### ryuga222's blog

By ryuga222, history, 7 months ago, , https://www.geeksforgeeks.org/count-number-triplets-product-equal-given-number/
Can someone provide a faster solution.
Constraints : 1 <= n <= 1e5
1 <= x,A[i] <= 1e7 for all 1<= i <=n Comments (9)
 » Auto comment: topic has been updated by ryuga222 (previous revision, new revision, compare).
 » Auto comment: topic has been updated by ryuga222 (previous revision, new revision, compare).
 » Hint:Let p, q, r be elements in array A such that p*q*r==x.p, q, r will be factors of x (including 1 & x).1 <= x <= 1e7Therefore, upper bound of number of factors of x is 2*square_root(1e7).
•  » » thanks. understood it.
 » 7 months ago, # | ← Rev. 2 →   it can be done in O(n*factors(x)*accesstime(frequency table)) but I guess there will be much more efficient solutions.Have three frequency tables(Maps/arrays/or other DS) for counting the frequency of a particular divisor of X. Table 1 will store number of ways to get value y(divisor of x) using just 1 number from the array. Table 2 will store number of ways to get value y using 2 numbers from the array. Table 3 will store number of ways to get value y using 3 numbers from the array.Initially all values in the table is 0.Now process the array from left to right(1..n) for an element a in the current position and for all possible divisors(d) of X we are going to update all the tables. if a doesn't divide d then the tables remain same else let p=d/a. Then table3[d]+=table2[p] i.e to get d using 3 divisors with one of them as a then we need to know the number of ways to form p using two divisor till the elements that we have seen previously. Similarly table2[d]+=table1[p] and table1[a]+=1.Make sure for each element first update the table 3,then table 2 then table 1 as 3 depends on 2 which again depends on 1.Now after processing all the elements of the array the answer is table3[X].
•  » » thanks, understood it.
 » 7 months ago, # | ← Rev. 4 →   SolutionWe can do it in (factors(x))^2 if we keep a hashmap or any similar one. You can google and find out that there are max 400 divisors for any number less than 1e7 which makes it even pass a (factors(x))^3 solution as my (factors(x))^3 solution was passing in the contest. Let me know if you have any problem in understanding any part of the code.
•  » » understood it. Do you have a link to the question where I can submit and try the solution??
 » Even though this isn't appropriate for a contest, there exists a randomized solution in by reducing the problem to 3SUM, since .