RemoveBoxes.java

/* (C) 2024 YourCompanyName */
package dynamic_programming;
/**
 * Created by gouthamvidyapradhan on 28/05/2019 Given several boxes with different colors
 * represented by different positive numbers. You may experience several rounds to remove boxes
 * until there is no box left. Each time you can choose some continuous boxes with the same color
 * (composed of k boxes, k >= 1), remove them and get k*k points. Find the maximum points you can
 * get.
 *
 * <p>Example 1: Input:
 *
 * <p>[1, 3, 2, 2, 2, 3, 4, 3, 1] Output: 23 Explanation: [1, 3, 2, 2, 2, 3, 4, 3, 1] ----> [1, 3,
 * 3, 4, 3, 1] (3*3=9 points) ----> [1, 3, 3, 3, 1] (1*1=1 points) ----> [1, 1] (3*3=9 points) ---->
 * [] (2*2=4 points) Note: The number of boxes n would not exceed 100.
 *
 * <p>Solution O(N ^ 4) For each sub-array [l, r] make a dp cache and calculate maximum of [l, i][1]
 * + [i + 1, r][1] or maximum of [l + 1, i - 1][n] + [i, r][1] where boxes[l] == boxes[i] where n is
 * the count of repetitions
 */
public class RemoveBoxes {

  int[][][] dp;

  public static void main(String[] args) {
    int[] boxes = {3, 3, 3};
    System.out.println(new RemoveBoxes().removeBoxes(boxes));
  }

  public int removeBoxes(int[] boxes) {
    dp = new int[boxes.length][boxes.length][boxes.length + 1];
    return calculate(0, boxes.length - 1, 1, boxes);
  }

  int calculate(int l, int r, int n, int[] boxes) {
    if (l > r) return 0;
    else {
      if (dp[l][r][n] != 0) return dp[l][r][n];
      dp[l][r][n] = (n * n) + calculate(l + 1, r, 1, boxes);
      for (int i = l + 1; i <= r; i++) {
        int center = 0, next = 0;
        if (boxes[l] == boxes[i]) {
          center = calculate(l + 1, i - 1, 1, boxes);
          next = calculate(i, r, n + 1, boxes);
        }
        dp[l][r][n] = Math.max(dp[l][r][n], center + next);
      }
    }
    return dp[l][r][n];
  }
}