Efficiently Solving the SUMFOUR Problem on SPOJ with Hashing

The SUMFOUR problem on SPOJ is a well-known challenge that requires finding the number of quadruplets in an array that sum up to a specific target. This article provides a step-by-step guide on how to solve the problem efficiently, leveraging hashing techniques to achieve On2 time complexity. We will walk through the problem statement, approach, and provide implementation details in Python.

Problem Statement

Given an integer array of size n and a target sum S, the task is to count the number of quadruplets a, b, c, d such that a b c d S and a, b, c, d are distinct indices in the array.

Constraints

The size of the array n can be up to 200, making a brute force solution with O(n^4) time complexity infeasible. We need an efficient solution that works within O(n^2) time complexity.

Approach

The key to solving this problem efficiently is to use hashing. By storing the sums of all pairs of numbers and their counts in a hash map, we can achieve the required time complexity.

Step 1: Store Pair Sums

We create a hash map to store the sums of all pairs of elements and their counts. This is done in a nested loop that iterates through each pair of indices (i, j) where i j and computes the sum sum_pair A[i] A[j]. The count of this sum is then stored in the hash map.

Step 2: Count Valid Quadruplets

Next, we iterate through each pair of indices (k, l) where k l. For each pair, we compute the required sum required_sum S - A[k] - A[l]. If this required sum exists in the hash map and the indices do not overlap, we count the number of valid quadruplets. It is crucial to avoid overcounting by ensuring that all indices are distinct.

Steps to Implement

Store Pair Sums: Create a hash map to store sums of all pairs of elements and their counts. Count Quadruplets: Iterate through each pair of indices and compute the required sum. Check if the required sum exists in the hash map and ensure the indices are distinct. Avoid Overcounting: Subtract cases where indices overlap to avoid double counting.

Example Code

Here is a sample implementation in Python:

def count_quadruplets(arr, S):
    n  len(arr)
    pair_sum  {}
    # Store the sums of all pairs
    for i in range(n):
        for j in range(i   1, n):
            sum_pair  arr[i]   arr[j]
            if sum_pair not in pair_sum:
                pair_sum[sum_pair]  0
            pair_sum[sum_pair]   1
    count  0
    # Find quadruplets
    for k in range(n):
        for l in range(k   1, n):
            required_sum  S - arr[k] - arr[l]
            if required_sum in pair_sum:
                count   pair_sum[required_sum]
            # Avoid overcounting
            for i in range(n):
                for j in range(i   1, n):
                    if i ! k and i ! l and j ! k and j ! l:
                        if arr[i]   arr[j]  required_sum:
                            count - 1
    return count // 2  # Each pair is counted twice

Example Usage

Here's how to use the function:

arr [1, 2, 3, 4, 5] S 10 print(count_quadruplets(arr, S)) 3

Complexity Analysis

Time Complexity: O(n^2) due to the nested loops for pairs. Space Complexity: O(n^2) in the worst case for storing the sums of pairs.

By using hashing to store and count pair sums, we significantly reduce the time complexity from O(n^4) to O(n^2), making the problem solvable within the provided constraints.

Conclusion

The use of hashing to store and count pair sums is a powerful technique that allows us to solve the SUMFOUR problem efficiently. While implementing this solution, it is essential to handle edge cases and ensure that the indices are distinct to avoid overcounting.