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 that is not possible to store in any dimensional array.
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 count// of the required sub-sequencesint 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 codeint main(){ int a[] = { -1, 2, -2, 1 }; int n = sizeof(a) / sizeof(a[0]); cout << countSubSeq(0, 0, 0, a, n); return 0;} |
Java
// Java implementation of the approachclass 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 |
Python3
# Python3 implementation of the approach# Function to return the count# of the required sub-sequencesdef 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 codea = [-1, 2, -2, 1]n = len(a)print(countSubSeq(0, 0, 0, a, n))# This code is contributed by mohit kumar |
C#
// C# implementation of the approachusing 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 |
PHP
<?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?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the count// of the required sub-sequencesfunction 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; } let 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 let a = [ -1, 2, -2, 1 ]; let n = a.length; document.write(countSubSeq(0, 0, 0, a, n));</script> |
Output:
3
Time Complexity: O(2N), as we are using recursion and T(N) = O(1) + 2*T(N-1) which will be equivalent to T(N) = (2^N – 1)*O(1). Where N is the number of elements in the array.
Auxiliary Space: O(N), where N is the recursion stack space.
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