Given an array arr[] of N integers, the task is to find the count of all the subsequences of the array that have the positive product.
Example:
Input: arr[] = {2, -3, -1}
Output: 3
{2}, {-3, -1} and {2, -3, -1} are the only possible subsequences.Input: arr[] = {2, 3, -1, 4, 5}
Output: 15
Naive approach: Generate all the subsequences of the array and compute the product of all the subsequences. If the product is positive, then increment the count by 1.
Efficient approach:
- Count the number of positive and negative elements in the array.
- Any number of positive elements can be chosen for the subsequence to maintain the positive product. The number of different combinations of subsequences with all the positive elements will be pow(2, count of positive elements).
- An even number of negative elements can be chosen for the subsequence to maintain the positive product. The number of different combinations of subsequences with an even number of negative elements will be pow(2, count of negative elements – 1).
- After that, remove 1 from the results for the empty subsequence.
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 all// the subsequences with positive productint cntSubSeq(int arr[], int n){ // To store the count of positive // elements in the array int pos_count = 0; // To store the count of negative // elements in the array int neg_count = 0; int result; for (int i = 0; i < n; i++) { // If the current element // is positive if (arr[i] > 0) pos_count++; // If the current element // is negative if (arr[i] < 0) neg_count++; } // For all the positive // elements of the array result = pow(2, pos_count); // For all the negative // elements of the array if (neg_count > 0) result *= pow(2, neg_count - 1); // For the empty subsequence result -= 1; return result;}// Driver codeint main(){ int arr[] = { 2, -3, -1, 4 }; int n = sizeof(arr) / sizeof(arr[0]); cout << cntSubSeq(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Function to return the count of all// the subsequences with positive productstatic int cntSubSeq(int arr[], int n){ // To store the count of positive // elements in the array int pos_count = 0; // To store the count of negative // elements in the array int neg_count = 0; int result; for (int i = 0; i < n; i++) { // If the current element // is positive if (arr[i] > 0) pos_count++; // If the current element // is negative if (arr[i] < 0) neg_count++; } // For all the positive // elements of the array result = (int) Math.pow(2, pos_count); // For all the negative // elements of the array if (neg_count > 0) result *= Math.pow(2, neg_count - 1); // For the empty subsequence result -= 1; return result;}// Driver codepublic static void main(String[] args){ int arr[] = { 2, -3, -1, 4 }; int n = arr.length; System.out.print(cntSubSeq(arr, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python 3 implementation of the approach import math# Function to return the count of all # the subsequences with positive product def cntSubSeq(arr, n): # To store the count of positive # elements in the array pos_count = 0; # To store the count of negative # elements in the array neg_count = 0 for i in range (n): # If the current element # is positive if (arr[i] > 0) : pos_count += 1 # If the current element # is negative if (arr[i] < 0): neg_count += 1 # For all the positive # elements of the array result = int(math.pow(2, pos_count)) # For all the negative # elements of the array if (neg_count > 0): result *= int(math.pow(2, neg_count - 1)) # For the empty subsequence result -= 1 return result# Driver code arr = [ 2, -3, -1, 4 ] n = len (arr); print (cntSubSeq(arr, n)) # This code is contributed by ANKITKUMAR34 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the count of all // the subsequences with positive product static int cntSubSeq(int []arr, int n) { // To store the count of positive // elements in the array int pos_count = 0; // To store the count of negative // elements in the array int neg_count = 0; int result; for (int i = 0; i < n; i++) { // If the current element // is positive if (arr[i] > 0) pos_count++; // If the current element // is negative if (arr[i] < 0) neg_count++; } // For all the positive // elements of the array result = (int) Math.Pow(2, pos_count); // For all the negative // elements of the array if (neg_count > 0) result *= (int)Math.Pow(2, neg_count - 1); // For the empty subsequence result -= 1; return result; } // Driver code public static void Main() { int []arr = { 2, -3, -1, 4 }; int n = arr.Length; Console.Write(cntSubSeq(arr, n)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of all// the subsequences with positive productfunction cntSubSeq(arr, n) { // To store the count of positive // elements in the array let pos_count = 0; // To store the count of negative // elements in the array let neg_count = 0; let result; for (let i = 0; i < n; i++) { // If the current element // is positive if (arr[i] > 0) pos_count++; // If the current element // is negative if (arr[i] < 0) neg_count++; } // For all the positive // elements of the array result = Math.pow(2, pos_count); // For all the negative // elements of the array if (neg_count > 0) result *= Math.pow(2, neg_count - 1); // For the empty subsequence result -= 1; return result;}// Driver codelet arr = [2, -3, -1, 4];let n = arr.length;document.write(cntSubSeq(arr, n));</script> |
Output:
7
Time Complexity: O(n)
Auxiliary Space: O(1)
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