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:
[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)
Approach #1: three dimensional DP. [C++] [Hard]
class Solution {
public:
int removeBoxes(vector<int>& boxes) {
int n = boxes.size();
m_ = vector<vector<vector<int>>>(n, vector<vector<int>>(n, vector<int>(n, 0)));
return dfs(boxes, 0, n-1, 0);
}
private:
vector<vector<vector<int>>> m_;
int dfs(const vector<int>& boxes, int l, int r, int k) {
if (l > r) return 0;
if (m_[l][r][k] != 0) return m_[l][r][k];
while (l < r && boxes[r-1] == boxes[r]) {--r; ++k;}
m_[l][r][k] = dfs(boxes, l, r-1, 0) + (k + 1) * (k + 1);
for (int i = l; i < r; ++i) {
if (boxes[i] == boxes[r])
m_[l][r][k] = max(m_[l][r][k], dfs(boxes, l, i, k + 1) + dfs(boxes, i + 1, r - 1, 0));
}
return m_[l][r][k];
}
};
Analysis:
dp[i][j][k] = max score of subarray b[i] ~ b[j] if there are k boxes that have the same color as b[j] following b[j]. Those k boxes are from box[j+1] ~ box[n], to simulate boxes with other colors are removed first.
Transition:
dp[i][j][k] = dp[i][j-1][0] + (k + 1)^2 # case 1
dp[i][p][k+1] + dp[p+1][j-1][0] # case 2
Case 1: drop box[j], remove k + 1 boxes.
Case 2: Try all breakpoints p, attach a[j] to a[p], i <= p < j, box[p] == box[j].
Ans:
dp[0][n-1][0]
Tim complexity: O(n^4) Space complexity: O(n^3)
Reference:
https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-546-remove-boxes/