LeetCode 312: Burst Balloons Solution in Python – A Step-by-Step Guide

What Is LeetCode 312: Burst Balloons?

In LeetCode 312: Burst Balloons, you’re given an array nums of integers representing balloon values. When you burst a balloon at index i, you earn nums[i-1] * nums[i] * nums[i+1] coins, and the balloon disappears, joining its neighbors. You repeat until all balloons are burst, aiming to maximize your total coins. For example, with [3, 1, 5, 8], bursting in the right order yields 167 coins. This problem tests your ability to optimize a sequence of actions with changing states.

Problem Statement

• Input: An integer array nums.
• Output: An integer—the maximum coins possible by bursting all balloons.
• Rules:
• Bursting balloon i earns nums[left] * nums[i] * nums[right] (left and right are adjacent unburst balloons).
• Add 1s as boundaries at both ends.
• All balloons must be burst.

Constraints

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

Examples

• Input: [3,1,5,8]
• Output: 167 (Order: 1, 5, 3, 8 → 1*1*5 + 1*5*3 + 3*3*8 + 3*8*1)
• Input: [1,5]
• Output: 10 (1*1*5 + 1*5*1)
• Input: [1]
• Output: 1 (1*1*1)

Understanding the Problem: Maximizing Coins

To solve LeetCode 312: Burst Balloons in Python, we need to find the optimal order to burst balloons, considering how each burst affects future scores. A naive approach—trying all permutations—is O(n!), far too slow. Instead, we’ll use:

• Best Solution (Dynamic Programming): O(n³) time, O(n²) space—bottom-up efficiency.
• Alternative Solution (Recursion with Memoization): O(n³) time, O(n²) space—top-down clarity.

Let’s start with the DP solution, explained in a beginner-friendly way.

Best Solution: Dynamic Programming (Bottom-Up)

Why This Is the Best Solution

The bottom-up DP approach is the top choice for LeetCode 312 because it’s efficient—O(n³) time, O(n²) space—and cleverly reframes the problem. Instead of choosing which balloon to burst first, it considers the last balloon burst in each subarray, building the solution from small segments to the whole. It’s like solving a puzzle by fitting the final pieces first—smart and scalable!

How It Works

Think of bursting balloons in reverse:

• Step 1: Add Boundaries:
• Pad nums with 1s: [1, nums, 1].
• Step 2: Define DP Table:
• dp[left][right]: Max coins for bursting all balloons between left and right (exclusive).
• Step 3: Fill Table:
• For each length, try each balloon as the last to burst.
• Coins = left_val * curr_val * right_val + dp[left][curr] + dp[curr][right].
• Step 4: Why It Works:
• Last balloon burst splits problem into independent subproblems.
• Bottom-up avoids recursion overhead.

It’s like building your score from the endgame up!

Step-by-Step Example

Example: nums = [3,1,5]

• Padded: [1, 3, 1, 5, 1]
• DP Table (5x5):
• Len 1: dp[0][2] = 1*3*1 = 3, dp[1][3] = 3*1*5 = 15, dp[2][4] = 1*5*1 = 5.
• Len 2: dp[0][3] = max(1*3*5 + dp[1][3], 1*1*5 + dp[0][2]) = max(15+0, 5+3) = 15.
• Len 3: dp[1][4] = max(3*1*1 + dp[2][4], 3*5*1 + dp[1][3]) = max(3+5, 15+0) = 15.
• Len 4: dp[0][4] = max(1*3*1 + dp[1][4], 1*1*1 + dp[0][3], 1*5*1 + dp[0][2]) = max(3+15, 1+15, 5+3) = 18.
• Result: dp[0][4] = 18.

