454.4Sum II

Tags: [map]

Link: https://leetcode.com/problems/4sum-ii/?tab=Description

Given four lists A, B, C, D of integer values, compute how many tuples(i, j, k, l)there are such thatA[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 -228to 228- 1 and the result is guaranteed to be at most 231- 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: hash_map

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
        """
        if A is None or B is None or C is None or D is None:
            raise ValueError('the input is invalid')

        arr_len = len(A)
        complement_map = {}

        for i in xrange(arr_len):
            for j in xrange(arr_len):
                sum = A[i] + B[j]
                if (-sum) in complement_map:
                    complement_map[-sum] += 1
                else:
                    complement_map[-sum] = 1

        counter = 0
        for i in xrange(arr_len):
            for j in xrange(arr_len):
                sum = C[i] + D[j]
                if sum in complement_map:
                    counter += complement_map[sum]

        return counter

Note:

• Time complexity = O(n^2), n is the number of the elements of the array.