20190718

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.

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 = 315 + 358 + 138 + 181 = 167

Solution

The first sight to see this problem, an idea came to my mind that: it should be dynamic programming.

But first I choose recursion to solve this problem, which is really slow and stucked at 24 / 70 test case.

Below is my first version code:

class Solution { public: int maxCoins(vector<int>& nums) { if(nums.size()==0) return 0; if(nums.size()==1) return nums[0]; int mm = 0; for(int i=0;i<nums.size();++i){ int n = nums[i]; // nums.erase(nums.begin()+i); if (i==0){ int tmp = n*nums[1]; nums.erase(nums.begin()+i); tmp += maxCoins(nums); mm = max(tmp,mm); nums.insert(nums.begin(),n); } else if(i==nums.size()-1){ int tmp = n*nums[i-1]; nums.erase(nums.begin()+i); tmp += maxCoins(nums); mm = max(tmp,mm); nums.insert(nums.end(),n); } else{ int tmp = nums[i-1]*n*nums[i+1]; nums.erase(nums.begin()+i); tmp += maxCoins(nums); mm = max(tmp,mm); nums.insert(nums.begin()+i,n); } } return mm; } };

today I am a little bit lazy, so I turn to discussion, and found a elegant solution:

it split the problem into sub problems: $maxCoins(nums[1:n])={\max }_{i}maxCoins(nums[1:i-1])+nums[i-1]\ast +maxCoins(nums[i+1:n])$ which means, it recurses on the last burst.

Below is the cpp code:

int maxCoins(vector<int>& nums) { int N = nums.size(); nums.insert(nums.begin(), 1); nums.insert(nums.end(), 1); // rangeValues[i][j] is the maximum # of coins that can be obtained // by popping balloons only in the range [i,j] vector<vector<int>> rangeValues(nums.size(), vector<int>(nums.size(), 0)); // build up from shorter ranges to longer ranges for (int len = 1; len <= N; ++len) { for (int start = 1; start <= N - len + 1; ++start) { int end = start + len - 1; // calculate the max # of coins that can be obtained by // popping balloons only in the range [start,end]. // consider all possible choices of final balloon to pop int bestCoins = 0; for (int final = start; final <= end; ++final) { int coins = rangeValues[start][final-1] + rangeValues[final+1][end]; // coins from popping subranges coins += nums[start-1] * nums[final] * nums[end+1]; // coins from final pop if (coins > bestCoins) bestCoins = coins; } rangeValues[start][end] = bestCoins; } } return rangeValues[1][N]; }

tbc