Code with Detailed Line-by-Line Explanation

Copy codeclass Solution:
    def maxCoins(self, nums: List[int]) -> int:
        # Add boundary 1s
        nums = [1] + nums + [1]
        n = len(nums)

        # Initialize DP table
        dp = [[0] * n for _ in range(n)]

        # Fill DP table for all lengths
        for length in range(2, n):
            for left in range(n - length):
                right = left + length
                # Try each balloon as last to burst
                for i in range(left + 1, right):
                    coins = (nums[left] * nums[i] * nums[right] + 
                             dp[left][i] + dp[i][right])
                    dp[left][right] = max(dp[left][right], coins)

        # Max coins for entire array
        return dp[0][n-1]

• Lines 3-4: Pad array, get new length.
• Line 6: Create n x n DP table.
• Lines 9-15: Loop:
• Length from 2 to n-1 (subarrays with balloons).
• For each left, compute right, try each i as last burst.
• Update dp[left][right] with max coins.
• Line 17: Return full range result.
• Time Complexity: O(n³)—three nested loops.
• Space Complexity: O(n²)—DP table.

This is like a balloon-bursting strategist—optimal and clear!

Alternative Solution: Recursion with Memoization (Top-Down)

Why an Alternative Approach?

The recursive approach with memoization also solves it in O(n³) time, O(n²) space, but starts top-down—breaking the problem into subproblems and caching results. It’s more intuitive for some, showing the thought process explicitly, though it’s slightly less efficient due to recursion overhead. It’s like planning your bursts step-by-step—educational and thorough!

How It Works

Break it down recursively:

• Step 1: Pad with 1s.
• Step 2: Define recursive function:
• Max coins for range (left, right) by trying each balloon as last.
• Step 3: Memoize to avoid recomputation.

Step-by-Step Example

Example: nums = [3,1]

• Padded: [1, 3, 1, 1]
• Call (0, 3):
• i=1: 1*3*1 + (0,1) + (1,3) = 3 + 0 + 0 = 3.
• i=2: 1*1*1 + (0,2) + (2,3) = 1 + 3 + 0 = 4.
• Result: 4.

Code for Recursive Approach

Copy codeclass Solution:
    def maxCoins(self, nums: List[int]) -> int:
        nums = [1] + nums + [1]
        n = len(nums)
        memo = {}

        def dp(left, right):
            if left + 1 >= right:
                return 0
            if (left, right) in memo:
                return memo[(left, right)]

            max_coins = 0
            for i in range(left + 1, right):
                coins = (nums[left] * nums[i] * nums[right] + 
                         dp(left, i) + dp(i, right))
                max_coins = max(max_coins, coins)

            memo[(left, right)] = max_coins
            return max_coins

        return dp(0, n-1)

• Time Complexity: O(n³)—each subproblem solved once.
• Space Complexity: O(n²)—memoization table.

It’s a thoughtful balloon planner!

Comparing the Two Solutions

• Bottom-Up DP (Best):
• Pros: O(n³) time, O(n²) space, no recursion overhead.
• Cons: Less intuitive initially.
• Top-Down Recursion (Alternative):
• Pros: O(n³) time, O(n²) space, clearer logic flow.
• Cons: Recursion stack.

Bottom-up wins for performance.

Additional Examples and Edge Cases

• [1,5]: 10.
• [3]: 3.
• [0,0,0]: 0.

Both handle these perfectly.

Complexity Breakdown

• Bottom-Up DP: Time O(n³), Space O(n²).
• Top-Down: Time O(n³), Space O(n²).

Bottom-up is slightly leaner.

Key Takeaways

• Bottom-Up DP: Last burst first—genius!
• Top-Down: Step-by-step—intuitive!
• Python: Arrays and loops shine—see [Python Basics](/python/basics).
• Strategy: Order matters in bursting.

Final Thoughts: Burst for Max Coins

LeetCode 312: Burst Balloons in Python is a dynamic programming delight. The bottom-up DP offers speed and elegance, while top-down recursion provides clarity. Want more DP challenges? Try LeetCode 300: Longest Increasing Subsequence or LeetCode 1143: Longest Common Subsequence. Ready to burst? Head to Solve LeetCode 312 on LeetCode and maximize those coins today!