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

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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, the answer equals the product of values pos_{i + 1} – pos_i (for all valid pairs) where pos_i is the position of i^th 1.

Below is the implementation of the above approach:

C++

// 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

// 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

Python3

# 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#

// 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
// 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
?>

Javascript

<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>

Output

Complexity Analysis:

Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(n), as extra space of size n was used

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