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 a string representing a sequence of colored balls on a board and a multiset of balls in hand. Your goal is to clear the board by inserting balls from your hand such that any group of three or more consecutive balls of the same color is removed, possibly triggering chain reactions. Which algorithmic approach guarantees finding the minimum number of insertions needed to clear the board optimally?

easy

A. Interval dynamic programming with memoization exploiting optimal substructure

B. Topological sorting of ball groups followed by greedy removals

C. Pure brute force recursion without memoization exploring all insertions

D. Greedy insertion of balls at the earliest possible position to remove groups

Solution

1. Step 1: Understand problem structure

   The problem requires minimizing insertions to clear the board, which involves exploring overlapping subproblems and optimal substructure.

2. Step 2: Identify suitable algorithm

   Interval DP with memoization efficiently explores all intervals and hand states, guaranteeing minimal moves by considering all splits and removals.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Interval DP is known for solving interval removal problems optimally [OK]

Hint: Interval DP handles overlapping subproblems optimally [OK]

Common Mistakes:

• Assuming greedy insertion always yields minimal moves
• Using brute force without memoization leads to exponential time
• Misapplying topological sort which is unrelated here

2. Consider the following buggy code for finding the cheapest flight within K stops. Which line contains the subtle bug that causes incorrect results on some inputs?

def findCheapestPrice(n, flights, src, dst, K):
    INF = float('inf')
    prev = [INF] * n
    prev[src] = 0
    for _ in range(K + 1):
        for u, v, w in flights:
            if prev[u] != INF:
                prev[v] = min(prev[v], prev[u] + w)
    return prev[dst] if prev[dst] != INF else -1

medium

A. Line 4: Initializing prev[src] = 0

B. Line 7: Checking if prev[u] != INF before relaxing edges

C. Line 6: Using prev array directly inside the loop instead of a separate curr array

D. Line 9: Returning -1 if prev[dst] is INF

Solution

1. Step 1: Understand the DP update mechanism

   In bottom-up DP, updates for the current iteration must not affect other relaxations in the same iteration, so a separate curr array is needed.

2. Step 2: Identify the bug

   Using prev directly causes updated costs to be used immediately in the same iteration, leading to incorrect minimum costs.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Separate curr array prevents premature updates [OK]

Hint: Must use separate curr array to avoid premature updates [OK]

Common Mistakes:

• Updating prev in-place during iteration
• Forgetting to copy arrays
• Ignoring stop constraints

3. What is the time complexity of the space-optimized bottom-up DP solution for the Maximal Square problem on an m x n matrix, and why might some candidates incorrectly think it is higher?

medium

A. O(m^3) because checking all squares requires nested loops

B. O(m * n * min(m,n)) because of checking all possible square sizes

C. O(m * n) because each cell is processed once with constant time updates

D. O(m + n) because only rows and columns are iterated separately

Solution

1. Step 1: Identify loops in the code

   Two nested loops iterate over rows and columns, each cell processed once.

2. Step 2: Understand DP update cost

   Each dp update is O(1), no nested checks for squares inside loops.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP avoids checking all squares explicitly [OK]

Hint: DP processes each cell once with constant work [OK]

Common Mistakes:

• Confusing brute force with DP complexity
• Assuming nested loops check all squares
• Ignoring constant time DP updates

4. What is the time and space complexity of the space-optimized bottom-up dynamic programming solution for the Minimum Path Sum problem on an m x n grid?

medium

A. Time: O(m*n), Space: O(m*n)

B. Time: O(m*n), Space: O(m)

C. Time: O(2^(m+n)), Space: O(m+n)

D. Time: O(m*n), Space: O(n)

Solution

1. Step 1: Identify time complexity

   The algorithm iterates over each cell once, so time is O(m*n).

2. Step 2: Identify space complexity

   Space optimized DP uses a single 1D array of length n, so space is O(n), not O(m*n) or O(m).

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Single dp array of size n updated row by row [OK]

Hint: Space optimized DP uses one row array, not full matrix [OK]

Common Mistakes:

• Assuming space is O(m*n) due to 2D dp
• Confusing recursion stack space with DP space
• Mistaking m for n in space complexity

5. What is the time complexity of the bottom-up dynamic programming solution for the Stone Game problem with n piles, and why?

medium

A. O(n^3) because of three nested loops over the piles

B. O(n^2) because the DP table of size nxn is filled once with constant work per cell

C. O(2^n) because all subsets of piles are considered

D. O(n) because only linear passes over the piles are needed

Solution

1. Step 1: Identify loops in bottom-up DP

   There are two nested loops: one for length from 2 to n, and one for start index i, total O(n^2) iterations.

2. Step 2: Constant work per dp[i][j]

   Each dp[i][j] is computed with a constant number of operations (max of two values), so total time is O(n^2).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP table size nxn filled once with O(1) work per cell [OK]

Hint: Two nested loops over intervals -> O(n^2) [OK]

Common Mistakes:

• Confusing recursion stack space with time
• Assuming triple nested loops cause O(n^3)