Burst Balloons

🎯

Burst Balloons

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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?

💡 This problem is challenging because the order of bursting balloons affects the coins earned, creating dependencies that are not straightforward. Beginners often struggle to identify the optimal substructure and how to break down the problem into manageable parts. Think of it like popping balloons in a sequence where each pop changes the neighbors, so the order really matters.

📋

Problem Statement

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

💡

Example

Input"[3,1,5,8]"

Output167

Burst balloons in order: 1 → 5 → 3 → 8. Coins collected: 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167.

• Single balloon [5] → output 5
• All balloons have the same number [1,1,1] → output 3
• Increasing sequence [1,2,3,4] → output 40
• Large balloon values [100,1,100] → output 20000

⚠️

Common Mistakes

Not adding virtual balloons at ends

Incorrect coin calculation at edges, leading to wrong results

✅ Add 1 at both ends of the nums array before processing

Incorrect base case in recursion

Infinite recursion or wrong results

✅ Return 0 when left + 1 == right, meaning no balloons to burst

Not memoizing all subproblems

Exponential time due to repeated calculations

✅ Use a 2D memo table indexed by left and right boundaries

Mixing up indices when bursting balloons

Wrong coins calculation and incorrect DP updates

✅ Carefully use left, i, right indices consistently with virtual balloons

Trying to optimize by bursting balloons in order instead of last

Fails to capture optimal substructure, leading to suboptimal results

✅ Think in terms of which balloon bursts last in a range, not first

🧠

Brute Force (Pure Recursion)

💡 Starting with brute force helps understand the problem's recursive nature and the explosion of possibilities, which motivates optimization. It shows how the problem can be broken down into smaller subproblems but also why naive recursion is inefficient.

Intuition

Try bursting each balloon last in the current range and recursively solve the subproblems on the left and right sides.

Algorithm

1. Define a recursive function that takes the current range of balloons.
2. For each balloon in the range, consider it as the last to burst.
3. Calculate coins gained by bursting this balloon last, plus coins from left and right subranges.
4. Return the maximum coins over all choices.

💡 The challenge is realizing that bursting a balloon last in a range isolates the problem into two smaller ranges, enabling recursion.

</>

Code

def maxCoins(nums):
    nums = [1] + nums + [1]
    n = len(nums)
    def dfs(left, right):
        if left + 1 == right:
            return 0
        max_coins = 0
        for i in range(left + 1, right):
            coins = nums[left] * nums[i] * nums[right]
            coins += dfs(left, i) + dfs(i, right)
            max_coins = max(max_coins, coins)
        return max_coins
    return dfs(0, n - 1)

# Example usage
if __name__ == '__main__':
    print(maxCoins([3,1,5,8]))  # Output: 167

Line Notes

nums = [1] + nums + [1]Add virtual balloons with value 1 at both ends to simplify edge calculations

def dfs(left, right):Define recursive function to compute max coins between indices left and right

if left + 1 == right:Base case: no balloons to burst between left and right

for i in range(left + 1, right):Try bursting each balloon i last in the current range

coins = nums[left] * nums[i] * nums[right]Coins gained by bursting balloon i last, considering neighbors

coins += dfs(left, i) + dfs(i, right)Add coins from bursting balloons in left and right subranges

max_coins = max(max_coins, coins)Keep track of maximum coins achievable

return max_coinsReturn the best result for this subproblem

public class BurstBalloons {
    public int maxCoins(int[] nums) {
        int n = nums.length + 2;
        int[] newNums = new int[n];
        newNums[0] = 1;
        newNums[n - 1] = 1;
        for (int i = 0; i < nums.length; i++) {
            newNums[i + 1] = nums[i];
        }
        return dfs(newNums, 0, n - 1);
    }
    private int dfs(int[] nums, int left, int right) {
        if (left + 1 == right) return 0;
        int maxCoins = 0;
        for (int i = left + 1; i < right; i++) {
            int coins = nums[left] * nums[i] * nums[right];
            coins += dfs(nums, left, i) + dfs(nums, i, right);
            maxCoins = Math.max(maxCoins, coins);
        }
        return maxCoins;
    }
    public static void main(String[] args) {
        BurstBalloons solution = new BurstBalloons();
        System.out.println(solution.maxCoins(new int[]{3,1,5,8})); // 167
    }
}

