454. 4Sum II¶

Edited by Charles

Medium · Python · HashTable

Problem

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

The idea of the solution is counting how many times/combinations num1 (in nums1) + num2 (in nums2) pops up and save them in a dictionary.

Then iterate all combinations of num3 and num4 in nums3 and nums4. If the opposite number exists in the dictionary, let count be added by the combination number in the dictionary.

Original study notes follow, including intermediate snippets and earlier attempts.

The idea of the solution is counting how many times/combinations num1 (in nums1) + num2 (in nums2) pops up and save them in a dictionary.

Then iterate all combinations of num3 and num4 in nums3 and nums4. If the opposite number exists in the dictionary, let count be added by the combination number in the dictionary.

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 5, 2026

Python solution¶

class Solution:
    def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        dict = {}
        count = 0

        for num1 in nums1:
            for num2 in nums2:
                sum_12 = num1 + num2
                if sum_12 not in dict:
                    dict[sum_12] = 1
                else:
                    dict[sum_12] += 1

        for num3 in nums3:
            for num4 in nums4:
                if -num3-num4 in dict:
                    count += dict[-num3-num4]

        return count

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