Remove Boxes

• Define a recursive function solve(current_boxes) that takes a list of integers representing the current state of boxes.
• The base case for the recursion is when current_boxes is empty, in which case it returns 0.
• Initialize a variable max_points to 0 to keep track of the maximum score achievable from the current state.
• Iterate through the current_boxes list from left to right. For each position i, identify the contiguous block of same-colored boxes starting at i. Let this block end at index j.
• The size of this block is k = j - i + 1.
• Create a new list, next_boxes, by removing the block from i to j from current_boxes.
• Make a recursive call solve(next_boxes) and add the score for the current removal, k*k, to its result.
• Update max_points = max(max_points, k*k + solve(next_boxes)).
• To avoid re-evaluating the same sub-block, advance the iterator i to j+1 after processing the block.
• The function returns max_points.

The core idea is to define a recursive function, say calculate(boxes_list), that takes the current list of boxes as input. In this function, we iterate through the list to find all continuous blocks of same-colored boxes. For each block found (e.g., k boxes starting at index i), we calculate the points for removing it (k*k). We then create a new list of boxes by removing this block and make a recursive call calculate(new_boxes_list) to find the maximum score for the rest of the boxes. The total score for this choice is k*k + calculate(new_boxes_list). We keep track of the maximum score found among all possible block removals at the current step. The base case for the recursion is an empty list of boxes, for which the score is 0. This method is very inefficient because it recalculates scores for the same sub-configurations of boxes multiple times.

1// This is a conceptual illustration. A direct implementation is complex due to list manipulations2// and would be extremely slow, likely causing a Time Limit Exceeded error.3// For this reason, a full, runnable code snippet is omitted.4/*5public int solve(List<Integer> boxes) {6    if (boxes.isEmpty()) {7        return 0;8    }9    int maxScore = 0;10    for (int i = 0; i < boxes.size(); ) {11        int j = i;12        while (j + 1 < boxes.size() && boxes.get(j + 1).equals(boxes.get(i))) {13            j++;14        }15        int k = j - i + 1;16        List<Integer> nextBoxes = new ArrayList<>(boxes.subList(0, i));17        nextBoxes.addAll(boxes.subList(j + 1, boxes.size()));18        19        int currentScore = k * k + solve(nextBoxes);20        maxScore = Math.max(maxScore, currentScore);21        22        i = j + 1;23    }24    return maxScore;25}26*/

• Initialize a 3D memoization table memo[n][n][n] to store results of subproblems and avoid re-computation.
• Define a recursive function solve(l, r, k) which computes the max score for boxes[l...r] with k boxes of color boxes[l] available to the left.
• Base Case: If l > r, the subarray is empty, return 0.
• Memoization Check: If memo[l][r][k] is already computed, return the stored value.
• Main Logic:
  • First, an optimization: greedily find the end of the contiguous block of color boxes[l]. Let this block be boxes[l...l']. Update k to include the count of boxes in this block.
  • Option 1 (Remove Now): Calculate the score for removing the current group of (k+1) boxes immediately. The score is (k+1)^2 plus the score from the rest of the array, solve(l'+1, r, 0).
  • Option 2 (Merge Later): Iterate through the rest of the subarray from m = l'+1 to r. If boxes[m] has the same color as boxes[l], consider merging.
    • This path's score is the sum of points from removing the intermediate part solve(l'+1, m-1, 0) and then solving for the remaining part solve(m, r, k+1), where k+1 is the new count of boxes to be merged.
  • The result for solve(l, r, k) is the maximum of Option 1 and all possibilities from Option 2.
• Store the computed result in memo[l][r][k] before returning.
• The initial call to start the process is solve(0, n-1, 0).

The state for our recursive function with memoization is solve(i, j, k), representing the maximum points obtainable from the subarray boxes[i...j], given that there are k boxes to the left of index i that have the same color as boxes[i].

• Base Case: If i > j, the subarray is empty, so we return 0 points.
• Recursive Step: When computing solve(i, j, k), we have two main choices regarding the first box, boxes[i], and its k preceding same-colored boxes.
  1. Remove the group now: We can remove the contiguous block of boxes starting at i (say, of length p) along with the k preceding boxes. The points gained would be (p+k)^2. The problem then reduces to solving for the remaining subarray.
  2. Merge with a later group: We can postpone removing the boxes[i] group and try to merge it with another block of the same color further down the array. To do this, we find another box boxes[m] (where m > i) of the same color. We then recursively calculate the maximum score obtainable by first clearing the boxes between i and m, and then solving the problem for the subarray starting at m, but now with an increased count of preceding boxes.

We take the maximum over all these possibilities. To avoid recomputing the same state (i, j, k), we store the results in a 3D array memo[n][n][n]. The initial call is solve(0, n-1, 0).

1class Solution {2    private int[][][] memo;3    private int[] boxes;4 5    public int removeBoxes(int[] boxes) {6        int n = boxes.length;7        this.boxes = boxes;8        this.memo = new int[n][n][n];9        return solve(0, n - 1, 0);10    }11 12    private int solve(int l, int r, int k) {13        if (l > r) {14            return 0;15        }16        if (memo[l][r][k] != 0) {17            return memo[l][r][k];18        }19 20        int original_l = l;21        int original_k = k;22 23        // Optimization: group all consecutive boxes of same color as boxes[l]24        while (l + 1 <= r && boxes[l + 1] == boxes[l]) {25            l++;26            k++;27        }28 29        // Option 1: Remove the current group of boxes[l]30        // The group has (l - original_l + 1) boxes from the array, plus original_k from the left.31        // Total count is k + 1.32        int res = (k + 1) * (k + 1) + solve(l + 1, r, 0);33 34        // Option 2: Try to merge with other groups of the same color35        for (int m = l + 1; m <= r; m++) {36            if (boxes[m] == boxes[l]) {37                // If we find another box of the same color at index m,38                // we can try to remove the intermediate part (l+1, m-1)39                // and then solve for the rest, now with (k+1) boxes of color boxes[l]40                // to be grouped with boxes[m...].41                res = Math.max(res, solve(l + 1, m - 1, 0) + solve(m, r, k + 1));42            }43        }44 45        return memo[original_l][r][original_k] = res;46    }47}