计算产品等于给定数量的三胞胎的数量，允许重复

给定正整数数组(可能包含重复项)，任务是查找乘积等于给定数t的三元组数。
例子：

Input: arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
        t = 93
Output: 18

Input: arr = [4, 2, 4, 2, 3, 1]
        t = 8
Output: 4  
[(4, 2, 1), (4, 2, 1), (2, 4, 1), (4, 2, 1)]

原始的方法：要解决这个最简单的方法是，如果他们的产品是等于T来比较T和增量次数每一个可能的三元组。
下面是上述方法的实现：

C++

// C++ program for above implementation
#include
 
using namespace std ;
 
int main()
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;
     
    int arr[] = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = sizeof(arr) /
                 sizeof(arr[0]) ;
     
    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;
     
    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                     j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target / (arr[i] * arr[j]) ;
             
                    for(int k = j + 1 ; k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }
cout << "Total number of triplets found : "
     << totalCount ;
     
return 0 ;
}
 
// This code is contributed by ANKITRAI1

Java

// Java program for above implementation
class GFG
{
public static void main(String[] args)
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;
     
    int[] arr = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = arr.length;
     
    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;
     
    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                    j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target /
                             (arr[i] * arr[j]);
             
                    for(int k = j + 1 ;
                            k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }
     
System.out.println("Total number of triplets found : " +
                                            totalCount);
}
}
 
// This code is contributed by mits

Python3

# Python program for above implementation
 
# The target value for which we have
# to find the solution
target = 93
 
arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
length = len(arr)
 
# This variable contains the total
# count of triplets found
totalCount = 0
 
# Loop from the first to the third
# last integer in the list
for i in range(length - 2):
 
    # Check if arr[i] is a factor of target
    # or not. If not, skip to the next element
    if target % arr[i] == 0:
        for j in range(i + 1, length - 1):
 
            # Check if the pair (arr[i], arr[j]) can be
            # a part of triplet whose product is equal
            # to the target
            if target % (arr[i] * arr[j]) == 0:
 
                # Find the remaining element of the triplet
                toFind = target // (arr[i] * arr[j])
                for k in range(j + 1, length):
 
                    # If element is found. increment the
                    # total count of the triplets
                    if arr[k] == toFind:
                        totalCount += 1
 
print ('Total number of triplets found: ', totalCount)

C#

// C# program for above implementation
 
using System;
class GFG
{
public static void Main()
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;
      
    int[] arr = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = arr.Length;
      
    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;
      
    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                    j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target /
                             (arr[i] * arr[j]);
              
                    for(int k = j + 1 ;
                            k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }
      
Console.Write("Total number of triplets found : " +
                                            totalCount);
}
}

输出：

Total number of triplets found:  18

时间复杂度： O( $n^3$ )
高效方法：

1. 从数组中删除不是t因子的数字。
2. 然后对数组进行排序，这样我们就不必验证每个数字的索引来避免对数的额外计数。
3. 然后存储每个数字出现在字典计数中的次数。
4. 使用两个循环来检查有效三元组的前两个数字，方法是检查它们的乘积是否除以t
5. 找到三元组的第三个数字，并检查我们是否已经看过三元组，以避免重复计算
6. 计算该三元组的总可能组合，以使它们以相同顺序出现(所有对都应遵循顺序(x，y，z)以避免重新排列)

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // This function returns the total number of
  // combinations of the triplet (x, y, z) possible in the
  // given list
  static int Combinations(int x, int y, int z,
                          HashMap count)
  {
 
    int valx = count.getOrDefault(x, 0);
    int valy = count.getOrDefault(y, 0);
    int valz = count.getOrDefault(z, 0);
 
    if (x == y) {
      if (y == z) {
        return (valx * (valy - 1) * (valz - 2)) / 6;
      }
      else {
        return ((((valy - 1) * valx) / 2) * valz);
      }
    }
    else if (y == z) {
      return valx * (((valz - 1) * valy) / 2);
    }
    else {
      return (valx * valy * valz);
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // Value for which solution has to be found
    int target = 93;
    int ar[] = { 1, 31, 3, 1, 93, 3, 31, 1, 93 };
 