Line Notes

int n = nums.length + 2;Create new array size with two extra virtual balloons

newNums[0] = 1; newNums[n - 1] = 1;Set virtual balloons at both ends to 1

for (int i = 0; i < nums.length; i++)Copy original balloons into new array offset by 1

if (left + 1 == right) return 0;Base case: no balloons between left and right

for (int i = left + 1; i < right; i++)Try bursting each balloon i last in current range

int coins = nums[left] * nums[i] * nums[right];Coins from bursting balloon i last

coins += dfs(nums, left, i) + dfs(nums, i, right);Add coins from left and right subproblems

maxCoins = Math.max(maxCoins, coins);Track maximum coins

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size() + 2;
        vector<int> newNums(n, 1);
        for (int i = 0; i < nums.size(); i++) {
            newNums[i + 1] = nums[i];
        }
        return dfs(newNums, 0, n - 1);
    }
private:
    int dfs(vector<int>& nums, int left, int right) {
        if (left + 1 == right) return 0;
        int maxCoins = 0;
        for (int i = left + 1; i < right; i++) {
            int coins = nums[left] * nums[i] * nums[right];
            coins += dfs(nums, left, i) + dfs(nums, i, right);
            maxCoins = max(maxCoins, coins);
        }
        return maxCoins;
    }
};

int main() {
    Solution sol;
    vector<int> nums = {3,1,5,8};
    cout << sol.maxCoins(nums) << endl; // 167
    return 0;
}

Line Notes

int n = nums.size() + 2;Add two virtual balloons to simplify edge cases

vector<int> newNums(n, 1);Initialize new array with 1s at ends

for (int i = 0; i < nums.size(); i++)Copy original balloons into new array offset by 1

if (left + 1 == right) return 0;Base case: no balloons to burst between left and right

for (int i = left + 1; i < right; i++)Try bursting each balloon i last in current range

int coins = nums[left] * nums[i] * nums[right];Coins from bursting balloon i last

coins += dfs(nums, left, i) + dfs(nums, i, right);Add coins from left and right subproblems

maxCoins = max(maxCoins, coins);Keep track of maximum coins

function maxCoins(nums) {
    nums = [1, ...nums, 1];
    const n = nums.length;
    function dfs(left, right) {
        if (left + 1 === right) return 0;
        let maxCoins = 0;
        for (let i = left + 1; i < right; i++) {
            let coins = nums[left] * nums[i] * nums[right];
            coins += dfs(left, i) + dfs(i, right);
            maxCoins = Math.max(maxCoins, coins);
        }
        return maxCoins;
    }
    return dfs(0, n - 1);
}

// Example usage
console.log(maxCoins([3,1,5,8])); // 167

Line Notes

nums = [1, ...nums, 1];Add virtual balloons with value 1 at both ends

function dfs(left, right) {Recursive function to compute max coins in range

if (left + 1 === right) return 0;Base case: no balloons between left and right

for (let i = left + 1; i < right; i++) {Try bursting each balloon i last in current range

let coins = nums[left] * nums[i] * nums[right];Coins from bursting balloon i last

coins += dfs(left, i) + dfs(i, right);Add coins from left and right subproblems

maxCoins = Math.max(maxCoins, coins);Track maximum coins achievable

return maxCoins;Return best result for this subproblem

⏱

Complexity

TimeO(n * 2^n)

SpaceO(n) recursion stack

Each recursive call tries all balloons in the current range, leading to exponential calls.

💡 For n=10, this means over 1000 calls, which quickly becomes infeasible for larger n.

Interview Verdict: TLE

This approach is too slow for large inputs but is essential to understand the problem's recursive structure.

🧠

Top-Down Memoization

💡 Memoization caches results of subproblems to avoid repeated calculations, drastically improving performance. It transforms the exponential brute force into a polynomial time solution by remembering answers.

Intuition

Store results of dfs(left, right) calls in a table so that repeated calls return instantly.

Algorithm

1. Initialize a 2D memo table with -1 indicating uncomputed states.
2. Modify the recursive function to check memo before computing.
3. If memo[left][right] is computed, return it immediately.
4. Otherwise, compute as before and store the result in memo.

💡 Memoization turns exponential recursion into polynomial by reusing results.

</>

Code

def maxCoins(nums):
    nums = [1] + nums + [1]
    n = len(nums)
    memo = [[-1] * n for _ in range(n)]
    def dfs(left, right):
        if left + 1 == right:
            return 0
        if memo[left][right] != -1:
            return memo[left][right]
        max_coins = 0
        for i in range(left + 1, right):
            coins = nums[left] * nums[i] * nums[right]
            coins += dfs(left, i) + dfs(i, right)
            max_coins = max(max_coins, coins)
        memo[left][right] = max_coins
        return max_coins
    return dfs(0, n - 1)

# Example usage
if __name__ == '__main__':
    print(maxCoins([3,1,5,8]))  # Output: 167

Line Notes

memo = [[-1] * n for _ in range(n)]Create memo table initialized to -1 to mark unvisited states

if memo[left][right] != -1:Return cached result if already computed

memo[left][right] = max_coinsStore computed result to avoid recomputation

return memo[left][right]Return memoized result for current subproblem

public class BurstBalloons {
    private int[][] memo;
    public int maxCoins(int[] nums) {
        int n = nums.length + 2;
        int[] newNums = new int[n];
        newNums[0] = 1;
        newNums[n - 1] = 1;
        for (int i = 0; i < nums.length; i++) {
            newNums[i + 1] = nums[i];
        }
        memo = new int[n][n];
        for (int i = 0; i < n; i++) {
            Arrays.fill(memo[i], -1);
        }
        return dfs(newNums, 0, n - 1);
    }
    private int dfs(int[] nums, int left, int right) {
        if (left + 1 == right) return 0;
        if (memo[left][right] != -1) return memo[left][right];
        int maxCoins = 0;
        for (int i = left + 1; i < right; i++) {
            int coins = nums[left] * nums[i] * nums[right];
            coins += dfs(nums, left, i) + dfs(nums, i, right);
            maxCoins = Math.max(maxCoins, coins);
        }
        memo[left][right] = maxCoins;
        return maxCoins;
    }
    public static void main(String[] args) {
        BurstBalloons solution = new BurstBalloons();
        System.out.println(solution.maxCoins(new int[]{3,1,5,8})); // 167
    }
}

Line Notes

memo = new int[n][n];Initialize memo table for caching subproblem results

Arrays.fill(memo[i], -1);Mark all states as uncomputed

if (memo[left][right] != -1) return memo[left][right];Return cached result if available

memo[left][right] = maxCoins;Store computed result for reuse

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

class Solution {
    vector<vector<int>> memo;
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size() + 2;
        vector<int> newNums(n, 1);
        for (int i = 0; i < nums.size(); i++) {
            newNums[i + 1] = nums[i];
        }
        memo.assign(n, vector<int>(n, -1));
        return dfs(newNums, 0, n - 1);
    }
private:
    int dfs(vector<int>& nums, int left, int right) {
        if (left + 1 == right) return 0;
        if (memo[left][right] != -1) return memo[left][right];
        int maxCoins = 0;
        for (int i = left + 1; i < right; i++) {
            int coins = nums[left] * nums[i] * nums[right];
            coins += dfs(nums, left, i) + dfs(nums, i, right);
            maxCoins = max(maxCoins, coins);
        }
        memo[left][right] = maxCoins;
        return maxCoins;
    }
};

int main() {
    Solution sol;
    vector<int> nums = {3,1,5,8};
    cout << sol.maxCoins(nums) << endl; // 167
    return 0;
}

Line Notes

memo.assign(n, vector<int>(n, -1));Initialize memo table with -1 to mark unvisited states

if (memo[left][right] != -1) return memo[left][right];Return cached result if computed

memo[left][right] = maxCoins;Store computed result for current subproblem

return memo[left][right];Return memoized result

function maxCoins(nums) {
    nums = [1, ...nums, 1];
    const n = nums.length;
    const memo = Array.from({length: n}, () => Array(n).fill(-1));
    function dfs(left, right) {
        if (left + 1 === right) return 0;
        if (memo[left][right] !== -1) return memo[left][right];
        let maxCoins = 0;
        for (let i = left + 1; i < right; i++) {
            let coins = nums[left] * nums[i] * nums[right];
            coins += dfs(left, i) + dfs(i, right);
            maxCoins = Math.max(maxCoins, coins);
        }
        memo[left][right] = maxCoins;
        return maxCoins;
    }
    return dfs(0, n - 1);
}

console.log(maxCoins([3,1,5,8])); // 167

Line Notes

const memo = Array.from({length: n}, () => Array(n).fill(-1));Create 2D memo array initialized to -1

if (memo[left][right] !== -1) return memo[left][right];Return cached result if available

memo[left][right] = maxCoins;Cache computed result for reuse

return memo[left][right];Return memoized result

⏱

Complexity

TimeO(n^3)

SpaceO(n^2) for memo table + O(n) recursion stack

There are O(n^2) subproblems and each tries O(n) balloons to burst last.

💡 For n=100, this means about 1 million operations, which is feasible.

Interview Verdict: Accepted

Memoization makes the solution efficient enough for typical constraints.

🧠

Bottom-Up Tabulation

💡 Tabulation builds the solution iteratively from smaller subproblems, avoiding recursion overhead. This approach is often easier to debug and understand once the recursive structure is clear.

