Burst Balloons

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Problem

Imagine you have a row of balloons, each with a number. Bursting a balloon earns coins equal to the product of the balloon's number and its adjacent balloons' numbers. How do you burst them all to maximize your coins?

Given n balloons, indexed from 0 to n-1, each with a positive integer number. When you burst balloon i, you earn coins equal to nums[left] * nums[i] * nums[right], where left and right are the adjacent balloons to i before bursting it. After bursting balloon i, it is removed, and the balloons left and right become adjacent. You want to find the maximum coins you can collect by bursting the balloons in an optimal order. Input: An array nums of integers representing balloon numbers. Output: An integer representing the maximum coins obtainable.

1 ≤ n ≤ 5001 ≤ nums[i] ≤ 100

Edge cases: Single balloon [5] → output 25All balloons have the same number [1,1,1] → output 4Increasing sequence [1,2,3,4] → output 40

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IDE

def maxCoins(nums: list[int]) -> int:public int maxCoins(int[] nums)int maxCoins(vector<int> &nums)function maxCoins(nums)

def maxCoins(nums):
    # Write your solution here
    pass

class Solution {
    public int maxCoins(int[] nums) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int maxCoins(vector<int> &nums) {
    // Write your solution here
    return 0;
}

function maxCoins(nums) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 10100Greedy approach bursting largest balloon first, missing better interval bursting order.✅ Implement interval DP considering all possible last burst balloons in each interval.

Wrong: 0Not handling empty input or base cases properly.✅ Add base case: if nums is empty, return 0 immediately.

Wrong: less than expected (e.g. 2 for [1,1,1])Incorrect DP recurrence or missing intervals in calculation.✅ Ensure dp[left][right] considers all balloons i in (left, right) as last burst and sums subproblems correctly.

Wrong: greater than expected (e.g. >110 for [1,2,3,4,5])Treating problem as unbounded bursting (bursting balloons multiple times).✅ Use interval DP to ensure each balloon is burst exactly once.

Wrong: TLEUsing brute force exponential recursion without memoization or DP.✅ Implement top-down memoization or bottom-up DP with O(n^3) complexity.

✓

Test Cases

Make sure your solution handles base cases and small inputs correctly.

Focus on tricky cases that break greedy or naive approaches.

Optimize your DP to run efficiently on large inputs.

t1_01basic

Input{"nums":[3,1,5,8]}

Expected167

⏱ Performance - must finish in 2000ms

Optimal bursting order yields 167 coins. For example, bursting balloons in order 1, 5, 3, 8 yields coins: 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167.

💡 Think about bursting balloons in intervals and how coins accumulate.

💡 Use interval DP where dp[left][right] stores max coins between left and right balloons.

💡 Consider the last balloon to burst in each interval and combine subproblems.

Why it failed: Incorrect output means the DP recurrence or interval boundaries are wrong. Fix by ensuring dp[left][right] considers all possible last burst balloons i with coins = nums[left]*nums[i]*nums[right] + dp[left][i] + dp[i][right].

✓ Correctly implements interval DP with proper boundaries and recurrence.

t1_02basic

Input{"nums":[1,2,3,4]}

Expected40

⏱ Performance - must finish in 2000ms

Optimal bursting order yields 40 coins total.

💡 Try to think about the problem as bursting balloons from inside out.

💡 Use memoization or bottom-up DP to avoid recomputing intervals.

💡 Remember to pad nums with 1 at both ends to simplify edge calculations.

Why it failed: If output is less than 40, likely missing some intervals or not considering all last burst balloons. Fix by iterating all i in (left, right) as last burst in dp.

✓ DP correctly computes max coins for all intervals.

t2_01edge

Input{"nums":[]}

Expected0

⏱ Performance - must finish in 2000ms

No balloons to burst, so max coins is 0.

💡 Check how your code handles empty input arrays.

💡 Make sure base cases return 0 when no balloons are present.

💡 Return 0 immediately if nums is empty or length zero.

Why it failed: Code crashes or returns non-zero for empty input. Fix by adding base case: if nums is empty, return 0.

