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[NEW ALGORITHM] : Burst Balloons #1870

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@ananydev

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Issue will be closed if:

  1. You mention more than one algorithm. You can create a separate issue for each algorithm once the current one is completed.
  2. You propose an algorithm that is already present or has been mentioned in a previous issue.
  3. You create a new issue without completing your previous issue.

Note: These actions will be taken seriously. Failure to follow the guidelines may result in the immediate closure of your issue.


Name: Burst Balloons

[NEW ALGORITHM]

About:

Propose a new algorithm to be added to the repository
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 = 315 + 358 + 138 + 181 = 167
Example 2:

Input: nums = [1,5]
Output: 10

Constraints:

n == nums.length
1 <= n <= 300
0 <= nums[i] <= 100

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