Count number of triples whose product is x [closed] - c++

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Given an array of integers and an integer x, I need to find the number of unordered triples in the array whose product is x. I have a brute-force O(n3) solution:
int solve(int a[], int n, int x) {
int ans = 0;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
for (int k = j+1; k < n; ++k)
if(a[i]*a[j]*a[k] == x)
ans++;
return ans;
}
Is there a faster way to do this, say in O(n log n) time?

Answer 1

You could
put all elements in a into a hashtable -- O(n)
Factorize x -- O(sqrt(x))
Build all possible triplets of divisors of x -- O(3^log(x))
Search for these triplets in your Hashtable -- O(1)
If a triplet exists, increase your counter -- O(1)
The runtime will now be dependent on the number of prime factors pf(x) of x, specifically O(3^pf(x)). Since the number of prime factors of x can be bounded by log(x) this yields O(3^log(x)) which could actually simplify further dependent on the exact base.

Answer 2

Code is in java but as you know, the idea remains the same.
Time complexity is O(n^1.5). I checked this with 10^5 elements and it works ok.
The idea is-
Put all elements of the list with their count in a map.
Now, if a[i]*a[j]*a[k] == x to be true, all these 3 have to be factors of x. Hence, if x%a[i] == 0, I hunt for checking all factors(pairs) that multiply to x/a[i] in sqrt(x/a[i]) time.
Since, we do it for every element in the list, complexity goes like this-
O(n) for adding elements to map.
O(n * n^0.5) for checking triplets.
So, total complexity = O(n) + O(n * n^0.5) = O(n^1.5) asymptotically.
ASSUMPTION - I am considering tuples as triplets having different indices in the array.
UPDATE - Current code considers different permutations of the same indices i,j,k as different tuples. You can fix it by having an additional map storing triplet indices as a string by joining them | (pipe) separated.
import java.util.*;
import java.io.*;
class Solution{
public static void main(String args[]) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String[] s = br.readLine().split("\\s");
int[] a = new int[n];
for(int i=0;i<n;++i){
a[i] = Integer.parseInt(s[i]);
}
int x = Integer.parseInt(br.readLine());
System.out.println(solve(a,n,x));
}
private static long solve(int a[], int n, int x) {
long tuples = 0;
Map<Integer,Integer> elements = new HashMap<>();
for(int i=0;i<a.length;++i){
elements.put(a[i],elements.getOrDefault(a[i],0) + 1);
}
for(int i=0;i<a.length;++i){
if(x%a[i] == 0){
tuples += factorPairs(x/a[i],elements,a[i]);
}
}
return tuples;
}
private static long factorPairs(int x,Map<Integer,Integer> elements,int factor1){
long pairs = 0;
for(int i=1;i*i<=x;++i){
int second = i;
int third = x/i;
if(second*third == x && elements.containsKey(second) && elements.containsKey(third)){
long second_cnt = elements.get(second) - (second == factor1 ? 1 : 0);
long third_cnt = elements.get(third) - (third == factor1 ? 1 : 0);
pairs += second_cnt * third_cnt;
}
}
return pairs;
}
}