计算形成最小产品三胞胎的方法

给定正整数数组。我们需要找到索引(i，j，k)的三倍数(i

Examples:
Input : 5
        1 3 2 3 4
Output : 2
The triplets are (1, 3, 2)
and (1, 2, 3)

Input : 5
        2 2 2 2 2
Output : 5
In this example we choose three 2s 
out of five, and the number of ways 
to choose them is 5C3.

Input : 6
        1 3 3 1 3 2
Output : 1
There is only one way (1, 1, 2).

在此问题中出现以下情况。

1. 所有三个最小元素都是相同的。例如{1、1、1、1、2、3、4}。这种情况的解决方案是ⁿ C ₃ 。
2. 两个元素是相同的。例如{1、2、2、2、3}或{1、1、2、2}。在这种情况下，第一个(或最小元素)的出现次数不能超过2。如果最小元素出现两次，则答案是第二个元素的计数(我们从第二个元素的所有出现中仅选择1个。如果最小值元素出现一次，计数为ⁿ C ₂ 。
3. 这三个要素都是截然不同的。例如{1、2、3、3、5}。在这种情况下，答案是第三元素(或ⁿ C ₁ )的出现次数。

我们首先以递增顺序对数组进行排序。然后从开始计算第3元素的3元素的频率。让频率为“计数”。出现以下情况。

• 如果第三个元素等于第一个元素，则为否。三元组的数量将为(count-2)*(count-1)*(count)/ 6，其中count是第3个元素的频率。
• 如果第三个元素等于第二个元素，则否。三元组将为(count-1)*(count)/ 2。否则没有。三元组将是计数值。

C++

// CPP program to count number of ways we can
// form triplets with minimum product.
#include 
using namespace std;
 
// function to calculate number of triples
long long noOfTriples(long long arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Count occurrences of third element
    long long count = 0;
    for (long long i = 0; i < n; i++)
        if (arr[i] == arr[2])
            count++;
     
    // If all three elements are same (minimum
    // element appears at least 3 times). Answer
    // is nC3.
    if (arr[0] == arr[2])
        return (count - 2) * (count - 1) * (count) / 6;
 
    // If minimum element appears once. 
    // Answer is nC2.
    else if (arr[1] == arr[2])
        return (count - 1) * (count) / 2;
  
    // Minimum two elements are distinct.
    // Answer is nC1.
    return count;
}
 
// Driver code
int main()
{
    long long arr[] = { 1, 3, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << noOfTriples(arr, n);
    return 0;
}

Java

// Java program to count number of ways we can
// form triplets with minimum product.
import java.util.Arrays;
 
class GFG {
         
    // function to calculate number of triples
    static long noOfTriples(long arr[], int n)
    {
         
        // Sort the array
        Arrays.sort(arr);
     
        // Count occurrences of third element
        long count = 0;
        for (int i = 0; i < n; i++)
            if (arr[i] == arr[2])
                count++;
         
        // If all three elements are same (minimum
        // element appears at least 3 times). Answer
        // is nC3.
        if (arr[0] == arr[2])
            return (count - 2) * (count - 1) *
                                      (count) / 6;
     
        // If minimum element appears once.
        // Answer is nC2.
        else if (arr[1] == arr[2])
            return (count - 1) * (count) / 2;
     
        // Minimum two elements are distinct.
        // Answer is nC1.
        return count;
    }
     
    //driver code
    public static void main(String arg[])
    {
         
        long arr[] = { 1, 3, 3, 4 };
        int n = arr.length;
         
        System.out.print(noOfTriples(arr, n));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to count number
# of ways we can form triplets
# with minimum product.
 
# function to calculate number of triples
def noOfTriples (arr, n):
 
    # Sort the array
    arr.sort()
     
    # Count occurrences of third element
    count = 0
    for i in range(n):
        if arr[i] == arr[2]:
            count+=1
     
    # If all three elements are same
    # (minimum element appears at l
    # east 3 times). Answer is nC3.
    if arr[0] == arr[2]:
        return (count - 2) * (count - 1) * (count) / 6
     
    # If minimum element appears once.
    # Answer is nC2.
    elif arr[1] == arr[2]:
        return (count - 1) * (count) / 2
     
    # Minimum two elements are distinct.
    # Answer is nC1.
    return count
     
# Driver code
arr = [1, 3, 3, 4]
n = len(arr)
print (noOfTriples(arr, n))
 
# This code is contributed by "Abhishek Sharma 44"

C#

// C# program to count number of ways
// we can form triplets with minimum product.
using System;
 
class GFG {
     
// function to calculate number of triples
static long noOfTriples(long []arr, int n)
{
    // Sort the array
    Array.Sort(arr);
  
    // Count occurrences of third element
    long count = 0;
    for (int i = 0; i < n; i++)
        if (arr[i] == arr[2])
            count++;
      
    // If all three elements are same (minimum
    // element appears at least 3 times). Answer
    // is nC3.
    if (arr[0] == arr[2])
        return (count - 2) * (count - 1) * (count) / 6;
  
    // If minimum element appears once. 
    // Answer is nC2.
    else if (arr[1] == arr[2])
        return (count - 1) * (count) / 2;
   
    // Minimum two elements are distinct.
    // Answer is nC1.
    return count;
}
 
//driver code
public static void Main()
{
    long []arr = { 1, 3, 3, 4 };
    int n = arr.Length;
    Console.Write(noOfTriples(arr, n));
}
}
 
//This code is contributed by Anant Agarwal.

Javascript

    // Javascript program to count number of ways we can
    // form triplets with minimum product.
     
    // function to calculate number of triples
    function noOfTriples(arr, n)
    {
        // Sort the array
        arr.sort();
 
        // Count occurrences of third element
        let count = 0;
        for (let i = 0; i < n; i++)
            if (arr[i] == arr[2])
                count++;
 
        // If all three elements are same (minimum
        // element appears at least 3 times). Answer
        // is nC3.
        if (arr[0] == arr[2])
            return (count - 2) * (count - 1) * (count) / 6;
 
        // If minimum element appears once. 
        // Answer is nC2.
        else if (arr[1] == arr[2])
            return (count - 1) * (count) / 2;
 
        // Minimum two elements are distinct.
        // Answer is nC1.
        return count;
    }
 
    let arr = [ 1, 3, 3, 4 ];
    let n = arr.length;
    document.write(noOfTriples(arr, n));
     
    // This code is contributed by mukesh07.

输出：

1