Intuition

Fill a DP table dp[left][right] representing max coins from bursting balloons between left and right, increasing the interval size.

Algorithm

1. Initialize dp table with zeros for all intervals.
2. Iterate over interval lengths from 2 to n.
3. For each interval (left, right), try bursting each balloon i last.
4. Update dp[left][right] with max coins from bursting i last plus dp[left][i] and dp[i][right].

💡 The key is to solve smaller intervals first so their results are ready when needed for larger intervals.

</>

Code

def maxCoins(nums):
    nums = [1] + nums + [1]
    n = len(nums)
    dp = [[0] * n for _ in range(n)]
    for length in range(2, n):
        for left in range(n - length):
            right = left + length
            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)
    return dp[0][n - 1]

# Example usage
if __name__ == '__main__':
    print(maxCoins([3,1,5,8]))  # Output: 167

Line Notes

dp = [[0] * n for _ in range(n)]Initialize DP table with zeros for all intervals

for length in range(2, n):Iterate over interval lengths starting from 2

for left in range(n - length):Set left boundary of interval

right = left + lengthCalculate right boundary of interval

for i in range(left + 1, right):Try bursting balloon i last in current interval

coins = nums[left] * nums[i] * nums[right] + dp[left][i] + dp[i][right]Calculate coins for bursting i last plus subproblems

dp[left][right] = max(dp[left][right], coins)Update DP table with maximum coins

public class BurstBalloons {
    public int maxCoins(int[] nums) {
        int n = nums.length + 2;
        int[] newNums = new int[n];
        newNums[0] = 1;
        newNums[n - 1] = 1;
        for (int i = 0; i < nums.length; i++) {
            newNums[i + 1] = nums[i];
        }
        int[][] dp = new int[n][n];
        for (int length = 2; length < n; length++) {
            for (int left = 0; left < n - length; left++) {
                int right = left + length;
                for (int i = left + 1; i < right; i++) {
                    int coins = newNums[left] * newNums[i] * newNums[right] + dp[left][i] + dp[i][right];
                    dp[left][right] = Math.max(dp[left][right], coins);
                }
            }
        }
        return dp[0][n - 1];
    }
    public static void main(String[] args) {
        BurstBalloons solution = new BurstBalloons();
        System.out.println(solution.maxCoins(new int[]{3,1,5,8})); // 167
    }
}

Line Notes

int[][] dp = new int[n][n];Create DP table to store max coins for intervals

for (int length = 2; length < n; length++)Iterate over interval lengths

for (int left = 0; left < n - length; left++)Set left boundary of interval

int right = left + length;Calculate right boundary

for (int i = left + 1; i < right; i++)Try bursting balloon i last in interval

int coins = newNums[left] * newNums[i] * newNums[right] + dp[left][i] + dp[i][right];Calculate coins for bursting i last plus subproblems

dp[left][right] = Math.max(dp[left][right], coins);Update DP table with max coins

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size() + 2;
        vector<int> newNums(n, 1);
        for (int i = 0; i < nums.size(); i++) {
            newNums[i + 1] = nums[i];
        }
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for (int length = 2; length < n; length++) {
            for (int left = 0; left < n - length; left++) {
                int right = left + length;
                for (int i = left + 1; i < right; i++) {
                    int coins = newNums[left] * newNums[i] * newNums[right] + dp[left][i] + dp[i][right];
                    dp[left][right] = max(dp[left][right], coins);
                }
            }
        }
        return dp[0][n - 1];
    }
};

int main() {
    Solution sol;
    vector<int> nums = {3,1,5,8};
    cout << sol.maxCoins(nums) << endl; // 167
    return 0;
}

Line Notes

vector<vector<int>> dp(n, vector<int>(n, 0));Initialize DP table with zeros

for (int length = 2; length < n; length++)Iterate over interval lengths

for (int left = 0; left < n - length; left++)Set left boundary

int right = left + length;Calculate right boundary

for (int i = left + 1; i < right; i++)Try bursting balloon i last

int coins = newNums[left] * newNums[i] * newNums[right] + dp[left][i] + dp[i][right];Calculate coins for bursting i last plus subproblems

dp[left][right] = max(dp[left][right], coins);Update DP table

function maxCoins(nums) {
    nums = [1, ...nums, 1];
    const n = nums.length;
    const dp = Array.from({length: n}, () => Array(n).fill(0));
    for (let length = 2; length < n; length++) {
        for (let left = 0; left < n - length; left++) {
            let right = left + length;
            for (let i = left + 1; i < right; i++) {
                let coins = nums[left] * nums[i] * nums[right] + dp[left][i] + dp[i][right];
                dp[left][right] = Math.max(dp[left][right], coins);
            }
        }
    }
    return dp[0][n - 1];
}

