Maneras de dividir una array binaria en sub-arrays de tal manera que cada sub-array contenga exactamente un 1

Proporcione una array de enteros arr[] que consta de elementos del conjunto {0, 1} . La tarea es imprimir el número de formas en que la array se puede dividir en sub-arrays, de modo que cada sub-array contenga exactamente un 1 .

Ejemplos: 

Entrada: arr[] = {1, 0, 1, 0, 1} 
Salida: 4 
A continuación se muestran las formas posibles: 

• {1, 0}, {1, 0}, {1}
• {1}, {0, 1, 0}, {1}
• {1, 0}, {1}, {0, 1}
• {1}, {0, 1}, {0, 1}

Entrada: arr[] = {0, 0, 0} 
Salida: 0 

Acercarse:

• Cuando todos los elementos de la array son 0 , el resultado será cero.
• De lo contrario, entre dos adyacentes, debe haber una sola separación. Entonces, la respuesta es igual al producto de los valores pos _{i + 1} – pos _i (para todos los pares válidos) donde pos _i es la posición de i ^th 1 .

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
int countWays(int arr[], int n)
{
 
    int pos[n], p = 0, i;
 
    // for loop for saving the positions of all 1s
    for (i = 0; i < n; i++) {
        if (arr[i] == 1) {
            pos[p] = i + 1;
            p++;
        }
    }
 
    // If array contains only 0s
    if (p == 0)
        return 0;
 
    int ways = 1;
    for (i = 0; i < p - 1; i++) {
        ways *= pos[i + 1] - pos[i];
    }
 
    // Return the total ways
    return ways;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 0, 1, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countWays(arr, n);
    return 0;
}

// Java implementation of the approach
class GFG
{
     
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
static int countWays(int arr[], int n)
{
    int pos[] = new int[n];
    int p = 0, i;
 
    // for loop for saving the
    // positions of all 1s
    for (i = 0; i < n; i++)
    {
        if (arr[i] == 1)
        {
            pos[p] = i + 1;
            p++;
        }
    }
 
    // If array contains only 0s
    if (p == 0)
        return 0;
 
    int ways = 1;
    for (i = 0; i < p - 1; i++)
    {
        ways *= pos[i + 1] - pos[i];
    }
 
    // Return the total ways
    return ways;
}
 
// Driver code
public static void main(String args[])
{
    int[] arr = { 1, 0, 1, 0, 1 };
    int n = arr.length;
    System.out.println(countWays(arr, n));
}
}
 
// This code is contributed
// by Akanksha Rai

# Python 3 implementation of the approach
 
# Function to return the number of ways
# the array can be divided into sub-arrays
# satisfying the given condition
def countWays(are, n):
    pos = [0 for i in range(n)]
    p = 0
 
    # for loop for saving the positions
    # of all 1s
    for i in range(n):
        if (arr[i] == 1):
            pos[p] = i + 1
            p += 1
 
    # If array contains only 0s
    if (p == 0):
        return 0
 
    ways = 1
    for i in range(p - 1):
        ways *= pos[i + 1] - pos[i]
 
    # Return the total ways
    return ways
 
# Driver code
if __name__ == '__main__':
    arr = [1, 0, 1, 0, 1]
    n = len(arr)
    print(countWays(arr, n))
     
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
static int countWays(int[] arr, int n)
{
    int[] pos = new int[n];
    int p = 0, i;
 
    // for loop for saving the positions
    // of all 1s
    for (i = 0; i < n; i++)
    {
        if (arr[i] == 1)
        {
            pos[p] = i + 1;
            p++;
        }
    }
 
    // If array contains only 0s
    if (p == 0)
        return 0;
 
    int ways = 1;
    for (i = 0; i < p - 1; i++)
    {
        ways *= pos[i + 1] - pos[i];
    }
 
    // Return the total ways
    return ways;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 0, 1, 0, 1 };
    int n = arr.Length;
    Console.Write(countWays(arr, n));
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the approach
 
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
function countWays($arr, $n)
{
    $pos = array_fill(0, $n, 0);
    $p = 0 ;
 
    // for loop for saving the positions
    // of all 1s
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] == 1)
        {
            $pos[$p] = $i + 1;
            $p++;
        }
    }
 
    // If array contains only 0s
    if ($p == 0)
        return 0;
 
    $ways = 1;
    for ($i = 0; $i < $p - 1; $i++)
    {
        $ways *= $pos[$i + 1] - $pos[$i];
    }
 
    // Return the total ways
    return $ways;
}
 
// Driver code
$arr = array(1, 0, 1, 0, 1);
$n = sizeof($arr);
echo countWays($arr, $n);
 
// This code is contributed by Ryuga
?>

<script>
      // JavaScript implementation of the approach
      // Function to return the number of ways
      // the array can be divided into sub-arrays
      // satisfying the given condition
      function countWays(arr, n) {
        var pos = new Array(n).fill(0);
        var p = 0, i;
 
        // for loop for saving the positions
        // of all 1s
        for (i = 0; i < n; i++) {
          if (arr[i] === 1) {
            pos[p] = i + 1;
            p++;
          }
        }
 
        // If array contains only 0s
        if (p === 0)
            return 0;
 
        var ways = 1;
        for (i = 0; i < p - 1; i++) {
          ways *= pos[i + 1] - pos[i];
        }
 
        // Return the total ways
        return ways;
      }
 
      // Driver code
      var arr = [1, 0, 1, 0, 1];
      var n = arr.length;
      document.write(countWays(arr, n));
</script>

4

Complejidad de tiempo: O(n), donde n es el tamaño de la array dada
Espacio auxiliar: O(n), ya que se usó un espacio adicional de tamaño n