LeetCode 454. 4Sum II

Problem Statement

(Source) 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

Solution

Naive solution would get an TLE, as the time complexity is $O(n^4)$. We can do some trade-off between time and space complexity.

class Solution(object):
    def fourSumCount(self, A, B, C, D):
        """
        :type A: List[int]
        :type B: List[int]
        :type C: List[int]
        :type D: List[int]
        :rtype: int
        """
        m = {}
        n = len(A)
        for i in xrange(n):
            for j in xrange(n):
                m[A[i] + B[j]] = m.get(A[i] + B[j], 0) + 1
        res = 0
        for i in xrange(n):
            for j in xrange(n):
                res += m.get(-(C[i] + D[j]), 0)
        return res

Complexity Analysis:

• Time Complexity: $O(n^2)$.
• Space Complexity: $O(n^2)$.