# 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 **((A _{i}, A_{j}, A_{k}) and (A_{j}, A_{i}, A_{k}) 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(N^{2}) 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.
- If all of the A
_{i}, A_{j}and A_{k}are unique, then number of combinations will be the prthe oduct of their frequencies. - If all of them are same, then we can only choose three of them, hence the formula stands at .
- If any of the two are same(let Ai and Aj), the count will be * frequency[A
_{k}]

- If all of the A

Below is the implementation of the above approach.

`// 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 does not increase 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

3

**Time Complexity:** O(N * log N)

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