Given an array arr[ ] of size N. The task is to find the no of subsets with sum as (sum of the array – 1) when 0 occurs in an array or return -1 if there is no subset possible.
Examples:
Input: arr[ ] : {3, 0, 5, 2, 4}
Output: -1
Explanation: sum of array arr[ ] is 14 and there is not any subset with sum 13.
Input: arr[ ] : {0, 0, 2, 1}
Output: 4
Explanation: sum of array arr[ ] is 3 and {0, 2}, {0, 2}, {0, 0, 2}, {2} are the four subsets with sum 2 .
Naive Approach: The task can be solved by generating all possible subsets of the array using recursion, and increment the count if a subset with sum as (sum of the array – 1) is encountered.
Below is the implementation of the above approach:
C++
// C++ program to print the count of// subsets with sum equal to the given value X#include <bits/stdc++.h>using namespace std;// Recursive function to return the count// of subsets with sum equal to the given valueint subsetSum(int arr[], int n, int i, int sum, int count){ // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0) { count++; } return count; } // Recursively calling the // function for two cases // Either the element can be // counted in the subset // If the element is counted, // then the remaining sum // to be checked is // sum - the selected element // If the element is not included, // then the remaining sum // to be checked is the total sum count = subsetSum(arr, n, i + 1, sum - arr[i], count); count = subsetSum(arr, n, i + 1, sum, count); return count == 0 ? -1 : count;}// Driver codeint main(){ int arr[] = { 0, 0, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); // Sum of array - 1 int sum = accumulate(arr, arr + n, 0) - 1; cout << subsetSum(arr, n, 0, sum, 0);} |
Java
// Java program to print the count of// subsets with sum equal to the given value Xpublic class GFG { // Recursive function to return the count // of subsets with sum equal to the given value static int subsetSum(int arr[], int n, int i, int sum, int count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0) { count++; } return count; } // Recursively calling the // function for two cases // Either the element can be // counted in the subset // If the element is counted, // then the remaining sum // to be checked is // sum - the selected element // If the element is not included, // then the remaining sum // to be checked is the total sum count = subsetSum(arr, n, i + 1, sum - arr[i], count); count = subsetSum(arr, n, i + 1, sum, count); return count == 0 ? -1 : count; } static int accumulate(int []arr) { int sum = 0; for(int i = 0; i < arr.length; i++) sum += arr[i]; return sum; } // Driver code public static void main (String[] args) { int arr[] = { 0, 0, 2, 1 }; int n = arr.length; // Sum of array - 1 int sum = accumulate(arr) - 1; System.out.println(subsetSum(arr, n, 0, sum, 0)); }}// This code is contributed by AnkThon |
Python3
# Python program to print the count of# subsets with sum equal to the given value X# Recursive function to return the count# of subsets with sum equal to the given valuedef subsetSum(arr, n, i, sum, count): # The recursion is stopped at N-th level # where all the subsets of the given array # have been checked if (i == n): # Incrementing the count if sum is # equal to 0 and returning the count if (sum == 0): count += 1 return count # Recursively calling the # function for two cases # Either the element can be # counted in the subset # If the element is counted, # then the remaining sum # to be checked is # sum - the selected element # If the element is not included, # then the remaining sum # to be checked is the total sum count = subsetSum(arr, n, i + 1, sum - arr[i], count) count = subsetSum(arr, n, i + 1, sum, count) return - 1 if count == 0 else countdef accumulate(arr): sum = 0 for i in range(len(arr)): sum += arr[i] return sum# Driver codearr = [0, 0, 2, 1]n = len(arr)# Sum of array - 1sum = accumulate(arr) - 1print(subsetSum(arr, n, 0, sum, 0))# This code is contributed by gfgking |
C#
// C# program to print the count of// subsets with sum equal to the given value Xusing System;public class GFG { // Recursive function to return the count // of subsets with sum equal to the given value static int subsetSum(int []arr, int n, int i, int sum, int count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0) { count++; } return count; } // Recursively calling the // function for two cases // Either the element can be // counted in the subset // If the element is counted, // then the remaining sum // to be checked is // sum - the selected element // If the element is not included, // then the remaining sum // to be checked is the total sum count = subsetSum(arr, n, i + 1, sum - arr[i], count); count = subsetSum(arr, n, i + 1, sum, count); return count == 0 ? -1 : count; } static int accumulate(int []arr) { int sum = 0; for(int i = 0; i < arr.Length; i++) sum += arr[i]; return sum; } // Driver code public static void Main (string[] args) { int []arr = { 0, 0, 2, 1 }; int n = arr.Length; // Sum of array - 1 int sum = accumulate(arr) - 1; Console.WriteLine(subsetSum(arr, n, 0, sum, 0)); }}// This code is contributed by AnkThon |
Javascript
<script> // JavaScript program to print the count of // subsets with sum equal to the given value X // Recursive function to return the count // of subsets with sum equal to the given value function subsetSum(arr, n, i, sum, count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0) { count++; } return count; } // Recursively calling the // function for two cases // Either the element can be // counted in the subset // If the element is counted, // then the remaining sum // to be checked is // sum - the selected element // If the element is not included, // then the remaining sum // to be checked is the total sum count = subsetSum(arr, n, i + 1, sum - arr[i], count); count = subsetSum(arr, n, i + 1, sum, count); return count == 0 ? -1 : count; } function accumulate(arr) { let sum = 0; for (let i = 0; i < arr.length; i++) { sum += arr[i]; } return sum; } // Driver code let arr = [0, 0, 2, 1]; let n = arr.length; // Sum of array - 1 let sum = accumulate(arr) - 1; document.write(subsetSum(arr, n, 0, sum, 0)); // This code is contributed by Potta Lokesh </script> |
Output:
4
Time complexity: O(2n), where n = array length
Auxiliary Space: O(n), recursion stack space
Efficient Approach: The above approach can be optimized by finding a possible subset with sum as (array_sum – 1) by removing 0s and one 1 from the array so that the subset-sum becomes equal to (sum of the array – 1 ).
