LeetCode [312] Burst Balloons

312. Burst Balloons

Hard

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

 =============

Eg.
nums = 3  1  5  8
                                     0  1  2  3  4  5
after extended nums = 1  3  1  5  8  1

bottom up

0  1  2  3  4  5
1  3  1  5  8  1
step 1: burst balloon 2 => 3*1*5 = 15 => dp[2][2] = dp[2][1]+dp[3][2]+15 = 15


0  1  2  3  4  5
1  3  1  5  8  1
step 2: burst balloon 3 => 3*5*8 = 120 => dp[2][3] = dp[2][2]+dp[4][3]+120 = 135


0  1  2  3  4  5
1  3  1  5  8  1
step 1: burst balloon 1 => 1*3*8= 24 => dp[1][3] = dp[1][0]+dp[2][3]+24 = 159

0  1  2  3  4  5
1  3  1  5  8  1
step 1: burst balloon 4 => 1*8*1 = 8 => dp[1][4] = dp[1][3]+dp[5][4]+8 = 167


 Ref
[1] https://leetcode.com/problems/burst-balloons/
[2] https://leetcode.com/discuss/72186/c-dynamic-programming-o-n-3-32-ms-with-comments

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//C++: TLE class Solution { public: int maxCoins(vector<int>& nums) { int n = nums.size(); if(n==0) return 0; int ret = 0; for(int i=0; i<n; ++i){ int tmp = (i==0?1:nums[i-1])*nums[i]*(i==n-1?1:nums[i+1]); vector<int> nums1 = nums; nums1.erase(nums1.begin()+i); tmp += maxCoins(nums1); ret = max(ret, tmp); } return ret; } }; class Solution { public: int maxCoins(vector<int>& nums) { int n = nums.size(); nums.insert(nums.begin(), 1); nums.insert(nums.end(), 1); //can be further optimized by removing all zeros //dp[s][e] is max coins by bursting all balloons from s to e vector<vector<int>> dp(n+2, vector<int>(n+2, 0)); for(int s = n; s>0; --s){ for(int e = s; e<=n; ++e){ int bestCoins = 0; for(int i = s; i<=e; ++i){ //coins is the max coins when balloon i is the last balloon int coins = dp[s][i-1] + dp[i+1][e] + nums[s-1]*nums[i]*nums[e+1]; bestCoins = max(bestCoins, coins); } dp[s][e] = bestCoins; } } return dp[1][n]; } };

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class Solution { public int maxCoins(int[] nums) { int n = nums.length; int[] numsExt = new int[n+2]; Arrays.fill(numsExt, 1); for(int i=1; i<=n; ++i) numsExt[i] = nums[i-1]; int[][] dp = new int[n+2][n+2]; for(int s=n; s>0; --s){ for(int e = s; e<=n; ++e){ int bestCoins = 0; for(int i=s; i<=e; ++i){ int coins = dp[s][i-1] + dp[i+1][e] + numsExt[s-1]*numsExt[i]*numsExt[e+1]; bestCoins = Math.max(bestCoins, coins); } dp[s][e] = bestCoins; } } return dp[1][n]; } }