计算获得包含最多一个负元素的正积的三胞胎的方法

给定大小为N ( 1≤N≤10 ⁵ )的数组arr [] ，任务是找到选择三元组i，j和k的方式数，以使i 和乘积arr [i] * arr [j] * arr [k]为正。
注意：每个三元组最多可以包含一个负数元素。

例子：

Input: arr[] = { 2, 5, -9, -3, 6} 
Output: 1 
Explanation: The total number ways to obtain a triplet i, j and k to satisfy given conditions is 1 {0, 1, 4}.

Input : arr[] = {2, 5, 6, -2, 5} 
Output : 4 
Explanation: The total number ways to obtain a triplet i, j and k to satisfy given conditions are 4 {0, 1, 2}, {0, 1, 4}, {1, 2, 4} and {0, 2, 4}.

方法：三元组的所有可能组合如下：

• ＃个负数元素或2个负数元素和1个正数元素。不能将这两种组合视为三元组中允许的最大负元素为1。
• 2个负( -ve )元素和1个正( + ve )元素。由于三元组的乘积将为负，因此无法考虑三元组。
• 3个积极因素。

请按照以下步骤解决问题：

• 遍历数组并计算正数组元素的频率，例如freq 。

• 使用公式PnC = ^N C ₃ =(N *(N – 1)*(N – 2))/ 6从数组元素的频率中选择有效三元组的方法计数。将获得的计数加到答案中。

• 打印获得的计数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to calculate
// possible number of triplets
long long int possibleTriplets(int arr[], int N)
{
    int freq = 0;
    // counting frequency of positive numbers
    // in array
 
    for (int i = 0; i < N; i++) {
 
        // If current array
        // element is positive
        if (arr[i] > 0) {
 
            // Increment frequency
            freq++;
        }
    }
 
    // Select a triplet from freq
    // elements such that i < j < k.
    return (freq * 1LL * (freq - 1)
            * (freq - 2))
           / 6;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 5, -9, -3, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << possibleTriplets(arr, N);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
 
// Function to calculate
// possible number of triplets
static int possibleTriplets(int arr[], int N)
{
    int freq = 0;
   
    // counting frequency of positive numbers
    // in array
    for (int i = 0; i < N; i++)
    {
 
        // If current array
        // element is positive
        if (arr[i] > 0)
        {
 
            // Increment frequency
            freq++;
        }
    }
 
    // Select a triplet from freq
    // elements such that i < j < k.
    return (int) ((freq * 1L * (freq - 1)
            * (freq - 2))
           / 6);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 5, -9, -3, 6 };
    int N = arr.length;
    System.out.print(possibleTriplets(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 Program to implement
# the above approach
 
# Function to calculate
# possible number of triplets
def possibleTriplets(arr, N):
    freq = 0
     
    # counting frequency of positive numbers
    # in array
    for i in range(N):
 
        # If current array
        # element is positive
        if (arr[i] > 0):
 
            # Increment frequency
            freq += 1
 
    # Select a triplet from freq
    # elements such that i < j < k.
    return (freq * (freq - 1) * (freq - 2)) // 6
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 5, -9, -3, 6]
    N = len(arr)
 
    print(possibleTriplets(arr, N))
 
    # This code is contributed by mohit kumar 29

C#

// C# Program to implement
// the above approach
using System;
 
public class GFG
{
 
// Function to calculate
// possible number of triplets
static int possibleTriplets(int []arr, int N)
{
    int freq = 0;
   
    // counting frequency of positive numbers
    // in array
    for (int i = 0; i < N; i++)
    {
 
        // If current array
        // element is positive
        if (arr[i] > 0)
        {
 
            // Increment frequency
            freq++;
        }
    }
 
    // Select a triplet from freq
    // elements such that i < j < k.
    return (int) ((freq * 1L * (freq - 1)
            * (freq - 2)) / 6);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 2, 5, -9, -3, 6 };
    int N = arr.Length;
    Console.Write(possibleTriplets(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

输出：

1

时间复杂度： O(N)
辅助空间： O(1)