Dados dos enteros n y m donde n es el número de 0s ym es el número de 1s . La tarea es imprimir todos los 0 y 1 en una sola fila de modo que no haya dos 0 juntos ni tres 1 juntos. Si no es posible organizar 0 y 1 de acuerdo con la condición, imprima -1 . Ejemplos:
Entrada: n = 1, m = 2
Salida: 011
Entrada: n = 4, m = 8
Salida: 110110110101
Enfoque: Solo tenemos respuestas cuando, ( (n – 1) ≤ m y m ≤ 2 * (n + 1) .
- Si (m == n – 1) entonces imprima el patrón 010101… comenzando desde 0 hasta que se hayan usado todos los 0 y 1 .
- Si (m > n) y m ≤ 2 * (n + 1) imprima el patrón 110110110… hasta que haya un exceso de 1 y cambie al patrón 0101010… cuando m sea igual a n – 1 .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the required pattern
void printPattern(int n, int m)
{
// When condition fails
if (m > 2 * (n + 1) || m < n - 1) {
cout << "-1";
}
// When m = n - 1
else if (abs(n - m) <= 1) {
while (n > 0 && m > 0) {
cout << "01";
n--;
m--;
}
if (n != 0) {
cout << "0";
}
if (m != 0) {
cout << "1";
}
}
else {
while (m - n > 1 && n > 0) {
cout << "110";
m = m - 2;
n = n - 1;
}
while (n > 0) {
cout << "10";
n--;
m--;
}
while (m > 0) {
cout << "1";
m--;
}
}
}
// Driver program
int main()
{
int n = 4, m = 8;
printPattern(n, m);
return 0;
}
Java
// Java implementation of the above approach
class GFG
{
// Function to print the required pattern
static void printPattern(int n, int m)
{
// When condition fails
if (m > 2 * (n + 1) || m < n - 1)
{
System.out.print("-1");
}
// When m = n - 1
else if (Math.abs(n - m) <= 1)
{
while (n > 0 && m > 0)
{
System.out.print("01");
n--;
m--;
}
if (n != 0)
{
System.out.print("0");
}
if (m != 0)
{
System.out.print("1");
}
}
else
{
while (m - n > 1 && n > 0)
{
System.out.print("110");
m = m - 2;
n = n - 1;
}
while (n > 0)
{
System.out.print("10");
n--;
m--;
}
while (m > 0)
{
System.out.print("1");
m--;
}
}
}
// Driver code
public static void main(String []args)
{
int n = 4, m = 8;
printPattern(n, m);
}
}
// This code is contributed by Ita_c.
Python3
# Python 3 implementation of the approach
# Function to print the required pattern
def printPattern(n, m):
# When condition fails
if (m > 2 * (n + 1) or m < n - 1):
print("-1", end = "")
# When m = n - 1
elif (abs(n - m) <= 1):
while (n > 0 and m > 0):
print("01", end = "");
n -= 1
m -= 1
if (n != 0):
print("0", end = "")
if (m != 0):
print("1", end = "")
else:
while (m - n > 1 and n > 0):
print("110", end = "")
m = m - 2
n = n - 1
while (n > 0):
print("10", end = "")
n -= 1
m -= 1
while (m > 0):
print("1", end = "")
m -= 1
# Driver Code
if __name__ == '__main__':
n = 4
m = 8
printPattern(n, m)
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to print the required pattern
static void printPattern(int n, int m)
{
// When condition fails
if (m > 2 * (n + 1) || m < n - 1)
{
Console.Write("-1");
}
// When m = n - 1
else if (Math.Abs(n - m) <= 1)
{
while (n > 0 && m > 0)
{
Console.Write("01");
n--;
m--;
}
if (n != 0)
{
Console.Write("0");
}
if (m != 0)
{
Console.Write("1");
}
}
else
{
while (m - n > 1 && n > 0)
{
Console.Write("110");
m = m - 2;
n = n - 1;
}
while (n > 0)
{
Console.Write("10");
n--;
m--;
}
while (m > 0)
{
Console.Write("1");
m--;
}
}
}
// Driver code
public static void Main()
{
int n = 4, m = 8;
printPattern(n, m);
}
}
// This code is contributed by Ryuga
PHP
<?php
// PHP implementation of the above approach
// Function to print the required pattern
function printPattern($n, $m)
{
// When condition fails
if ($m > 2 * ($n + 1) || $m < $n - 1)
{
echo("-1");
}
// When m = n - 1
else if (abs($n - $m) <= 1)
{
while ($n > 0 && $m > 0)
{
System.out.print("01");
$n--;
$m--;
}
if ($n != 0)
{
echo("0");
}
if ($m != 0)
{
echo("1");
}
}
else
{
while ($m - $n > 1 && $n > 0)
{
echo("110");
$m = $m - 2;
$n = $n - 1;
}
while ($n > 0)
{
echo("10");
$n--;
$m--;
}
while ($m > 0)
{
echo("1");
$m--;
}
}
}
// Driver code
$n = 4; $m = 8;
printPattern($n, $m);
// This code is contributed by
// Mukul Singh.
?>
Javascript
<script>
// JavaScript implementation of the above approach
// Function to print the required pattern
function printPattern( n, m)
{
// When condition fails
if (m > 2 * (n + 1) || m < n - 1) {
document.write("-1");
}
// When m = n - 1
else if (Math.abs(n - m) <= 1) {
while (n > 0 && m > 0) {
document.write("01");
n--;
m--;
}
if (n != 0) {
document.write("0");
}
if (m != 0) {
document.write("1");
}
}
else {
while (m - n > 1 && n > 0) {
document.write("110");
m = m - 2;
n = n - 1;
}
while (n > 0) {
document.write("10");
n--;
m--;
}
while (m > 0) {
document.write("1");
m--;
}
}
}
var n = 4, m = 8;
printPattern(n, m);
</script>
Producción
110110110101
Publicación traducida automáticamente
Artículo escrito por himanshu_rathore y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA