Array Partition

class Solution {
public:
    const static int OFFSET = 10000;
        
    int pairSum(vector<int>& data) {
        int histogram[2 * OFFSET + 1] = {0};
        for (int num : data) {
            histogram[num + OFFSET]++;
        }
        
        int pairSumResult = 0;
        bool evenPosition = true;
        for (int index = 0; index <= 2 * OFFSET; index++) {
            while (histogram[index]) {
                if (evenPosition) pairSumResult += index - OFFSET;
                evenPosition = !evenPosition;
                histogram[index]--;
            }
        }
        return pairSumResult;
    }
};

The provided C++ code describes a solution to the problem of maximizing the sum of minima from pairs in an array, defined in the pairSum function within the Solution class.

Focus on how the function computes the sum:

• An array histogram is utilized to count occurrences of each integer. The offset allows handling of negative numbers by mapping the range of possible values to non-negative indices.
• Using a for loop, you scan through the entire histogram. Within this loop, a second while loop decrements each histogram count until it reaches zero, ensuring all occurrences of a number are considered.
• A boolean flag evenPosition is used to control the pairing. On true, which indicates the first number of a potential pair, the value (index - OFFSET) is added to the result pairSumResult. The flag toggles with each iteration to alternate between the first and second elements of the pairs.
• Finally, the total computed in pairSumResult is returned as the output of the function.

This method benefits from a histogram-based counting sort, optimizing pairing of elements and ensuring the time complexity is managed efficiently by reducing repetitive comparison operations.

class PairSolution {
    final static int OFFSET = 10000;
    
    public int pairSumming(int[] array) {
        // This array will track occurrences of each number
        int[] countArray = new int[2 * OFFSET + 1];
        for (int num : array) {
            countArray[num + OFFSET]++;
        }
        
        // Aggregate the sum in this variable
        int aggregateSum = 0;
        boolean takeNumber = true;
        for (int idx = 0; idx <= 2 * OFFSET; idx++) {
            while (countArray[idx] > 0) {
                // If the index is even, add the adjusted original element
                aggregateSum += (takeNumber ? idx - OFFSET : 0);
                // Toggle the status for next number
                takeNumber = !takeNumber;
                // Count down the occurrences
                countArray[idx]--;
            }
        }
        return aggregateSum;
    }
}

The provided solution in Java addresses the problem of finding the aggregate sum of pairs with an optimal grouping strategy to maximize the sum when paired. The algorithm functions by offsetting numbers to handle negative values, counting occurrences using an array, and then iterating to calculate the paired sum.

• Define a class PairSolution with a method pairSumming that accepts an integer array and returns an integer.
• Initialize a fixed offset of 10000 to adjust the range of numbers for possible negative values.
• Create countArray to track the occurrence of each number, adjusted by the offset.
• Iterate through the input array, updating the count for each number in countArray.
• Initialize aggregateSum to zero and set a boolean takeNumber to true for conditional addition in the upcoming loop.
• Process countArray to pair values optimally:
  • For each index with a non-zero count, alternate the addition of the number's original value (adjusted by subtracting the offset) to aggregateSum every other occurrence, thus following optimal pairing logic to maximize the sum.
  • Toggle the takeNumber boolean to control pairing.

The array processing strategy ensures that each number, adjusted for negative indexing, contributes optimally to the paired sum by leveraging an alternate addition based on their occurrences. The method returns the computed aggregateSum after parsing all numbers, ensuring a time-efficient solution through direct indexing and controlled iteration.

class Solution:
    def pairSum(self, values: List[int]) -> int:
        max_val = 10000
        # Frequency count of elements
        freq = [0] * (2 * max_val + 1)
        for num in values:
            freq[num + max_val] += 1
        
        # Initialize the result to zero
        result_sum = 0
        even_pos = True
        for idx in range(2 * max_val + 1):
            while freq[idx] > 0:
                if even_pos:
                    result_sum += idx - max_val
                even_pos = not even_pos
                freq[idx] -= 1
        return result_sum

The given Python script solves the problem of finding the maximum sum of paired integers in an array where each integer can only be used once. The function pairSum within the Solution class achieves this by organizing a sorted array into the max pairs from the smallest values up, accumulating the sum of the first (smaller) element in each pair. Here's the step-by-step breakdown of how it achieves this:

1. Define max_val to ensure all possible values of integers are covered within the index range of a frequency list by offsetting negative values.

2. Create a freq list initialized to zero to keep track of the frequency of each number in the array, where the index represents the number adjusted by max_val.

3. Loop through the input list values, updating the frequency of each number in the freq list.

4. Initialize result_sum to accumulate the sum of the first elements of the pairs sorted by value and even_pos as a boolean to toggle between positions in a pair (first or second).

5. Iterate over all possible values represented in the freq array. For each value present (freq[idx] > 0), add the number (adjusted back by subtracting max_val) to result_sum if it's in the first position of the pair, then toggle even_pos.

6. The result after processing is stored in result_sum, which is the sum of all minimum values from each pair.

This method ensures you efficiently pair and sum elements to get the maximum sum possible under the constraints, using an O(N) space complexity for frequency tracking and O(K) time complexity to iterate through values, where K could be much larger than N but still makes the method efficient for sufficiently large ranges of input values.