finding triplets

Question

I know this type of question has been posted before, but i want to discuss something new.So, i am posting it.

Given an unsorted array of integers, find all triplets tha

Answer 1

First of all. Finding all triplets in worst case is O(n^3). Suppose you have n=3k numbers. K of them are 3, k are 4 and k are 5.

3,....,3,4,....,4,5,.....5

There are k^3 = n^3/27 = O(n^3) such triplets. Just printing them takes O(n^3) time.

Next will be explaining of 3SUM problem in such form:

There are numbers s_1, ..., s_n each of them in range [-u;u]. How many triplets a,b,c there are that a+b=c?

1. transforming. Get 2*u numbers a_-u, ..., a_0, a_1, ... a_u. a_i is amount of numbers s_j, that s_j = i. This is done in O(n+u)

2. res = a_0 * sum(i=-u..u, i=/=0, C(a_i, 2)) + C(a_0,3) a_0 = 0

3. Build a polynom P(x) = sum(i = -u...u, a_i*x^(i+u).

4. Find Q(x) = P(x)*P(x) using FFT.

5. Note that Q(x) = sum(i=-2u..2u, b_i*x^(i+2*u)), where b_i is number of pairs s_u,s_k that s_u+s_k = i (This includes using same number twice).

6. For all even i do b_i = b_i - a_(i/2). This will remove using same number twice.

7. Sum all b_i*a_i/2 - add this to res.

Example: to be more simple, I will assume that range for numbers is [0..u] and won't use any +u in powers of x. Suppose that we have numbers 1,2,3 - a_0 = 0, a_1 = 1, a_2 = 1, a_3 = 1

• res = 0

• P(x) = x + x^2 + x^3

• Q(x) = x^2 +2x^3+3x^4+2x^5+x^6

• After subtracting b_2 = 0, b_3 = 2, b_4 = 2, b_5 = 2, b_6 = 0

• res += 0*1/2 + 2*1/2 + 2*0/2 + 2*0/2 + 6*0/2 = 1