Partition Array for Maximum Sum

class Solution {
public:
    int maximumSumWithPartitions(vector<int>& array, int maxLen) {
        int len = array.size();
    
        int memo[len + 1];
        memset(memo, 0, sizeof(memo));
            
        for (int startIdx = len - 1; startIdx >= 0; startIdx--) {
            int maxVal = 0;
            int limit = min(len, startIdx + maxLen);
    
            for (int j = startIdx; j < limit; j++) {
                maxVal = max(maxVal, array[j]);
                // Compare and store the maximum sum possible with partitioning.
                memo[startIdx] = max(memo[startIdx], memo[j + 1] + maxVal * (j - startIdx + 1));
            }
        }
        return memo[0];
    }
};

The C++ solution provided addresses the challenge of determining the maximum sum obtained from partitioning an array with partitions restricted by a specified maximum length. The implementation primarily functions through dynamic programming to keep track and calculate the optimal partitions:

• Declare and initialized a memo array to cache results for subproblems, where each index represents the maximum sum from that start index to the end of the array.
• Iterate over the array elements from end to start to fill the memo dynamically, allowing calculation based on previously computed results for subarrays.
• During each iteration for a position startIdx, another loop explores potential partitions up to the defined maximum length (maxLen). Keep track of the maximum value (maxVal) in the current partition to update the memo for that starting index.
• The inner loop calculates the score of possible partitions from startIdx and updates the maximum sum for this position, based on earlier computed results and current partition values.
• Finally, return the value in memo[0] which holds the maximum sum achievable for the entire array using valid partitions.

This dynamic approach ensures all possibilities are correctly evaluated by breaking down the problem into sub-parts, and efficiently using previously stored information to solve each subproblem, leading to an overall optimal solution for the original problem.

class Solution {
    public int maximumPartitionSum(int[] array, int maxLen) {
        int arraySize = array.length;
    
        int[] dpArray = new int[arraySize + 1];
        Arrays.fill(dpArray, 0);
    
        for (int idx = arraySize - 1; idx >= 0; idx--) {
            int maxElement = 0;
            int endLimit = Math.min(arraySize, idx + maxLen);
    
            for (int pointer = idx; pointer < endLimit; pointer++) {
                maxElement = Math.max(maxElement, array[pointer]);
                dpArray[idx] = Math.max(dpArray[idx], dpArray[pointer + 1] + maxElement * (pointer - idx + 1));
            }
        }
        return dpArray[0];
    }
}

This Java solution tackles the problem of finding the maximum sum partition of an array with constraints on the maximum length of each partition.

• Begin by defining the maximumPartitionSum method which takes an array and an integer maxLen representing the maximum length of a partition.
• Initialize arraySize to store the length of the provided array.
• Create a dynamic programming array dpArray with an additional space to handle index offsets smoothly. Initialize all elements to zero using Arrays.fill(dpArray, 0).
• Process the array from the end to the beginning using a loop. This reverse traversal helps in solving subproblems that depend on the results of the next elements.
• For each element in the array, calculate the possible maximum values for subsequences starting at the current index and having lengths up to maxLen.
• Inside the loop, maintain a variable maxElement to track the maximum value found in the current subarray. It helps in calculating the potential maximum sum for the partition that extends up to maxLen elements from the current index.
• Use another loop nested inside the main loop to explore each partition length and update the dpArray with the maximum possible sum calculated using the formula dpArray[pointer + 1] + maxElement * (pointer - idx + 1).
• After iterating through all possible partitions and lengths, the first element of dpArray will hold the maximum sum achievable under the given constraints.

This algorithm efficiently calculates the solution by utilizing dynamic programming principles, ensuring that each sub-problem is solved only once and using previous solutions to construct the answer for the entire problem.