# Solving Leetcode Interviews in Seconds with AI: Burst Balloons


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "312" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the solution to the balloon bursting problem:

*   **High-level approach:** The problem is solved using dynamic programming. The core idea is to consider each balloon as the last one to be burst within a subrange. By iterating over different subrange lengths and last balloons to burst, the optimal solution is built bottom-up, avoiding redundant calculations. Padding the input array with 1s simplifies boundary condition handling.
*   **Complexity:** Runtime complexity is O(n^3) and storage complexity is O(n^2), where n is the number of balloons.

	
	# Code
	```python
	def maxCoins(nums):
    """
    Calculates the maximum coins you can collect by bursting the balloons wisely.

    Args:
        nums: A list of integers representing the numbers on the balloons.

    Returns:
        The maximum coins you can collect.
    """

    n = len(nums)
    nums = [1] + nums + [1]  # Pad with 1s for boundary conditions
    dp = [[0] * (n + 2) for _ in range(n + 2)]

    # Iterate over subrange lengths
    for length in range(2, n + 2):
        # Iterate over starting indices of subranges
        for left in range(n - length + 2):
            right = left + length
            # Iterate over possible last balloons to burst within the subrange
            for i in range(left + 1, right):
                dp[left][right] = max(dp[left][right],
                                       dp[left][i] + nums[left] * nums[i] * nums[right] + dp[i][right])

    return dp[0][n + 1]
	```