Given an array arr[] of N integers, the task is to find the count of all the subsequences of the array that have a negative products.
Examples:
Input: arr[] = {1, -5, -6}
Output: 4
Explanation
{-5}, {-6}, {1, -5} and {1, -6} are the only possible subsequencesInput: arr[] = {2, 3, 1}
Output: 0
Explanation
There is no such possible subsequence with negative product
Naive Approach:
Generate all the subsequences of the array and compute the product of all the subsequences. If the product is negative then increment the count by 1.
Efficient Approach:
- Count the number of positive and negative elements in the array
- An odd number of negative elements can be chosen for the subsequence to maintain the negative product. The number of different combinations of subsequences with an odd number of negative elements will be pow(2, count of negative elements – 1)
- Any number of positive elements can be chosen for the subsequence to maintain the negative product. The number of different combinations of subsequences with all the positive elements will be pow(2, count of positive elements)
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;
// Function to return the count of all// the subsequences with negative 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);
else
result = 0;
return result;
}// Driver codeint main()
{ int arr[] = { 3, -4, -1, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << cntSubSeq(arr, n);
return 0;
} |
Java
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to return the count of all // the subsequences with negative 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 *= Math.pow(2, neg_count - 1);
else
result = 0 ;
return result;
} // Driver code public static void main(String[] args)
{ int arr[] = { 3,-4, -1, 6 };
int n = arr.length;
System.out.print(cntSubSeq(arr, n));
} } // This code is contributed by ANKITKUMAR34 |
Python3
# Python 3 implementation of the approach import math
# Function to return the count of all # the subsequences with negative 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))
else:
result = 0
return result
# Driver code arr = [ 2, -3, -1, 4 ]
n = len (arr);
print (cntSubSeq(arr, n))
|
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the count of all // the subsequences with negative 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);
else
result = 0 ;
return result;
} // Driver code public static void Main(String[] args)
{ int []arr = { 3,-4, -1, 6 };
int n = arr.Length;
Console.Write(cntSubSeq(arr, n));
} }// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of all// the subsequences with negative productfunction cntSubSeq(arr, n)
{ // To store the count of positive
// elements in the array
var pos_count = 0;
// To store the count of negative
// elements in the array
var neg_count = 0;
var result;
for(var 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);
else
result = 0;
return result;
}// Driver codevar arr = [ 3, -4, -1, 6 ];
var n = arr.length;
document.write(cntSubSeq(arr, n));// This code is contributed by noob2000</script> |
Output:
8
Time Complexity: O(n)
Auxiliary Space: O(1)
Another Approach:
We can also count the number of subsequences with a negative product by subtracting total number of subsequences with positive subsequences from the total number of subsequences.
To find the total number of subsequences with a positive product using the approach discussed in this article.