    // length of the array
    int N = ar.length;
 
    // Create a list of integers from arr which
    // contains only factors of the target
    // Using list comprehension
    ArrayList list = new ArrayList<>();
    for (int i = 0; i < N; i++)
      if (ar[i] != 0 && target % ar[i] == 0)
        list.add(ar[i]);
 
    // Sort the list
    Collections.sort(list);
    int length = list.size();
 
    // ArrayList to Array Conversion
    int[] arr
      = list.stream().mapToInt(i -> i).toArray();
 
    // Initialize the Map with the default value
    HashMap count = new HashMap<>();
    HashSet tripletSeen = new HashSet<>();
 
    // Count the number of times a value is present in
    // the list and store it in a Map for further
    // use
    for (int val : list)
      count.put(val, count.getOrDefault(val, 0) + 1);
 
    // Used to store the total number of triplets
    int totalCount = 0;
 
    for (int i = 0; i < length - 2; i++) {
      for (int j = i + 1; j < length - 1; j++) {
 
        // Check if the pair (arr[i], arr[j]) can be
        // a part of triplet whose product is equal
        // to the target
        if (target % (arr[i] * arr[j]) == 0) {
 
          int toFind = target / (arr[i] * arr[j]);
 
          // This condition makes sure that a
          // solution is not repeated
          int a[] = { arr[i], arr[j], toFind };
          Arrays.sort(a);
          String str
            = (a[0] + "#" + a[1] + "#" + a[2]);
 
          if (toFind >= arr[i] && toFind >= arr[j]
              && tripletSeen.contains(str)
              == false) {
 
            tripletSeen.add(str);
            totalCount += Combinations(
              arr[i], arr[j], toFind, count);
          }
        }
      }
    }
 
    System.out.println(
      "Total number of triplets found: "
      + totalCount);
  }
}
 
// This code is contributed by Kingash.

Python3

# Python3 code to find the number of triplets
# whose product is equal to a given number
# in quadratic time
 
# This function is used to initialize
# a dictionary with a default value
from collections import defaultdict
 
# Value for which solution has to be found
target = 93
arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
 
# Create a list of integers from arr which
# contains only factors of the target
# Using list comprehension
arr = [x for x in arr if x != 0 and target % x == 0]
 
# Sort the list
arr.sort()
length = len(arr)
 
# Initialize the dictionary with the default value
tripletSeen = defaultdict(lambda : False)
count = defaultdict(lambda : 0)
 
# Count the number of times a value is present in
# the list and store it in a dictionary for further use
for key in arr:
    count[key] += 1
 
# Used to store the total number of triplets
totalCount = 0
 
# This function returns the total number of combinations
# of the triplet (x, y, z) possible in the given list
def Combinations(x, y, z):
 
    if x == y:
        if y == z:
            return (count[x]*(count[y]-1)*(count[z]-2)) // 6
        else:
            return ((((count[y]-1)*count[x]) // 2)*count[z])
             
    elif y == z:
        return count[x]*(((count[z]-1)*count[y]) // 2)
     
    else:
        return (count[x] * count[y] * count[z])
 
for i in range(length - 2):
    for j in range(i + 1, length - 1):
 
        # Check if the pair (arr[i], arr[j]) can be a
        # part of triplet whose product is equal to the target
        if target % (arr[i] * arr[j]) == 0:
            toFind = target // (arr[i] * arr[j])
 
            # This condition makes sure that a solution is not repeated
            if (toFind >= arr[i] and toFind >= arr[j] and
                tripletSeen[(arr[i], arr[j], toFind)] == False):
                     
                tripletSeen[(arr[i], arr[j], toFind)] = True
                totalCount += Combinations(arr[i], arr[j], toFind)
 
print ('Total number of triplets found: ', totalCount)

输出：

Total number of triplets found:  18

时间复杂度： O( $n^2$ )