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# Burst Balloons LeetCode Solution

## Problem – Burst Balloons

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] -->  --> []
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`

### Burst Balloons LeetCode Solution in Java

``````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 = 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
``````

### Burst Balloons LeetCode Solution in Python

``````class Solution:
def maxCoins(self, nums):
A =  + nums + 

@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 )

return dfs(0, len(A) - 1)
``````

### Burst Balloons LeetCode Solution in C++

``````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[N];
}
``````
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##### Conclusion:

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