Put Marbles in Bags

class Solution {
public:
    long long distributeMarbles(vector<int>& marbleWeights, int count) {
        int totalMarbles = marbleWeights.size();
        vector<int> sumWeights(totalMarbles - 1, 0);
        for (int index = 0; index < totalMarbles - 1; ++index) {
            sumWeights[index] += marbleWeights[index] + marbleWeights[index + 1];
        }
    
        sort(sumWeights.begin(), sumWeights.end());
    
        long long result = 0;
        for (int j = 0; j < count - 1; ++j) {
            result += sumWeights[totalMarbles - 2 - j] - sumWeights[j];
        }
    
        return result;
    }
};

In the "Put Marbles in Bags" solution using C++, the given function distributeMarbles efficiently allocates marble weights into a certain number of bags, aiming for an optimal distribution based on the sum of weights. The solution accomplishes this through the following steps:

1. Determine the total number of marbles using marbleWeights.size().
2. Initialize a vector sumWeights to store cumulative weights of consecutive marbles.
3. Loop through the marbleWeights array, summing weights of consecutive marbles and storing these in sumWeights.
4. Sort the sumWeights vector in ascending order to facilitate optimal pair selection.
5. Initialize a result accumulator (result) to store the final calculated difference between the selected weight sums.
6. Iterate through the sorted sumWeights, from highest to lowest, accumulating differences between complementary sums in the array.
7. Return the accumulated differences as the result of the function.

This method ensures a strategic distribution of weights, where differences in bag weights are minimized based on the sorting and selection approach. This approach manipulates the properties of sorted arrays to quickly achieve the desired result, applicable particularly in scenarios where bag count optimization based on weight is critical.

class Solution {
    public long calculateDifference(int[] weights, int maxPairs) {
        int length = weights.length;
        int[] combinedWeights = new int[length - 1];
        for (int i = 0; i < length - 1; ++i) {
            combinedWeights[i] = weights[i] + weights[i + 1];
        }
            
        // Sorting the calculated sums for the defined range
        Arrays.sort(combinedWeights, 0, length - 1);
            
        long result = 0l;
        for (int i = 0; i < maxPairs - 1; ++i) {
            result += combinedWeights[length - 2 - i] - combinedWeights[i];
        }
    
        return result;
    }
}

The Java solution provided addresses the problem of finding the difference between the heaviest and lightest pairs from a list of weight sums. The process involves summing adjacent weights and then calculating the differences between the highest and lowest sums after sorting. Here's how the solution functions:

• Start by calculating the sum of adjacent weights in the weights[] array and store these sums in a combinedWeights[] array.
• Sort the combinedWeights[] array.
• Calculate the total difference by considering sums from the higher-end and the lower-end of the combinedWeights[] array up to maxPairs - 1, ensuring the total pairs considered are one less than the provided maxPairs.
• Return the resultant difference.

Apply these steps:

1. Sum the weights of adjacent marbles.
2. Store these sums in a new array.
3. Sort this array of sums.
4. Calculate the difference by subtracting the sorted array's smallest summed values from the largest, ensuring the count of the pairs considered aligns with the specified maximum pairs minus one.
5. Return the total calculated difference as your result.

This approach guarantees that you efficiently find the desired difference based on the conditions specified, using simple array manipulations and sort operations.