- If there are k zeroes in the array, then there are 2^k (k is the number of zeroes) ways to remove 0 from the array.
- The multiplication of these two (no of ways to remove 0 and no of ways to remove 1) results in the no of possible subsets.
Below is the implementation of the above code:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the no of// possible subsetsint subsetCount(int arr[], int n){ // Count the no of 0s and 1s // in array int zeros = 0, ones = 0; for (int i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; } else if (arr[i] == 1) { ones++; } } // Store no of ways to remove 0 int no_of_ways_0 = pow(2, zeros); // Store no of ways to remove 1 int no_of_ways_1 = ones; // Store the total count of subsets int count_subset = no_of_ways_0 * no_of_ways_1; // If there is no subset possible // with required sum, return -1 if (count_subset == 0) return -1; return count_subset;}// Driver Codeint main(){ int arr[] = { 0, 0, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << subsetCount(arr, n);} |
Java
// Java implementation of the above approachimport java.util.*;public class GFG{// Function to find the no of// possible subsetsstatic int subsetCount(int []arr, int n){ // Count the no of 0s and 1s // in array int zeros = 0, ones = 0; for (int i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; } else if (arr[i] == 1) { ones++; } } // Store no of ways to remove 0 int no_of_ways_0 = (int)Math.pow(2, zeros); // Store no of ways to remove 1 int no_of_ways_1 = ones; // Store the total count of subsets int count_subset = no_of_ways_0 * no_of_ways_1; // If there is no subset possible // with required sum, return -1 if (count_subset == 0) return -1; return count_subset;}// Driver Codepublic static void main(String args[]){ int []arr = { 0, 0, 2, 1 }; int n = arr.length; System.out.println(subsetCount(arr, n));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python 3 implementation of the above approach# Function to find the no of# possible subsetsdef subsetCount(arr, n): # Count the no of 0s and 1s # in array zeros = 0 ones = 0 for i in range(n): if (arr[i] == 0): zeros += 1 elif (arr[i] == 1): ones += 1 # Store no of ways to remove 0 no_of_ways_0 = pow(2, zeros) # Store no of ways to remove 1 no_of_ways_1 = ones # Store the total count of subsets count_subset = no_of_ways_0 * no_of_ways_1 # If there is no subset possible # with required sum, return -1 if (count_subset == 0): return -1 return count_subset# Driver Codeif __name__ == "__main__": arr = [0, 0, 2, 1] n = len(arr) print(subsetCount(arr, n)) # This code is contributed by ukasp. |
C#
// C# implementation of the above approachusing System;class GFG{// Function to find the no of// possible subsetsstatic int subsetCount(int []arr, int n){ // Count the no of 0s and 1s // in array int zeros = 0, ones = 0; for (int i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; } else if (arr[i] == 1) { ones++; } } // Store no of ways to remove 0 int no_of_ways_0 = (int)Math.Pow(2, zeros); // Store no of ways to remove 1 int no_of_ways_1 = ones; // Store the total count of subsets int count_subset = no_of_ways_0 * no_of_ways_1; // If there is no subset possible // with required sum, return -1 if (count_subset == 0) return -1; return count_subset;}// Driver Codepublic static void Main(){ int []arr = { 0, 0, 2, 1 }; int n = arr.Length; Console.Write(subsetCount(arr, n));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript implementation of the above approach// Function to find the no of// possible subsetsfunction subsetCount(arr, n){ // Count the no of 0s and 1s // in array let zeros = 0, ones = 0; for (let i = 0; i < n; i++) { if (arr[i] == 0) { zeros++; } else if (arr[i] == 1) { ones++; } } // Store no of ways to remove 0 let no_of_ways_0 = Math.pow(2, zeros); // Store no of ways to remove 1 let no_of_ways_1 = ones; // Store the total count of subsets let count_subset = no_of_ways_0 * no_of_ways_1; // If there is no subset possible // with required sum, return -1 if (count_subset == 0) return -1; return count_subset;}// Driver Codelet arr = [ 0, 0, 2, 1 ];let n = arr.length;document.write(subsetCount(arr, n));// This code is contributed by Samim Hossain Mondal.</script> |
Output
4
Time complexity: O(n), where n = array length
Auxiliary Space: O(1)
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