Burst Balloons

Description

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.

• 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

Example 1:

Input：[4, 1, 5, 10]
Output：270
Explanation：
nums = [4, 1, 5, 10] burst 1, get coins 4 * 1 * 5 = 20
nums = [4, 5, 10]   burst 5, get coins 4 * 5 * 10 = 200
nums = [4, 10]    burst 4, get coins 1 * 4 * 10 = 40
nums = [10]    burst 10, get coins 1 * 10 * 1 = 10
Total coins 20 + 200 + 40 + 10 = 270

Example 2:

Input：[3,1,5]
Output：35
Explanation：
nums = [3, 1, 5] burst 1, get coins 3 * 1 * 5 = 15
nums = [3, 5] burst 3, get coins 1 * 3 * 5 = 15
nums = [5] burst 5, get coins 1 * 5 * 1 = 5
Total coins 15 + 15 + 5  = 35

public int maxCoins(int[] AA) {
        // Write your code here
        int n = AA.length;
        if (n == 0) {
            return 0;
        }

        int i, j, k, len;
        int[] A = new int[n + 2];
        A[0] = A[n + 1] = 1;
        for (i = 1; i <= n; ++i) {
            A[i] = AA[i - 1];
        }

        n += 2;
        int[][] f = new int[n][n];
        for (i = 0; i < n - 1; ++i) {
            f[i][i+1] = 0; // balloon balloon
        }

        for (len = 3; len <= n; ++len) {
            for (i = 0; i <= n - len; ++i) {
                j = i + len - 1;
                f[i][j] = 0;
                for (k = i + 1; k < j; ++k) {
                    f[i][j] = Math.max(f[i][j], f[i][k] + f[k][j] + A[i] * A[k] * A[j]);
                }
            }
        }

        return f[0][n - 1];
    }