Count number of triplets with product equal to given number with duplicates allowed | Set-2

Given an array of positive integers(may contain duplicates) and a number ‘m’, find the number of unordered triplets ((Ai, Aj, Ak) and (Aj, Ai, Ak) and other permutations are counted as one only) with product equal to ‘m’.

Examples:

Input: arr[] = { 1, 4, 6, 2, 3, 8}, M = 24
Output: 3
The triplets are {1, 4, 6} {1, 3, 8} {4, 2, 3}

Input: arr[] = { 0, 4, 6, 2, 3, 8}, M = 18
Output: 0
There are no triplets in this case

A solution with O(N2) has been discussed in the previous post. In this post a better approach with lesser complexity has been discussed.

Approach: The below algorithm is followed to solve the above problem.

  • Use a hash-map to count the frequency of every element in the given array.
  • Declare a set which can store triplets, so that only unordered triplets are taken to count.
  • Iterate from 1 to sqrt(m) in a loop(let variable be i), since the maximum number by which M is divisible is sqrt(M) leaving out M.
  • Check if M is divisible by i or not and i is present in the array of integers or not, if it is, then again loop from 1 to M/i.(let the loop variable be j).
  • Again Check if M is divisible by j or not and j is present in the array of integers or not, if it is then check if the remaining number that is ( (M / i) / j) is present or not.
  • If it is present, then a triplet has been formed. To avoid duplicate triplets, insert them in the set in sorted order.
  • Check if the set the size increases after the insertion of triplet if it does then use combinatorics to find the number of triplets.
  • To find the number of triplets, the following conditions will be there.
    1. If all of the Ai, Aj and Ak are unique, then number of combinations will be the product of their frequencies.
    2. If all of them are same, then we can only choose three of them, hence the formula stands at frequency \choose 3.
    3. If any of the two are same(let Ai and Aj), the count will be frequency[Ai] \choose 2 * frequency[Ak]

Below is the implementation of the above approach.

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// C++ program to find the
// number of triplets in array
// whose product is equal to M
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the triplets
int countTriplets(int a[], int m, int n)
{
  
    // hash-map to store the frequency of every number
    unordered_map<int, int> frequency;
  
    // set to store the unique triplets
    set<pair<int, pair<int, int> > > st;
  
    // count the number of times
    // every elememt appears in a map
    for (int i = 0; i < n; i++) {
        frequency[a[i]] += 1;
    }
  
    // stores the answer
    int ans = 0;
  
    // iterate till sqrt(m) since tnum2t is the
    // mamimum number tnum2t can divide M except itself
    for (int i = 1; i * i <= m; i++) {
  
        // if divisible and present
        if (m % i == 0 and frequency[i]) {
  
            // remaining number after division
            int num1 = m / i;
  
            // iterate for the second number of the triplet
            for (int j = 1; j * j <= num1; j++) {
  
                // if divisible and present
                if (num1 % j == 0 and frequency[j]) {
  
                    // remaining number after division
                    int num2 = num1 / j;
  
                    // if the third number is present in array
                    if (frequency[num2]) {
  
                        // a temp array to store the triplet
                        int temp[] = { num2, i, j };
  
                        // sort the triplets
                        sort(temp, temp + 3);
  
                        // get the size of set
                        int setsize = st.size();
  
                        // insert the triplet in ascending order
                        st.insert({ temp[0], { temp[1], temp[2] } });
  
                        // if the set size increases after insertion,
                        //  it means a new triplet is found
                        if (setsize != st.size()) {
  
                            // if all the number in triplets are unique
                            if (i != j and j != num2)
                                ans += frequency[i] * frequency[j] * frequency[num2];
  
                            // if Ai and Aj are same among triplets
                            else if (i == j && j != num2)
                                ans += (frequency[i] * (frequency[i] - 1) / 2)
                                       * frequency[num2];
  
                            // if Aj and Ak are same among triplets
                            else if (j == num2 && j != i)
                                ans += (frequency[j] * (frequency[j] - 1) / 2)
                                       * frequency[i];
  
                            // if three of them are
                            // same among triplets
                            else if (i == j and j == num2)
                                ans += (frequency[i] * (frequency[i] - 1) * (frequency[i] - 2) / 6);
  
                            // if Ai and Ak are same among triplets
                            else
                                ans += (frequency[i] * (frequency[i] - 1) / 2)
                                       * frequency[j];
                        }
                    }
                }
            }
        }
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int a[] = { 1, 4, 6, 2, 3, 8 };
    int m = 24;
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << countTriplets(a, m, n);
  
    return 0;
}

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Output:

3

Time Complexity: O(N * log N)



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