4Sum II Tuple Count
Reported by candidates from SambaNova Systems's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The detail that matters in this SambaNova Systems problem is the 64-bit return type. Reported in June 2022, 4Sum II Tuple Count hands you four arrays of up to 500 elements each and asks how many index tuples sum to zero. Brute force is 500^4, which is dead on arrival. This is a hash-table counting problem, and it's a classic. If you've seen the meet-in-the-middle trick, it takes ten minutes. If you blank under the clock, StealthCoder is the safety net running invisibly during the live OA, but the idea below is short enough to memorize tonight.
The problem
Given four integer arrays, count index tuples (i, j, k, l) such that nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0. Equal values at different indices represent different tuples. Return the count as a 64-bit integer. Function fourSumCount(nums1: int[], nums2: int[], nums3: int[], nums4: int[]) → long Examples Example 1 nums1 = [1,2] nums2 = [-2,-1] nums3 = [-1,2] nums4 = [0,2] return = 2 There are two complementary pair combinations. Example 2 nums1 = [0] nums2 = [0] nums3 = [0] nums4 = [0] return = 1 The single tuple sums to zero. Example 3 nums1 = [1,1] nums2 = [-1] nums3 = [0] nums4 = [0] return = 2 The equal ones occur at two distinct indices. Constraints 1 <= nums1.length, nums2.length, nums3.length, nums4.length <= 500. -10^9 <= numsX[i] <= 10^9. The returned count fits in a signed 64-bit integer.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Split the four arrays into two pairs. Compute every sum of nums1[i] + nums2[j] and store counts in a hash map, sum to frequency. That's up to 250,000 entries. Then loop over every nums3[k] + nums4[l], and add map[-(c+d)] to your answer. Total work is O(n^2) time and O(n^2) space. The pitfalls are small but real. Don't dedupe values, because equal numbers at different indices are different tuples, so you count frequencies, not presence. Use a 64-bit accumulator, since the count can reach 500^4 in the worst case. Use getOrDefault or equivalent so missing keys don't throw. Values reach 10^9, so a pair sum overflows 32-bit in some languages. If you freeze on the live OA, StealthCoder can surface this two-map approach so you just type it out and check the examples.
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4Sum II Tuple Count FAQ
What's the trick for 4Sum II Tuple Count?+
Meet in the middle. Hash all pair sums from the first two arrays with their counts, then for each pair sum from the last two arrays, look up the negation. It drops the work from n^4 to n^2, which is about 250,000 operations per side here.
How hard is this one really?+
Medium, and easy once you know the hash-map split. The hard part is seeing that you can separate the arrays into two halves. Without that insight you'll waste time on sorting and four-pointer ideas that don't fit.
Why does the problem stress a 64-bit return?+
With four arrays of 500 elements, the tuple count can hit 500^4, which is 62.5 billion. That overflows a 32-bit int. Declare your counter as long or the equivalent in your language, or large tests will fail silently.
Do duplicate values count as separate tuples?+
Yes. Example 3 shows it: two ones at different indices give two tuples. So store frequencies in the map and add the frequency on each match, not just 1. A set would give you wrong answers.
How do I prepare for this in 48 hours?+
Write the two-map solution from scratch twice, then test it against the three examples. Then do two related hash-map counting problems like two sum variants. Focus on the complement lookup pattern and the overflow handling, since both show up in many OA questions.