Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note:

  • You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
  • 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Input: [3,1,5,8]
Output: 167
Explanation: nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
            coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

Solution

Time complexity : O(n^3)
Space complexity : O(n^2)

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int N = nums.size();
        nums.insert(nums.begin(), 1);
        nums.push_back(1);

        vector<vector<int>> dp(N+2, vector<int>(N+2, 0));
        for (int len=1; len<=N; ++len) {
            for (int i=1; i+len-1<=N; ++i) {
                int j = i+len-1;
                for (int k=i; k<=j; ++j) {
                    dp[i][j] = max(dp[i][j], dp[i][k-1]+nums[i-1]*nums[k]*nums[j+1]+dp[k+1][j]);
                }
            }
        }

        return dp[1][N];
    }
};

定義 dp[i][j]為範圍 i~j 的氣球，可獲得的最大收益。

k 代表在 i~j 的範圍中，第 k 個氣球為最後引爆的。

故第 k 顆氣球的收益由以下組成：

  1. dp[i][k-1] (範圍中，左邊爆光的最大收益)
  2. nums[i-1]*nums[k]*nums[j+1] (中間k引爆的收益)
  3. dp[k+1][j] (範圍中，右邊爆光的最大收益)

範圍長度由最少1個 ~ 最多N個(陣列長度)。

最後的答案為dp[1][N]，即範圍1~N可獲得的最大收益。