README_EN.md

comments	difficulty	edit_url	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0546.Remove%20Boxes/README_EN.md	Memoization Array Dynamic Programming

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

Python3

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

Java

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;
    }
}

C++

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);
    }
};

Go

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)
}