How do you solve Burst Balloons on LeetCode?

Burst Balloons (LeetCode #312) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n²) space complexity. Key insight: Understanding the problem requires recognizing the impact of the order of bursting balloons.

What is the brute force solution for Burst Balloons?

The brute force approach for Burst Balloons is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach involves trying every possible order of bursting the balloons and calculating the coins collected for each order. This is straightforward but inefficient as it explores all permutations. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Burst Balloons?

Burst Balloons uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Burst Balloons?

Explain your thought process clearly; interviewers appreciate understanding your reasoning. Consider edge cases, such as the smallest and largest inputs, during your explanation. Practice writing clean, efficient code under time constraints.

What are common mistakes when solving Burst Balloons?

Forgetting to account for the virtual balloons at the edges. Not properly updating the DP table or misunderstanding its indices.

#312

Burst Balloons

Hard

ArrayDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n³)
Space	O(n)	O(n²)

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Intuition

Time O(n³)Space O(n²)

The optimal solution uses dynamic programming to avoid recalculating results for subproblems. We build a DP table where each entry represents the maximum coins that can be collected by bursting balloons between two indices.

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Algorithm

3 steps

1Step 1: Create a DP array where dp[i][j] represents the maximum coins obtainable by bursting all balloons between indices i and j.
2Step 2: Iterate over all possible lengths of balloon segments from 1 to n.
3Step 3: For each segment, calculate the maximum coins by considering each balloon as the last one to burst in that segment and update the DP table accordingly.

solution.py10 lines

1def maxCoins(nums):
2    nums = [1] + nums + [1]
3    n = len(nums)
4    dp = [[0] * n for _ in range(n)]
5    for length in range(1, n - 1):
6        for left in range(1, n - length):
7            right = left + length - 1
8            for i in range(left, right + 1):
9                dp[left][right] = max(dp[left][right], dp[left][i - 1] + dp[i + 1][right] + nums[left - 1] * nums[i] * nums[right + 1])
10    return dp[1][n - 2]

ℹ

Complexity note: The time complexity is O(n³) due to the three nested loops: one for the length of the segment, one for the left boundary, and one for the last balloon to burst. The space complexity is O(n²) for storing the DP table.

1Understanding the problem requires recognizing the impact of the order of bursting balloons.
2Dynamic programming is effective for problems with overlapping subproblems and optimal substructure.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.