What is the time complexity of Burst Balloons?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve Burst Balloons?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Burst Balloons?

Insight 1: Standard top-down recursion with memoization leads to TLE (Time Limit Exceeded). We need to use bottom-up dynamic programming. Insight 2: The problem is best solved by reversing the thought process. Instead of thinking about which balloon to burst next, consider which balloon was the *last* one to be burst within a given range.

Which companies ask LeetCode 312. Burst Balloons in interviews?

This problem is frequently asked at Flipkart, PhonePe, QBurst, Samsung, Snap.

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Burst Balloons - LeetCode 312 Solution

Q: What is the space complexity of Burst Balloons?

The space complexity is O(n^2).

Burst Balloons - Complete Solution Guide

Burst Balloons is LeetCode problem 312, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i th 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 Explanati

Detailed Explanation

The "Burst Balloons" problem challenges us to find the maximum coins we can collect by bursting a sequence of balloons. Each balloon has a number written on it. When we burst a balloon at index `i`, we get `nums[i - 1] * nums[i] * nums[i + 1]` coins. The balloons at the boundaries (leftmost and rightmost) are considered to have a value of 1 if we try to access indices out of bounds. The goal is to find the optimal order of bursting the balloons to maximize the total coins collected.

Solution Approach

The provided solution uses bottom-up dynamic programming to solve the problem. A 2D DP table `dp[i][j]` stores the maximum coins obtainable by bursting all the balloons between index `i` and index `j` (exclusive) in the extended `new_nums` array. The algorithm iterates through all possible lengths of intervals, and for each interval, it tries every possible last balloon to burst (`k`). The maximum coins for the interval `[i, j]` is then calculated as the sum of the maximum coins for the intervals `[i, k]` and `[k, j]`, plus the coins gained by bursting balloon `k` last, which is `new_nums[i] * new_nums[k] * new_nums[j]`.

Step-by-Step Algorithm

Step 1: Create a new array `new_nums` by adding 1 at the beginning and end of the original `nums` array. This handles the boundary conditions where the 'neighboring' balloon doesn't exist.
Step 2: Initialize a 2D array `dp` of size `n x n` (where n is the length of `new_nums`) with all values set to 0. `dp[i][j]` will store the maximum coins obtained by bursting balloons within the range (i, j) in `new_nums`.
Step 3: Iterate through different lengths of subproblems, starting from length 2 (since length 1 gives 0 coins).
Step 4: For each length, iterate through all possible starting positions `i` for that length.
Step 5: Calculate the ending position `j` of the interval as `i + length`.
Step 6: For each interval `[i, j]`, iterate through all possible last balloons to burst `k` within the interval (i.e., `i + 1 <= k < j`).
Step 7: Calculate the coins obtained by bursting `k` last: `dp[i][k] + dp[k][j] + new_nums[i] * new_nums[k] * new_nums[j]`
Step 8: Update `dp[i][j]` with the maximum coins obtained so far: `dp[i][j] = max(dp[i][j], coins)`.
Step 9: After all iterations, `dp[0][n - 1]` will contain the maximum coins that can be obtained by bursting all the balloons.
Step 10: Return `dp[0][n - 1]`.

Key Insights

Insight 1: Standard top-down recursion with memoization leads to TLE (Time Limit Exceeded). We need to use bottom-up dynamic programming.
Insight 2: The problem is best solved by reversing the thought process. Instead of thinking about which balloon to burst next, consider which balloon was the *last* one to be burst within a given range.
Insight 3: Insert padding of 1 at the beginning and end of the nums array. This simplifies boundary handling and avoids out-of-bounds errors when calculating coin values.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Flipkart, PhonePe, QBurst, Samsung, Snap.

