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546. Remove Boxes

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

• Java
• C++
• Python
• Go

• 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);
      }
  };
  

• 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
  
  

• 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)
  }
  

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