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 conditionint 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 codeint 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 approachclass GFG{ // Function to return the number of ways// the array can be divided into sub-arrays// satisfying the given conditionstatic 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 codepublic 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 conditiondef 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 codeif __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 approachusing System;class GFG{ // Function to return the number of ways// the array can be divided into sub-arrays// satisfying the given conditionstatic 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 codepublic 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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