546. Remove Boxes - Detailed Explanation

Problem Statement

Description:
You are given several boxes with different colors represented by positive integers in an array boxes. You can remove boxes according to the following rule:

• Choose a contiguous group of boxes with the same color and remove them.
• When you remove a group of k boxes, you gain k * k points.
• After removal, the remaining boxes on both sides become adjacent.

Your goal is to remove all the boxes in such an order that maximizes the total points. Return the maximum points you can achieve.

Examples:

1. Example 1:

   • Input: boxes = [1, 3, 2, 2, 2, 3, 4, 3, 1]
   • Output: 23
   • Explanation:
     One optimal removal sequence is:
     1. Remove the three 2's → points = 3×3 = 9, remaining boxes: [1, 3, 3, 4, 3, 1].
     2. Remove the three 3's (by strategically merging them later) → points = 3×3 = 9, remaining boxes: [1, 4, 1].
     3. Remove the two 1's → points = 2×2 = 4, remaining boxes: [4].
     4. Remove the single 4 → points = 1×1 = 1.
        Total = 9 + 9 + 4 + 1 = 23.

2. Example 2:

   • Input: boxes = [1, 1, 1]
   • Output: 9
   • Explanation:
     Remove all three 1's together, and gain 3×3 = 9 points.

Constraints:

• (1 \leq \text{boxes.length} \leq 100)
• (1 \leq \text{boxes}[i] \leq 100)

Hints Before the Solution

1. Order Matters:
   Unlike many removal problems, the order in which you remove the boxes greatly affects the total score. Sometimes it is beneficial to delay the removal of a group in order to merge it with later boxes of the same color.

2. State Definition:
   Think about defining a state that represents not only a subarray of boxes but also extra boxes of the same color adjacent to it. For example, define dp(i, j, k) as the maximum points obtainable from the subarray boxes[i...j] given that there are k extra boxes (of the same color as boxes[i]) attached to the left.

3. Recurrence Idea:
   Consider two main options:

   • Remove the first group (including the extra k boxes) immediately.
   • Try to merge the first box with a similar colored box later in the array to form a larger group before removal.

Approaches

Approach: Dynamic Programming with Recursion and Memoization

Idea:
Define a function (dp(i, j, k)) where:

• (i, j) are indices in the boxes array, and
• (k) represents the number of boxes immediately to the left of index (i) that have the same color as boxes[i] (they have been “accumulated” for a larger removal).

The recurrence can be described as follows:

1. Base Case:
   When (i > j), there are no boxes to remove, so return 0.

2. Merge Same Color at Start:
   While (i+1 \leq j) and boxes[i+1] is the same as boxes[i], you can merge them. Increase (k) and move (i) forward.
   Let (count = k + 1) represent the total count of boxes of this color now.

3. Remove the Group Directly:
   One option is to remove the (count) boxes immediately and gain ((count)^2) points, then solve the remaining subproblem: [ dp(i, j, k) = (count)^2 + dp(i+1, j, 0) ]

4. Merge with Future Same Colored Boxes:
   For any index (m) where (i < m \leq j) and boxes[m] equals boxes[i], try to merge: [ dp(i, j, k) = \max \bigl( dp(i, j, k),\; dp(i+1, m-1, 0) + dp(m, j, count) \bigr) ] This means you remove the boxes between (i+1) and (m-1) first, then merge boxes[i] with boxes[m] (and the extra boxes) to get a higher score.

5. Memoization:
   Use a 3D memoization table (or a dictionary) keyed by ((i, j, k)) to avoid recomputing overlapping subproblems.

Why It Works:

• The recurrence considers both immediate removal and the possibility of merging with later boxes.
• Memoization ensures that each state is computed only once, reducing the exponential number of possibilities.

Code Implementations

Python Implementation