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, the answer equals the 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++
// 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
4
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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