546. Remove Boxes

Description

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 getk*kpoints.
Find the maximum points you can get.

Example 1:
Input:

[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 boxesnwould not exceed 100.

Discussion

Method 1

Memorization

unordered_map<set, points>

Method 2

区间dp

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

private int removeBoxesSub(int[] boxes, int i, int j, int k, int[][][] dp) {
    if (i > j) return 0;
    if (dp[i][j][k] > 0) return dp[i][j][k];

    for (; i + 1 <= j && boxes[i + 1] == boxes[i]; i++, k++); // optimization: all boxes of the same color counted continuously from the first box should be grouped together
    int res = (k + 1) * (k + 1) + removeBoxesSub(boxes, i + 1, j, 0, dp);

    for (int m = i + 1; m <= j; m++) {
        if (boxes[i] == boxes[m]) {
            res = Math.max(res, removeBoxesSub(boxes, i + 1, m - 1, 0, dp) + removeBoxesSub(boxes, m, j, k + 1, dp));
        }
    }

    dp[i][j][k] = res;
    return res;
}