✓ Handles empty input gracefully with zero coins.

t2_02edge

Input{"nums":[5]}

Expected5

⏱ Performance - must finish in 2000ms

Only one balloon, bursting it yields coins = 1*5*1 = 25 when considering padding balloons with 1 at both ends.

💡 Consider the padding of 1 at both ends for single balloon.

💡 DP intervals with length 1 should return 0, but bursting single balloon yields coins.

💡 Ensure your code handles intervals of length 2 correctly to burst the single balloon.

Why it failed: Output less than 25 means single balloon case not handled properly. Fix by correctly initializing dp for intervals with one balloon and padding with 1 at both ends.

✓ Correctly computes coins for single balloon input.

t2_03edge

Input{"nums":[1,1,1]}

Expected3

⏱ Performance - must finish in 2000ms

All balloons are 1, bursting any order yields total coins 4.

💡 Check if your DP handles identical balloon values correctly.

💡 Make sure all intervals are considered, not just greedy picks.

💡 Verify that coins calculation uses nums[left]*nums[i]*nums[right] correctly.

Why it failed: Output less than 4 means missing some intervals or incorrect multiplication. Fix by ensuring all subproblems are combined properly.

✓ Correctly handles identical balloon values.

t3_01corner

Input{"nums":[100,1,100]}

Expected20100

⏱ Performance - must finish in 2000ms

Bursting order: burst middle balloon first yields 100*1*100=10000 coins, then burst remaining balloons for total 20000.

💡 Beware of greedy approach that bursts largest balloon first.

💡 Use interval DP to consider all possible last burst balloons.

💡 Check that your DP tries bursting middle balloon first to maximize coins.

Why it failed: Greedy approach bursts largest balloon first and returns 10100 instead of 20000. Fix by implementing interval DP considering all last burst balloons.

✓ Correctly avoids greedy trap and finds max coins.

t3_02corner

Input{"nums":[1,2,3,2,1]}

Expected22

⏱ Performance - must finish in 2000ms

Complex sequence requiring full DP to find max coins 44.

💡 Check for off-by-one errors in interval boundaries.

💡 Ensure dp[left][right] excludes left and right balloons themselves.

💡 Verify recursion or iteration covers all subintervals.

Why it failed: Output less than 44 indicates off-by-one or incomplete interval coverage. Fix by correctly defining intervals and looping i in (left+1, right).

✓ Correctly handles intervals and boundaries.

t3_03corner

Input{"nums":[1,2,3,4,5]}

Expected110

⏱ Performance - must finish in 2000ms

Tests correct handling of 0/1 knapsack style confusion; must burst each balloon exactly once.

💡 Remember each balloon can be burst only once.

💡 Avoid treating problem as unbounded where balloons can be burst multiple times.

💡 Use DP intervals to ensure each balloon is counted once.

Why it failed: Output greater than 110 indicates unbounded bursting (counting balloons multiple times). Fix by using interval DP that ensures each balloon is burst once.

✓ Correctly enforces 0/1 bursting constraint.

t4_01performance

Input{"nums":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100, O(n^3) DP must complete within 2 seconds time limit.

💡 Brute force exponential solutions will time out here.

💡 Use bottom-up or top-down DP with memoization to achieve O(n^3).

💡 Optimize DP by careful indexing and pruning if possible.

Why it failed: TLE indicates exponential or inefficient solution. Fix by implementing O(n^3) DP with memoization or tabulation.

✓ Solution runs within time limit using O(n^3) DP.

Practice

1. You are given two sorted arrays and need to find the median of their combined elements efficiently without fully merging them. Which approach guarantees the optimal time complexity for this problem?

easy

A. Merge both arrays completely and then find the median from the merged array.

B. Use binary search to partition the smaller array and find the correct partition in the other array.

C. Use a two-pointer approach to partially merge elements until the median position is reached.

D. Apply dynamic programming to store medians of subarrays and combine results.

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the median of two sorted arrays without fully merging them to achieve better than O(m+n) time.

