Ways to divide a binary array into sub-arrays such that each sub-array contains exactly one 1

Give an integer array arr[] consisting of elements from the set {0, 1}. The task is to print the number of ways the array can be divided into sub-arrays such that each sub-array contains exactly one 1.

Examples:

Input: arr[] = {1, 0, 1, 0, 1}
Output: 4
Below are the possible ways:



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

Input: arr[] = {0, 0, 0}
Output: 0

Approach:

  • When all the elements of the array are 0 then the result will be zero.
  • Else, between two adjacent ones we must have only one separation. So, answer equals to product of values posi + 1 – posi (for all valid pairs) where posi is the position of ith 1.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java














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// 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 



















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Python3














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# 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 



















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C#














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// 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 



















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PHP














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<?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 


?> 



















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Output:


4







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