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Introduction

LeetCode 312: Burst Balloons asks us to maximize the number of coins we can collect by bursting all the balloons in a specific order. Each balloon has a value, and when a balloon is burst, it releases coins equal to the product of the values of the adjacent balloons. The goal is to find the optimal order to burst the balloons to maximize the sum of the coins collected.

This problem can be efficiently solved using dynamic programming (DP), which allows us to break down the problem into smaller subproblems and build up the solution.

Problem Statement

You are given an array nums representing coins in balloons. You are initially given a row of balloons, each balloon has a certain value. Once you burst a balloon, you will get coins equal to the product of the values of the adjacent balloons.

If the balloon at index i is burst, the coins you get will be:

• nums[i-1] * nums[i] * nums[i+1] (if it is not on the boundary).

Your task is to burst all the balloons in an optimal order to maximize the total coins you can collect.

Note: You may not burst any balloon at the boundaries (first and last), but you can treat the array as if it has virtual balloons with value 1 at both ends.

Example

python

# Example 1
Input: nums = [3, 1, 5, 8]
Output: 167
Explanation: 
- nums = [3, 1, 5, 8], assume there are 1's on the boundaries, so we have:
  [1, 3, 1, 5, 1, 8, 1]
  Burst the balloons in the following order:
  5 * 1 * 8 = 40
  3 * 1 * 5 = 15
  1 * 3 * 1 = 3
  Maximum coins = 40 + 15 + 3 + 109 = 167.

# Example 2
Input: nums = [1, 5]
Output: 10

Step-by-Step Solution

🧠 Intuition

At first glance, this problem might seem greedy, but it is more suitable for dynamic programming because the order in which we burst balloons matters. Once a balloon is burst, it affects the values of the adjacent balloons, which makes the problem inherently overlapping and optimal substructure.

We can solve this problem by considering all possible subproblems of bursting balloons within any range and calculating the maximum coins we can collect for each subproblem. The key idea is to consider bursting balloons between every possible pair of indices and recursively solving the subproblems.

✅ Approach

1. Add Virtual Boundaries: Since we can’t burst the first and last balloons directly, we treat the nums array as if it has virtual balloons (with value 1) at both ends. This simplifies the logic and avoids edge cases.
2. Define Subproblem: Let dp[i][j] represent the maximum coins we can collect by bursting all balloons between indices i and j (exclusive).
3. Recursive Relation:

• For each subarray of balloons between i and j, we try every balloon k between i and j as the last balloon to burst. When k is burst, the coins we collect are nums[i] * nums[k] * nums[j] (the value of the balloon and its neighbors), plus the optimal coins collected from the two subarrays dp[i][k] and dp[k][j].
• The relation is:
• dp[i][j]=max⁡(dp[i][j],dp[i][k]+dp[k][j]+nums[i]∗nums[k]∗nums[j]) for all k where i<k<jdp[i][j] = \max(dp[i][j], dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]) \text{ for all } k \text{ where } i < k < jdp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]+nums[i]∗nums[k]∗nums[j]) for all k where i<k<j

1. Final Answer: The answer is stored in dp[0][n+1], where n is the length of the original nums array. This represents the maximum coins we can collect from all the balloons between the first and last virtual balloons.

✅ Python Code

python

class Solution:
    def maxCoins(self, nums: list[int]) -> int:
        # Add 1 at the beginning and end of the array
        nums = [1] + nums + [1]
        n = len(nums)
        
        # dp[i][j] will store the maximum coins we can collect between i and j
        dp = [[0] * n for _ in range(n)]
        
        # Consider subarrays of length 2 to n
        for length in range(2, n):
            for i in range(n - length):
                j = i + length
                # Try all positions k to burst the balloon last between i and j
                for k in range(i + 1, j):
                    dp[i][j] = max(dp[i][j], dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j])
        
        # The result is stored in dp[0][n-1]
        return dp[0][n-1]

Explanation of the Code

1. Add Virtual Boundaries:

• We start by adding 1 at both ends of the nums array. This simplifies handling the boundary conditions and ensures that the problem can be solved uniformly.

1. Dynamic Programming Table (dp):

• We create a 2D table dp where dp[i][j] stores the maximum number of coins we can collect by bursting all balloons between indices i and j.

1. Iterate Over Subarray Lengths:

• We loop over all possible subarray lengths from 2 to n. For each subarray, we compute the maximum coins that can be collected by bursting the last balloon at each possible position k within that subarray.

1. Recursive Calculation:

• For each subarray between i and j, we try every possible balloon k to burst last. The recurrence relation calculates the maximum coins by considering the contributions from both subarrays (dp[i][k] and dp[k][j]) and the coins earned by bursting k.

1. Final Result:

• The final result is stored in dp[0][n-1], which represents the maximum coins we can collect by bursting all the balloons from the first to the last (including the virtual balloons at the boundaries).

⏱️ Time and Space Complexity

MetricValueTime ComplexityO(n^3)Space ComplexityO(n^2)

• Time Complexity:
• The solution involves filling a 2D DP table of size n x n. For each cell, we iterate over k (the possible last balloon to burst), leading to a time complexity of O(n^3), where n is the length of the nums array with the virtual boundaries.
• Space Complexity:
• We use a 2D array dp of size n x n to store intermediate results, so the space complexity is O(n^2).

🔍 Edge Cases

1. Single Balloon:
2. If nums contains only one balloon, the maximum coins we can collect is simply the value of that balloon (after adding the virtual boundaries).
3. All Balloons Are Zero:
4. If all balloons have a value of zero, the result will be zero, as there is no way to earn any coins.
5. Small Arrays:
6. For very small arrays (e.g., length 2 or 3), the solution still works efficiently because the DP table handles these cases in constant time.

✅ Conclusion

LeetCode 312: Burst Balloons is a dynamic programming problem that requires finding the optimal order of bursting balloons to maximize the coins collected. By breaking down the problem into overlapping subproblems and using DP, we can efficiently solve the problem with a time complexity of O(n^3).