546

546. Remove Boxes (Hard)

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.

 

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)

 

Constraints:

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

Related Topics:
Dynamic Programming, Depth-first Search

Similar Questions:

• Strange Printer (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/remove-boxes/
// Author: github.com/lzl124631x
// Time: O(N^4)
// Space: O(N^3)
// Ref: https://leetcode.com/problems/remove-boxes/discuss/101310/Java-top-down-and-bottom-up-DP-solutions
class Solution {
    vector<vector<vector<int>>> dp;
    int solve(vector<int> &A, int i, int j, int k) {
        if (i > j) return 0;
        if (dp[i][j][k] > 0) return dp[i][j][k];
        for (; i + 1 <= j && A[i + 1] == A[i]; ++i, ++k);
        int ans = (k + 1) * (k + 1) + solve(A, i + 1, j, 0);
        for (int m = i + 1; m <= j; ++m) {
            if (A[i] != A[m]) continue;
            ans = max(ans, solve(A, i + 1, m - 1, 0) + solve(A, m, j, k + 1));
        }
        return dp[i][j][k] = ans;
    }
public:
    int removeBoxes(vector<int>& A) {
        int N = A.size();
        dp.assign(N, vector<vector<int>>(N, vector<int>(N)));
        return solve(A, 0, N - 1, 0);
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/remove-boxes/
// Author: github.com/lzl124631x
// Time: O(N^4)
// Space: O(N^3)
// Ref: https://leetcode.com/problems/remove-boxes/discuss/101310/Java-top-down-and-bottom-up-DP-solutions
class Solution {
public:
    int removeBoxes(vector<int>& A) {
        int N = A.size();
        vector<vector<vector<int>>> dp(N, vector<vector<int>>(N, vector<int>(N)));
        for (int i = 0; i < N; ++i) {
            for (int k = 0; k <= i; ++k) dp[i][i][k] = (k + 1) * (k + 1);
        }
        for (int len = 1; len < N; ++len) {
            for (int j = len; j < N; ++j) {
                int i = j - len;
                for (int k = 0; k <= i; ++k) {
                    int ans = (k + 1) * (k + 1) + dp[i + 1][j][0];
                    for (int m = i + 1; m <= j; ++m) {
                        if (A[m] != A[i]) continue;
                        ans = max(ans, dp[i + 1][m - 1][0] + dp[m][j][k + 1]);
                    }
                    dp[i][j][k] = ans;
                }
            }
        }
        return dp[0][N - 1][0];
    }
};