package cloud.dqn.leetcode

/**
 * https://leetcode.com/problems/4sum-ii/description/
    Given four lists A, B, C, D of integer values, compute how many
    tuples (i, j, k, l) there are such that
    A[i] + B[j] + C[k] + D[l] is zero.

    To make problem a bit easier, all A, B, C, D have same length
    of N where 0 ≤ N ≤ 500.
    All integers are in the range of -2^28 to 2^28 - 1 and
    the result is guaranteed to be at most 2^31 - 1.

Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
 */
class FourSum2Kt {
    companion object {
        val MAX_VALUE = Math.pow(2.0, 28.0).toInt()
        val MIN_VALUE = (MAX_VALUE * -1) -1
    }
    class Solution {
        fun fourSumCount(A: IntArray, B: IntArray, C: IntArray, D: IntArray): Int {
            var count = 0
            val diffAB = HashMap<Int, Int>(A.size * B.size)

            A.forEach { aValue ->
                B.forEach { bValue ->
                    val diff = 0 - (aValue + bValue)
                    diffAB[diff] = (diffAB[diff] ?: 0) + 1
                }
            }

            C.forEach { cValue ->
                D.forEach { dValue ->
                    diffAB[cValue + dValue]?.let {
                        count += it
                    }
                }
            }

            return count
        }
    }
}