Question

546. Remove Boxes

中文文档

Description

You are 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 (i.e., composed of k boxes, k >= 1), remove them and get k * k points.

Return _the maximum points you can get_.

 

Example 1:

Input: boxes = [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)

Example 2:

Input: boxes = [1,1,1]
Output: 9

Example 3:

Input: boxes = [1]
Output: 1

 

Constraints:

• 1 <= boxes.length <= 100
• 1 <= boxes[i] <= 100

Solutions

Solution 1

class Solution:
  def removeBoxes(self, boxes: List[int]) -> int:
    @cache
    def dfs(i, j, k):
      if i > j:
        return 0
      while i < j and boxes[j] == boxes[j - 1]:
        j, k = j - 1, k + 1
      ans = dfs(i, j - 1, 0) + (k + 1) * (k + 1)
      for h in range(i, j):
        if boxes[h] == boxes[j]:
          ans = max(ans, dfs(h + 1, j - 1, 0) + dfs(i, h, k + 1))
      return ans

    n = len(boxes)
    ans = dfs(0, n - 1, 0)
    dfs.cache_clear()
    return ans

class Solution {
  private int[][][] f;
  private int[] b;

  public int removeBoxes(int[] boxes) {
    b = boxes;
    int n = b.length;
    f = new int[n][n][n];
    return dfs(0, n - 1, 0);
  }

  private int dfs(int i, int j, int k) {
    if (i > j) {
      return 0;
    }
    while (i < j && b[j] == b[j - 1]) {
      --j;
      ++k;
    }
    if (f[i][j][k] > 0) {
      return f[i][j][k];
    }
    int ans = dfs(i, j - 1, 0) + (k + 1) * (k + 1);
    for (int h = i; h < j; ++h) {
      if (b[h] == b[j]) {
        ans = Math.max(ans, dfs(h + 1, j - 1, 0) + dfs(i, h, k + 1));
      }
    }
    f[i][j][k] = ans;
    return ans;
  }
}

class Solution {
public:
  int removeBoxes(vector<int>& boxes) {
    int n = boxes.size();
    vector<vector<vector<int>>> f(n, vector<vector<int>>(n, vector<int>(n)));
    function<int(int, int, int)> dfs;
    dfs = [&](int i, int j, int k) {
      if (i > j) return 0;
      while (i < j && boxes[j] == boxes[j - 1]) {
        --j;
        ++k;
      }
      if (f[i][j][k]) return f[i][j][k];
      int ans = dfs(i, j - 1, 0) + (k + 1) * (k + 1);
      for (int h = i; h < j; ++h) {
        if (boxes[h] == boxes[j]) {
          ans = max(ans, dfs(h + 1, j - 1, 0) + dfs(i, h, k + 1));
        }
      }
      f[i][j][k] = ans;
      return ans;
    };
    return dfs(0, n - 1, 0);
  }
};

func removeBoxes(boxes []int) int {
  n := len(boxes)
  f := make([][][]int, n)
  for i := range f {
    f[i] = make([][]int, n)
    for j := range f[i] {
      f[i][j] = make([]int, n)
    }
  }
  var dfs func(i, j, k int) int
  dfs = func(i, j, k int) int {
    if i > j {
      return 0
    }
    for i < j && boxes[j] == boxes[j-1] {
      j, k = j-1, k+1
    }
    if f[i][j][k] > 0 {
      return f[i][j][k]
    }
    ans := dfs(i, j-1, 0) + (k+1)*(k+1)
    for h := i; h < j; h++ {
      if boxes[h] == boxes[j] {
        ans = max(ans, dfs(h+1, j-1, 0)+dfs(i, h, k+1))
      }
    }
    f[i][j][k] = ans
    return ans
  }
  return dfs(0, n-1, 0)
}