Leetcode: Burst Balloons

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: 
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

    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

这道题先参见dietpepsi的讲解：https://leetcode.com/discuss/72216/share-some-analysis-and-explanations

Be Naive First

When I first get this problem, it is far from dynamic programming to me. I started with the most naive idea the backtracking.

We have n balloons to burst, which mean we have n steps in the game. In the i th step we have n-i balloons to burst, i = 0~n-1. Therefore we are looking at an algorithm of O(n!). Well, it is slow, probably works for n < 12 only.

Of course this is not the point to implement it. We need to identify the redundant works we did in it and try to optimize.

Well, we can find that for any balloons left the maxCoins does not depends on the balloons already bursted. This indicate that we can use memorization (top down) or dynamic programming (bottom up) for all the cases from small numbers of balloon until n balloons. How many cases are there? For k balloons there are C(n, k) cases and for each case it need to scan the k balloons to compare. The sum is quite big still. It is better than O(n!) but worse than O(2^n).

Better idea

We then think can we apply the divide and conquer technique? After all there seems to be many self similar sub problems from the previous analysis.

Well, the nature way to divide the problem is burst one balloon and separate the balloons into 2 sub sections one on the left and one one the right. However, in this problem the left and right become adjacent and have effects on the maxCoins in the future.

Then another interesting idea come up. Which is quite often seen in dp problem analysis. That is reverse thinking. Like I said the coins you get for a balloon does not depend on the balloons already burst. Therefore instead of divide the problem by the first balloon to burst, we divide the problem by the last balloon to burst.

Why is that? Because only the first and last balloons we are sure of their adjacent balloons before hand!

For the first we have nums[i-1]*nums[i]*nums[i+1] for the last we have nums[-1]*nums[i]*nums[n].

OK. Think about n balloons if i is the last one to burst, what now?

We can see that the balloons is again separated into 2 sections. But this time since the balloon i is the last balloon of all to burst, the left and right section now has well defined boundary and do not affect each other! Therefore we can do either recursive method with memoization or dp.

Final

Here comes the final solutions. Note that we put 2 balloons with 1 as boundaries and also burst all the zero balloons in the first round since they won't give any coins. The algorithm runs in O(n^3) which can be easily seen from the 3 loops in dp solution.

状态转移方程：

dp[l][r] = max(dp[l][r], nums[l] * nums[m] * nums[r] + dp[l][m] + dp[m][r])

dp[l][r]表示扎破(l, r)范围内所有气球获得的最大硬币数，不含边界；

l与r的跨度k从2开始逐渐增大；

三重循环依次枚举范围跨度k，左边界l，中点m；右边界r = l + k；

因为m是（l,r）区间内最后剩下的气球，打爆它最后的coin数是 num[l]*nums[m]*nums[r], 之前已经打爆的左区间（l,m）和右区间（m,r）最大coin数是dp[l][m] 和dp[m][r], 递推式出来了，注意l,r都是在边界外的

最后注意一点，一开始要先把为0的气球打破，这是最优的做法

public class Solution {
    public int maxCoins(int[] nums) {
        int[] arr = new int[nums.length+2];
        int n = 1;
        for (int num : nums) {
            if (num > 0) 
                arr[n++] = num;
        }
        arr[0] = arr[n++] = 1;
        int[][] dp = new int[n][n];
        for (int len=2; len<=n-1; len++) {
            for (int l=0; l<=n-1-len; l++) {
                int r = l+len;
                for (int m=l+1; m<r; m++) {
                    dp[l][r] = Math.max(dp[l][r], arr[l]*arr[r]*arr[m]+dp[l][m]+dp[m][r]);
                }
            }
        }
        return dp[0][n-1];
    }
}

还有divide and conquer的做法：（未深究）

public int maxCoins(int[] iNums) {
    int[] nums = new int[iNums.length + 2];
    int n = 1;
    for (int x : iNums) if (x > 0) nums[n++] = x;
    nums[0] = nums[n++] = 1;


    int[][] memo = new int[n][n];
    return burst(memo, nums, 0, n - 1);
}

public int burst(int[][] memo, int[] nums, int left, int right) {
    if (left + 1 == right) return 0;
    if (memo[left][right] > 0) return memo[left][right];
    int ans = 0;
    for (int i = left + 1; i < right; ++i)
        ans = Math.max(ans, nums[left] * nums[i] * nums[right] 
        + burst(memo, nums, left, i) + burst(memo, nums, i, right));
    memo[left][right] = ans;
    return ans;
}