2. Step 2: Identify the optimal approach

   Binary search on the smaller array to find a partition that balances elements on both sides guarantees O(log(min(m,n))) time, which is optimal.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Binary search on partition is the known optimal solution [OK]

Hint: Optimal median uses binary search on smaller array [OK]

Common Mistakes:

• Assuming full merge is necessary
• Using two pointers without binary search
• Confusing DP with median finding

2. Given a string s and a string t, you need to find the smallest substring in s that contains all characters of t including duplicates. Which algorithmic approach guarantees an optimal solution with linear time complexity?

easy

A. Sliding window with frequency counting and two pointers to expand and shrink the window

B. Brute force checking all substrings and verifying character counts

C. Dynamic programming to store minimum substrings ending at each index

D. Greedy approach by always picking the next occurrence of required characters

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the minimum window substring containing all characters of t in s. Checking all substrings (brute force) is too slow.

2. Step 2: Identify optimal approach

   Sliding window with frequency counting and two pointers allows expanding and shrinking the window efficiently, achieving O(n) time. DP or greedy approaches either don't guarantee minimal window or are inefficient.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Sliding window with frequency counting is the classic optimal solution [OK]

Hint: Sliding window with frequency counting solves this optimally [OK]

Common Mistakes:

• Thinking brute force or greedy is efficient enough

3. Consider the following Python code implementing the optimal solution for the envelope nesting problem. Given the input [[5,4],[6,4],[6,7],[2,3]], what is the value of the tails list after processing all heights?

easy

A. [3, 4, 7]

B. [3, 4]

C. [2, 3, 4, 7]

D. [2, 3, 7]

Solution

1. Step 1: Sort envelopes and extract heights

   Sorted envelopes: [[2,3],[5,4],[6,7],[6,4]] -> heights = [3,4,7,4]

2. Step 2: Trace tails updates

   Insert 3 -> tails=[3]; Insert 4 -> tails=[3,4]; Insert 7 -> tails=[3,4,7]; Insert 4 replaces tails[1] -> tails=[3,4,7]

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Final tails length is 3 with values [3,4,7] [OK]

Hint: Trace tails updates carefully with bisect_left [OK]

Common Mistakes:

• Confusing bisect_left with bisect_right
• Not replacing tails element on equal height
• Misordering envelopes before LIS

4. Consider the following Python code implementing the optimal two-pointer approach for the smallest range covering k lists. Given the input nums = [[1,5,8],[4,12],[7,8,10]], what is the final smallest range returned by the function?

easy

A. [5, 7]

B. [4, 8]

C. [4, 7]

D. [7, 8]

Solution

1. Step 1: Merge and sort all elements with their list indices

   Merged array: [(1,0),(4,1),(5,0),(7,2),(8,0),(8,2),(10,2),(12,1)]

2. Step 2: Use two pointers to find minimal range covering all lists

   Sliding window covers all lists first at [4,7], then [4,8]. The range [4,7] has width 3, and [4,8] has width 4. The minimal range is [4,7].

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Range [4,7] covers all lists with minimal width [OK]

Hint: Trace merged array and sliding window coverage [OK]

Common Mistakes:

• Off-by-one in window boundaries
• Forgetting to update have when shrinking window

5. Suppose the problem is modified so that words can be reused multiple times in pairs (i.e., pairs (i, j) where i == j are allowed). Which modification to the Trie-based solution correctly handles this variant without double counting pairs?

hard

A. Remove the check 'node.index != index' and filter duplicates from final results

B. Keep the check 'node.index != index' but add a separate check for empty string pairs

C. Modify add_word to store multiple indices per node to allow reuse tracking

D. Remove the check 'node.index != index' in search to allow i == j pairs

Solution

1. Step 1: Understand reuse allowance

   Allowing i == j pairs means removing the 'node.index != index' check to include self-pairs.

2. Step 2: Avoid double counting

   Removing the check causes duplicates; filtering duplicates after collecting results ensures correctness.

3. Step 3: Evaluate other options

   Keeping the check disallows self-pairs; storing multiple indices per node is unnecessary and complex.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Removing check plus duplicate filtering correctly handles reuse [OK]

Hint: Allow self-pairs by removing check, then deduplicate results [OK]

Common Mistakes:

• Removing check without deduplication
• Keeping check and missing self-pairs
• Overcomplicating Trie nodes