Number of subsequences with zero sum

Given an array arr[] of N integers. The task is to count the number of sub-sequences whose sum is 0.

Examples:

Input: arr[] = {-1, 2, -2, 1}
Output: 3
All possible sub-sequences are {-1, 1}, {2, -2} and {-1, 2, -2, 1}



Input: arr[] = {-2, -4, -1, 6, -2}
Output: 2

Approach: The problem can be solved using recursion. Recursively, we start from the first index, and either select the number to be added in the subsequence or we do not select the number at an index. Once the index exceeds N, we need to check if the sum evaluated is 0 or not and the count of numbers taken in subsequence should be a minimum of one. If it is, then we simply return 1 which is added to the number of ways.

Dynamic Programming cannot be used to solve this problem because of the sum value which can be anything which is not possible to store in any dimensional array.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of the required sub-sequences
int countSubSeq(int i, int sum, int cnt,
                int a[], int n)
{
  
    // Base case
    if (i == n) {
  
        // Check if the sum is 0
        // and at least a single element
        // is in the sub-sequence
        if (sum == 0 && cnt > 0)
            return 1;
        else
            return 0;
    }
    int ans = 0;
  
    // Do not take the number in
    // the current sub-sequence
    ans += countSubSeq(i + 1, sum, cnt, a, n);
  
    // Take the number in the
    // current sub-sequence
    ans += countSubSeq(i + 1, sum + a[i],
                       cnt + 1, a, n);
  
    return ans;
}
  
// Driver code
int main()
{
    int a[] = { -1, 2, -2, 1 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countSubSeq(0, 0, 0, a, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG
{
  
    // Function to return the count
    // of the required sub-sequences
    static int countSubSeq(int i, int sum, int cnt,
                                    int a[], int n) 
    {
  
        // Base case
        if (i == n)
        {
  
            // Check if the sum is 0
            // and at least a single element
            // is in the sub-sequence
            if (sum == 0 && cnt > 0)
            {
                return 1;
            
            else
            {
                return 0;
            }
        }
        int ans = 0;
  
        // Do not take the number in
        // the current sub-sequence
        ans += countSubSeq(i + 1, sum, cnt, a, n);
  
        // Take the number in the
        // current sub-sequence
        ans += countSubSeq(i + 1, sum + a[i],
                                cnt + 1, a, n);
  
        return ans;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = {-1, 2, -2, 1};
        int n = a.length;
        System.out.println(countSubSeq(0, 0, 0, a, n));
    }
  
// This code has been contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
  
# Function to return the count
# of the required sub-sequences
def countSubSeq(i, Sum, cnt, a, n):
  
    # Base case
    if (i == n):
  
        # Check if the Sum is 0
        # and at least a single element
        # is in the sub-sequence
        if (Sum == 0 and cnt > 0):
            return 1
        else:
            return 0
    ans = 0
  
    # Do not take the number in
    # the current sub-sequence
    ans += countSubSeq(i + 1, Sum, cnt, a, n)
  
    # Take the number in the
    # current sub-sequence
    ans += countSubSeq(i + 1, Sum + a[i], 
                           cnt + 1, a, n)
  
    return ans
  
# Driver code
a = [-1, 2, -2, 1]
n = len(a)
print(countSubSeq(0, 0, 0, a, n))
  
# This code is contributed by mohit kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
  
    // Function to return the count
    // of the required sub-sequences
    static int countSubSeq(int i, int sum, 
                           int cnt, int []a, int n) 
    {
  
        // Base case
        if (i == n)
        {
  
            // Check if the sum is 0
            // and at least a single element
            // is in the sub-sequence
            if (sum == 0 && cnt > 0)
            {
                return 1;
            
            else
            {
                return 0;
            }
        }
          
        int ans = 0;
  
        // Do not take the number in
        // the current sub-sequence
        ans += countSubSeq(i + 1, sum, cnt, a, n);
  
        // Take the number in the
        // current sub-sequence
        ans += countSubSeq(i + 1, sum + a[i],
                                  cnt + 1, a, n);
  
        return ans;
    }
  
    // Driver code
    public static void Main()
    {
        int []a = {-1, 2, -2, 1};
        int n = a.Length;
        Console.Write(countSubSeq(0, 0, 0, a, n));
    }
  
// This code is contributed by Akanksha Rai

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach 
  
// Function to return the count 
// of the required sub-sequences 
function countSubSeq($i, $sum, $cnt, $a, $n
  
    // Base case 
    if ($i == $n
    
  
        // Check if the sum is 0 
        // and at least a single element 
        // is in the sub-sequence 
        if ($sum == 0 && $cnt > 0) 
            return 1; 
        else
            return 0; 
    
    $ans = 0; 
  
    // Do not take the number in 
    // the current sub-sequence 
    $ans += countSubSeq($i + 1, $sum
                        $cnt, $a, $n); 
  
    // Take the number in the 
    // current sub-sequence 
    $ans += countSubSeq($i + 1, $sum + $a[$i], 
                        $cnt + 1, $a, $n); 
  
    return $ans
  
// Driver code 
$a = array( -1, 2, -2, 1 ); 
$n = count($a) ;
  
echo countSubSeq(0, 0, 0, $a, $n); 
  
// This code is contributed by Ryuga
?>

chevron_right


Output:

3


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.