int a[] = { 10, 1, 0, 11, 10 };
int N = a.length;
formBinaryStr(N, a);
}
}
//contributed by Arnab Kundu
Python3
# Python3 implementation of the approach # Function that returns false if the # number passed as argument contains # digit(s) other than '0' or '1' def isBinary(n): while n != 0: temp = n % 10 if temp != 0 and temp != 1: return False n = n // 10 return True # Function that checks whether the # binary string can be formed or not def formBinaryStr(n, a): flag = True # Empty string for storing # the binary number s = "" for i in range(0, n): # check if a[i] can be a # part of the binary string if isBinary(a[i]) == True: # Conversion of int into string s += str(a[i]) else: # if a[i] can't be a part # then break the loop flag = False break # possible to create binary string if flag == True: print(s) # impossible to create binary string else: cout << "-1\n" # Driver code if __name__ == "__main__": a = [10, 1, 0, 11, 10] N = len(a) formBinaryStr(N, a) # This code is contributed by Rituraj Jain
C#
// C# implementation of the approach
using System;
public class Solution
{
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
public static bool isBinary(int n)
{
while (n != 0)
{
int temp = n % 10;
if (temp != 0 && temp != 1)
{
return false;
}
n = n / 10;
}
return true;
}
//Function that checks whether the
//binary String can be formed or not
public static void formBinaryStr(int n, int[] a)
{
bool flag = true;
// Empty String for storing
// the binary number
string s = "";
for (int i = 0; i < n; i++)
{
// check if a[i] can be a
// part of the binary String
if (isBinary(a[i]))
{
// Conversion of int into String
s += "" + a[i];
}
else
{
// if a[i] can't be a part
// then break the loop
flag = false;
break;
}
}
// possible to create binary String
if (flag)
{
Console.Write(s + "\n");
}
// impossible to create binary String
else
{
Console.Write("-1\n");
}
}
// Driver code
public static void Main(string[] args)
{
int[] a = new int[] {10, 1, 0, 11, 10};
int N = a.Length;
formBinaryStr(N, a);
}
}
// This code is contributed by Shrikant13
PHP
<?php
// PHP implementation of the approach
// Function that returns false if the
// number passed as argument contains
// digit(s) other than '0' or '1'
function isBinary($n)
{
while ($n != 0)
{
$temp = $n % 10;
if ($temp != 0 && $temp != 1)
{
return false;
}
$n = intval($n / 10);
}
return true;
}
// Function that checks whether the
// binary string can be formed or not
function formBinaryStr($n, &$a)
{
$flag = true;
// Empty string for storing
// the binary number
$s = "";
for ($i = 0; $i < $n; $i++)
{
// check if a[i] can be a
// part of the binary string
if (isBinary($a[$i]))
// Conversion of int into string
$s = $s.strval($a[$i]);
else
{
// if a[i] can't be a part
// then break the loop
$flag = false;
break;
}
}
// possible to create binary string
if ($flag)
echo $s . "\n";
// impossible to create binary string
else
echo "-1\n";
}
// Driver code
$a = array( 10, 1, 0, 11, 10 );
$N = sizeof($a) / sizeof($a[0]);
formBinaryStr($N, $a);
// This code is contributed by ita_c
?>
Javascript
<script>
// Javascript implementation of the approach
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
function isBinary(n)
{
while (n != 0) {
var temp = n % 10;
if (temp != 0 && temp != 1) {
return false;
}
n = parseInt(n / 10);
}
return true;
}
// Function that checks whether the
// binary String can be formed or not
function formBinaryStr(n , a) {
var flag = true;
// Empty String for storing
// the binary number
var s = "";
for (i = 0; i < n; i++) {
// check if a[i] can be a
// part of the binary String
if (isBinary(a[i]))
// Conversion of var into String
s += "" + a[i];
else {
// if a[i] can't be a part
// then break the loop
flag = false;
break;
}
}
// possible to create binary String
if (flag)
document.write(s + "\n");
// impossible to create binary String
else
document.write("-1\n");
}
// Driver code
var a = [ 10, 1, 0, 11, 10 ];
var N = a.length;
formBinaryStr(N, a);
// This code contributed by Rajput-Ji
</script>
Producción:
10101110
Complejidad de tiempo: O(N*log(MAX)), donde N es la longitud de la array y MAX es el número máximo en la array
Espacio auxiliar: O(M), donde M es la longitud de la string