Partition Array for Maximum Sum | Front Partition

1. Overview and Problem Statement

When given an integer array and an integer $k$, the Partition Array for Maximum Sum problem requires us to partition the array into contiguous subarrays of at most length $k$. After partitioning, every element within an individual subarray transforms into the maximum value present in that specific subarray. Our ultimate goal is to find a valid partitioning configuration that maximizes the total sum of the array elements after these transformations.

• Input Example: nums = [1, 15, 7, 9, 2, 5, 10], $k = 3$
• Constraint: Each partition can have a length between $1$ and $k$ (inclusive). Lengths exceeding $k$ are strictly forbidden.

2. Core Mental Model: The Real-World Transformation Analogy

Imagine you are managing an array of numerical items where you group items into boxes (partitions). Each box can hold at most $k$ items.

• The moment you seal a box, a strict rule applies: every single item inside that box instantly adopts the value of the highest item currently residing in that box.
• Your financial reward is the sum of all items across all boxes.
• Therefore, you must strategically decide where to close one box and start the next so that when every item is boosted to its box's maximum, the grand total summation is as large as possible.

3. Exploring Partitioning Choices (Naive vs. Better Approach)

If we evaluate the example array [1, 15, 7, 9, 2, 5, 10] with $k = 3$:

• Partition Choice A:

  • Subarrays: [1, 15] (len 2), [7, 9, 2] (len 3), [5, 10] (len 2)
  • Transformed: [15, 15], [9, 9, 9], [10, 10]
  • Summation: $30 + 27 + 20 = 77$

• Partition Choice B:

  • Subarrays: [1, 15, 7] (len 3), [9] (len 1), [2, 5, 10] (len 3)
  • Transformed: [15, 15, 15], [9], [10, 10, 10]
  • Summation: $45 + 9 + 30 = 84$

Since Choice B yields $84$ while Choice A yields $77$, Choice B is superior. Because multiple overlapping partition configurations exist, we cannot rely on a greedy choice alone; we must explore all valid partitions using recursion and select the maximum possible sum.

4. Step-by-Step Evolution: The Front Partition Framework

Following the pattern established in problems like Palindrome Partitioning II, we apply the Front Partition technique:

1. Express everything in terms of an index: Define f(index) as the maximum sum obtainable from the subarray starting at index up to the end of the array (nums.size() - 1).
2. Try every valid partition from the current index: Use a loop to form partitions of length $1, 2, \dots, k$.
3. Take the best partition: Calculate the sum for the current partition plus the recursive result of the remaining array, and take the maximum across all choices.

5. Mathematical and Algorithmic Recurrence

For a state f(index):

• Base Case: When index == nums.size(), no elements remain, so return 0.
• Transitions: Initialize maxAns = Integer.MIN_VALUE, len = 0, and maxElement = Integer.MIN_VALUE. Iterate variable j from index to min(nums.size() - 1, index + k - 1):
  • Increment len++.
  • Update maxElement = max(maxElement, nums[j]).
  • Calculate current partition sum: sum = (len * maxElement) + f(j + 1).
  • Update maxAns = max(maxAns, sum).
• Return maxAns.

6. Complete Flow Walkthrough (Recursion & Memoization)

As recursive calls branch out, identical subproblems (calling f(index) with the same index multiple times) repeat. We optimize this by applying Memoization using a 1D DP array of size $N$ initialized to -1.

Memoized Recursive Implementation (Java)

class Solution {
    public int maxSumAfterPartitioning(int[] arr, int k) {
        int n = arr.length;
        int[] dp = new int[n];
        Arrays.fill(dp, -1);
        return solve(0, arr, k, n, dp);
    }

    private int solve(int index, int[] arr, int k, int n, int[] dp) {
        // Base Case: Reached the end of the array
        if (index == n) {
            return 0;
        }

        // Return memoized result if already computed
        if (dp[index] != -1) {
            return dp[index];
        }

        int maxAns = Integer.MIN_VALUE;
        int len = 0;
        int maxElement = Integer.MIN_VALUE;

        // Try every valid partition of length up to k
        for (int j = index; j < Math.min(n, index + k); j++) {
            len++;
            maxElement = Math.max(maxElement, arr[j]);
            int sum = (len * maxElement) + solve(j + 1, arr, k, n, dp);
            maxAns = Math.max(maxAns, sum);
        }

        return dp[index] = maxAns;
    }
}

7. Converting to Tabulation (Bottom-Up DP)

To transition from memoization to iterative tabulation:

1. Base Case Translation: dp[n] = 0.
2. Loop Direction: Since memoization moves from index $0$ to $n-1$, tabulation iterates from index $n-1$ down to $0$.
3. State Copy: Replicate the exact recurrence logic inside the outer loop.

Tabulation Implementation (Java)

class Solution {
    public int maxSumAfterPartitioning(int[] arr, int k) {
        int n = arr.length;
        int[] dp = new int[n + 1]; // Size n + 1 to handle base case dp[n] = 0
        
        // Base case dp[n] = 0 is already initialized by default in Java

        // Fill DP table from right to left (bottom-up)
        for (int index = n - 1; index >= 0; index--) {
            int maxAns = Integer.MIN_VALUE;
            int len = 0;
            int maxElement = Integer.MIN_VALUE;

            for (int j = index; j < Math.min(n, index + k); j++) {
                len++;
                maxElement = Math.max(maxElement, arr[j]);
                int sum = (len * maxElement) + dp[j + 1];
                maxAns = Math.max(maxAns, sum);
            }
            dp[index] = maxAns;
        }

        return dp[0];
    }
}

8. Complexity Analysis

Summary and Key Takeaways

• Problem Classification: Belongs to the family of Front Partition Dynamic Programming problems (similar to Palindrome Partitioning).
• State Definition: dp[index] represents the maximum sum achievable from the given index to the end of the array.
• Partition Constraint: At each state, the inner loop explores at most $k$ elements (index + k), ensuring bounded transitions.
• Subarray Transformation: The cost of any partition is calculated dynamically as (length of partition) * (maximum element in partition).
• Optimal Substructure: Overlapping subproblems allow memoization and tabulation to drastically reduce exponential time down to polynomial $O(N \times K)$ performance.
• Bottom-Up Conversion: Reversing the traversal direction from $n-1$ down to $0$ converts top-down memoized recursion into a clean, safe iterative tabulation solution.