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 -228 to 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
My code:
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
"""
length=len(A)
if length==0:
return 0
sum_dict1={}
for i in range(length):
for j in range(length):
sum_value=A[i]+B[j]
if sum_value not in sum_dict1:
sum_dict1[sum_value]=1
else:
sum_dict1[sum_value]+=1
sum_dict2={}
for i in range(length):
for j in range(length):
sum_value=C[i]+D[j]
if sum_value not in sum_dict2:
sum_dict2[sum_value]=1
else:
sum_dict2[sum_value]+=1
overall=0
for key in sum_dict1:
key2=-key
if key2 in sum_dict2:
overall+=sum_dict1[key]*sum_dict2[key2]
#print(sum_dict1)
#print(sum_dict2)
return overall
Solution: https://leetcode.com/problems/4sum-ii/