Is 4Sum II easy or hard?

4Sum II is rated Medium difficulty. The brute force idea is straightforward, but recognizing the pair-sum hash map optimization requires familiarity with hash-based counting patterns.

4Sum II Python/Java solution

Python and Java implementations follow the same strategy: compute pair sums of the first two arrays using a dictionary or HashMap, then iterate through the other two arrays and accumulate matches using complement lookups.

How to solve 4Sum II in O(n^2)?

Precompute all pair sums from arrays A and B and store them in a hash map with their frequencies. Then iterate over pairs from C and D and check if the complement -(c + d) exists in the map. Each lookup counts how many valid tuples complete the zero sum.

What is the best approach for 4Sum II?

The hash map pair-sum approach is the most efficient. Compute all sums of A[i] + B[j] and store their frequencies in a hash map. Then iterate through pairs from C and D and look up the complement -(c + d). This reduces the complexity from O(n^4) to O(n^2).

Is 4Sum II asked at Google/Amazon/Meta?

Variants of multi-sum and pair-sum counting problems appear in interviews at companies like Amazon, Google, and Meta. The problem tests hash map usage, frequency counting, and the ability to reduce a higher-dimensional search space.

What data structure is used in 4Sum II?

A hash map (hash table) is the primary data structure. It stores frequencies of pair sums so the algorithm can perform constant-time lookups while scanning the remaining arrays.

What is the time complexity of 4Sum II?

The optimal solution runs in O(n^2) time. It builds a hash map of pair sums from the first two arrays in O(n^2), then scans pair sums from the remaining arrays with constant-time hash lookups.

Description Solution Video 12:02

454. 4Sum II

Medium57.9% acceptance4 companies

Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

0 <= i, j, k, l < n
nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1

Constraints:

n == nums1.length
n == nums2.length
n == nums3.length
n == nums4.length
1 <= n <= 200
-2²⁸ <= nums1[i], nums2[i], nums3[i], nums4[i] <= 2²⁸

Asked at 4 companies

Google Amazon Microsoft Meta

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Array Hash Table

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454. 4Sum II

Asked at 4 companies

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