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Dorchester Center, MA 02124
You are given n
balloons, indexed from 0
to n - 1
. Each balloon is painted with a number on it represented by an array nums
. You are asked to burst all the balloons.
If you burst the ith
balloon, you will get nums[i - 1] * nums[i] * nums[i + 1]
coins. If i - 1
or i + 1
goes out of bounds of the array, then treat it as if there is a balloon with a 1
painted on it.
Return the maximum coins you can collect by bursting the balloons wisely.
Example 1:
Input: nums = [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
Example 2:
Input: nums = [1,5]
Output: 10
Constraints:
n == nums.length
1 <= n <= 300
0 <= nums[i] <= 100
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;
}
// 12 ms
class Solution:
def maxCoins(self, nums):
A = [1] + nums + [1]
@lru_cache(None)
def dfs(i, j):
return max([A[i]*A[k]*A[j] + dfs(i,k) + dfs(k,j) for k in range(i+1, j)] or [0])
return dfs(0, len(A) - 1)
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];
}
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