Burst Balloons | LeetCode 312

Hard

Array Dynamic Programming

Time: O(n^3)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int maxCoins(int[] nums) {
        int n = nums.length;
        int[] new_nums = new int[n + 2];
        new_nums[0] = 1;
        new_nums[n + 1] = 1;
        for (int i = 0; i < n; i++) {
            new_nums[i + 1] = nums[i];
        }

        int[][] dp = new int[n + 2][n + 2];

        for (int length = 2; length < n + 2; length++) {
            for (int i = 0; i < n + 2 - length; i++) {
                int j = i + length;
                for (int k = i + 1; k < j; k++) {
                    int coins = dp[i][k] + dp[k][j] + new_nums[i] * new_nums[k] * new_nums[j];
                    dp[i][j] = Math.max(dp[i][j], coins);
                }
            }
        }

        return dp[0][n + 1];
    }
}

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Problem Info

Description

Problem 312: Burst Balloons - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

FlipkartPhonePeQBurstSamsungSnap

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:312

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Create a new array `new_nums` by adding 1 at the beginning and end of the original `nums` array. This handles the boundary conditions where the 'neighboring' balloon doesn't exist.
2
Step 2: Initialize a 2D array `dp` of size `n x n` (where n is the length of `new_nums`) with all values set to 0. `dp[i][j]` will store the maximum coins obtained by bursting balloons within the range (i, j) in `new_nums`.
3
Step 3: Iterate through different lengths of subproblems, starting from length 2 (since length 1 gives 0 coins).
4
Step 4: For each length, iterate through all possible starting positions `i` for that length.
5
Step 5: Calculate the ending position `j` of the interval as `i + length`.
6
Step 6: For each interval `[i, j]`, iterate through all possible last balloons to burst `k` within the interval (i.e., `i + 1 <= k < j`).
7
Step 7: Calculate the coins obtained by bursting `k` last: `dp[i][k] + dp[k][j] + new_nums[i] * new_nums[k] * new_nums[j]`
8
Step 8: Update `dp[i][j]` with the maximum coins obtained so far: `dp[i][j] = max(dp[i][j], coins)`.
9
Step 9: After all iterations, `dp[0][n - 1]` will contain the maximum coins that can be obtained by bursting all the balloons.
10
Step 10: Return `dp[0][n - 1]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^3)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^3) because there are three nested loops. The outer loop iterates through lengths (O(n)), the middle loop iterates through starting positions (O(n)), and the inner loop iterates through possible last balloons to burst (O(n)). The space complexity is O(n^2) because the `dp` table has dimensions `n x n`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Top-Down Dynamic Programming (Memoization)

A recursive approach with memoization can also be used. Define a recursive function `solve(i, j)` which returns the maximum coins obtainable by bursting balloons between indices i and j. Use a 2D array to store computed values to avoid redundant calculations.

Time: O(n^3)Space: O(n^2)

Trade-offs: While theoretically having the same time complexity, in practice, the bottom-up DP solution might be slightly faster due to avoiding recursion overhead. However, the top-down approach can be more intuitive for some.

✨

Key Insights

1Insight 1: Standard top-down recursion with memoization leads to TLE (Time Limit Exceeded). We need to use bottom-up dynamic programming.
2Insight 2: The problem is best solved by reversing the thought process. Instead of thinking about which balloon to burst next, consider which balloon was the *last* one to be burst within a given range.
3Insight 3: Insert padding of 1 at the beginning and end of the nums array. This simplifies boundary handling and avoids out-of-bounds errors when calculating coin values.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array `nums`. The algorithm correctly handles this case because `new_nums` will be `[1, 1]` and the loops won't execute significantly, returning 0.
Edge case 2: Input array with a single element. The algorithm correctly inserts 1s at both ends, and the DP logic handles this single-element case properly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to add the padding '1' at the beginning and end of the input array. This leads to incorrect calculations of coin values at the boundaries.

✗Common mistake 2: Not understanding the DP recurrence relation. The key is to think about the *last* balloon to burst, and divide the problem into smaller subproblems based on that last burst.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing resource allocation. Imagine tasks with dependencies. Bursting a balloon is like completing a task, and the coins represent the overall value gained. The goal is to find the optimal sequence of task completion to maximize the value.

🏦Real-world scenario 2: Financial trading strategies. Consider different investment options. The order of investing and divesting affects the overall profit. The balloon bursting problem models finding the optimal order to maximize profit.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Burst Balloons"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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