4Sum II

class Solution {
public:
    int sumOfFour(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
        vector<vector<int>> listGroups = {A, B, C, D};
        int n = int(listGroups.size());

        auto computeCounts = [&](int startIdx, int endIdx) {
            map<int, int> counts({{0, 1}});
            for (int i = startIdx; i < endIdx; i++) {
                map<int, int> tempCounts;
                for (int element : listGroups[i]) {
                    for (auto it = counts.begin(); it != counts.end(); ++it) {
                        tempCounts[element + it->first] += it->second;
                    }
                }
                counts = tempCounts;
            }
            return counts;
        };

        map<int, int> half1Counts = computeCounts(0, n / 2);
        map<int, int> half2Counts = computeCounts(n / 2, n);
        int totalCount = 0;
        for (auto it = half1Counts.begin(); it != half1Counts.end(); ++it) {
            totalCount += it->second * half2Counts[-it->first];
        }
        return totalCount;
    }
};

The provided C++ solution effectively solves the "4Sum II" problem by calculating the total number of tuples (i, j, k, l) from four lists A, B, C, and D such that the sum A[i] + B[j] + C[k] + D[l] is zero. The approach used in the solution involves dividing the problem into two parts using a two-point technique and leveraging the map data structure to store intermediary sums and their counts, which significantly reduces the complexity compared to a naive approach.

The solution uses a nested function, computeCounts, defined within sumOfFour to facilitate the computation of sum counts between given index ranges in subarrays. Here's an overview of how the code accomplishes this:

• Split the lists into two halves; compute sum counts for each half.
• computeCounts operates by iterating over the selected subarray elements and updates map entries where keys are the possible sums and values are their occurrence counts.
• This function uses nested loops to iterate over the elements of the lists and the current sums in the map to generate new possible sums, updating the count each time a new sum is generated.
• For combining the results, iterate through the sums computed from the first half of the lists. For each sum, find the corresponding negative sum in the second half's computed results to check how many ways the total sum can be zero. Multiply the occurrence counts from both halves for each valid sum pair to get the total number of tuples that sum to zero.

In summary, the code effectively reduces the computational complexity by breaking down the problem into manageable parts, utilizing maps for efficient sum lookup and aggregation. This approach ensures a much faster execution time compared to a brute-force method. The separation of concerns and clear logical flow also aids in understanding and maintaining the code.

import java.util.HashMap;

class Solver {
    private int[][] numbersArray;

    public int computeFourSumCount(int[] A, int[] B, int[] C, int[] D) {
        numbersArray = new int[][]{A, B, C, D};
        int length = numbersArray.length;
        Map<Integer, Integer> firstHalf = calculateSums(0, length / 2);
        Map<Integer, Integer> secondHalf = calculateSums(length / 2, length);
        int result = 0;
        for (int sum : firstHalf.keySet())
            result += firstHalf.get(sum) * secondHalf.getOrDefault(-sum, 0);
        return result;
    }

    private Map<Integer, Integer> calculateSums(int start, int end) {
        Map<Integer, Integer> countMap = new HashMap<>();
        countMap.put(0, 1);
        for (int i = start; i < end; i++) {
            Map<Integer, Integer> tempMap = new HashMap<>();
            for (int element : numbersArray[i]) {
                for (int sum : countMap.keySet()) {
                    tempMap.put(sum + element, tempMap.getOrDefault(sum + element, 0) + countMap.get(sum));
                }
            }
            countMap = tempMap;
        }
        return countMap;
    }
}

The given Java code presents a method to solve the "4Sum II" problem, which involves finding tuples from four separate arrays (A, B, C, D) such that their sum totals to zero. The solution utilizes a hashmap-based approach to efficiently count combinations of sums in two divided half segments of the combined arrays.

• Create a private 2D integer array numbersArray to store input arrays A, B, C, and D.
• Define a method computeFourSumCount which:
  • Initializes a Map named firstHalf built by calculateSums for the first two arrays.
  • Initializes a Map named secondHalf built by calculateSums for the last two arrays.
  • Combines results from these maps, adding product of matched sums from firstHalf against their negatives in secondHalf (reflecting the required zero sum).
• Implement calculateSums, a helper method which:
  • Initializes a hashmap countMap counting ways to achieve various sums.
  • Iteratively explores possible sum combinations through nested loops and updates countMap using another temporary map tempMap.

By decomposing the problem into two parts and using hashmaps to store intermediate results, this solution improves efficiency and manages complexity compared to a straightforward four-array nested loop approach, significantly optimizing the calculation of four sum combinations to zero.

from collections import Counter

class FourNumberSum:
    def computeFourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
        def computeSumCounts(lists: List[List[int]]) -> Counter:
            result = Counter({0: 1})
            for group in lists:
                temp_counter = Counter()
                for value in group:
                    for existing_sum in result:
                        temp_counter[existing_sum + value] += result[existing_sum]
                result = temp_counter
            return result

        grouped_lists = [nums1, nums2, nums3, nums4]
        half = len(grouped_lists) // 2
        left_part, right_part = computeSumCounts(grouped_lists[:half]), computeSumCounts(grouped_lists[half:])
        return sum(left_part[sum_val] * right_part[-sum_val] for sum_val in left_part)

The given Python3 code defines a class FourNumberSum which is used to solve the problem of finding the count of tuples that can be formed from four given lists of integers (nums1, nums2, nums3, nums4). Each tuple's sum must be zero. The solution employs a divide and conquer approach along with the use of hash tables to efficiently sum combinations and count occurrences.

• The class contains a method computeFourSumCount, which is the main function to compute the count of valid four-number tuples.

• Inside this method, computeSumCounts is a helper function used to compute hash maps (Counter from the collections module) representing all possible sums of subdivided list combinations and their respective counts.

• The initial list of numbers is divided into two halves (grouped_lists[:half] and grouped_lists[half:]). Each half is processed separately to create two counters (left_part and right_part) that store all possible sums and their counts.

• Finally, a comprehensive count is determined by traversing through each possible sum in the left_part and checking if the negative of that sum exists in the right_part. The result is the multiplication of corresponding counts (number of ways the sum from the first half can be matched with sums from the second half to result in zero).

This efficient combination of hash tables and divide-and-conquer reduces the complexity compared to the straightforward method (O(n^4)). Here, handling subdivided tasks separately and only focusing on matching sums greatly improves performance.