console.log(maxCoins([3,1,5,8])); // 167

Line Notes

const dp = Array.from({length: n}, () => Array(n).fill(0));Initialize 2D DP array with zeros

for (let length = 2; length < n; length++) {Iterate over interval lengths

for (let left = 0; left < n - length; left++) {Set left boundary

let right = left + length;Calculate right boundary

for (let i = left + 1; i < right; i++) {Try bursting balloon i last

let coins = nums[left] * nums[i] * nums[right] + dp[left][i] + dp[i][right];Calculate coins for bursting i last plus subproblems

dp[left][right] = Math.max(dp[left][right], coins);Update DP table with max coins

⏱

Complexity

TimeO(n^3)

SpaceO(n^2)

Three nested loops: interval length, left boundary, and balloon to burst last.

💡 For n=100, this is about 1 million operations, efficient for interview constraints.

Interview Verdict: Accepted

Bottom-up DP is often preferred for clarity and avoiding recursion limits.

📊

All Approaches - One-Glance Tradeoffs

💡 In interviews, coding the memoized or bottom-up DP approach is best to balance clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n * 2^n)	O(n) recursion stack	Yes (deep recursion)	N/A	Mention only - never code
2. Top-Down Memoization	O(n^3)	O(n^2) memo + O(n) recursion stack	Possible for large n	Yes	Good to code if comfortable with recursion
3. Bottom-Up Tabulation	O(n^3)	O(n^2)	No	Yes	Preferred approach for clarity and safety

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Clarify the problem and constraints.Explain the brute force recursive approach to show understanding of problem structure.Introduce memoization to optimize recursion.Present bottom-up DP for iterative solution.Discuss time and space complexities and tradeoffs.Write clean, tested code.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test & Optimize: 7min. Total ~25min

What the Interviewer Tests

Ability to identify overlapping subproblems, implement recursion with memoization, and convert to bottom-up DP. Also tests understanding of interval DP and careful indexing.

Common Follow-ups

• What if balloons have zero or negative values? → Handle by filtering or adjusting logic.
• Can you optimize space usage? → Possible but complex due to interval dependencies.

💡 These follow-ups test robustness and deeper optimization understanding.

🔍

Pattern Recognition

When to Use

1) Problem involves intervals or ranges; 2) Optimal substructure depends on subintervals; 3) Order of operations affects result; 4) Reverse thinking (last operation first) helps.

Signature Phrases

'burst balloon last''adjacent balloons''maximum coins'

NOT This Pattern When

Problems that only require greedy or simple DP without interval dependencies.

Similar Problems

Matrix Chain Multiplication - similar interval DP structurePalindrome Partitioning - interval DP with substring ranges

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. The following code attempts to implement the space-optimized DP solution for Maximal Square. Identify the line containing the subtle bug that causes incorrect results on some inputs.

def maximalSquare(matrix):
    if not matrix or not matrix[0]:
        return 0
    rows, cols = len(matrix), len(matrix[0])
    dp = [0] * (cols + 1)
    max_side = 0
    prev = 0
    for i in range(rows):
        for j in range(1, cols + 1):
            temp = dp[j]
            if matrix[i][j-1] == '1':
                dp[j] = 1 + min(dp[j], dp[j-1], prev)  # Bug here
                max_side = max(max_side, dp[j])
            else:
                dp[j] = 0
            prev = temp
    return max_side * max_side

medium

A. Line with 'dp[j] = 1 + min(dp[j], dp[j-1], dp[j-1])'

B. Line with 'temp = dp[j]'

C. Line with 'prev = temp'

D. Line with 'max_side = max(max_side, dp[j])'

Solution

1. Step 1: Identify the DP recurrence

   The recurrence requires min(dp[j], dp[j-1], prev), but code uses dp[j-1] twice, missing 'prev'.

2. Step 2: Understand impact

   Using dp[j-1] twice ignores the diagonal value 'prev', causing incorrect dp values and max square size.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct recurrence uses three distinct values [OK]

Hint: Check if all three DP states are used in min() [OK]

Common Mistakes:

• Forgetting to use 'prev' in min()
• Overwriting dp before using dependencies
• Incorrect dp initialization

3. The following code attempts to solve the Unique Paths II problem using a space-optimized DP approach. Identify the line containing the subtle bug that causes incorrect results when the start cell is an obstacle.

def uniquePathsWithObstacles(obstacleGrid):
    m, n = len(obstacleGrid), len(obstacleGrid[0])
    dp = [0]*n
    dp[0] = 1  # Bug here
    for i in range(m):
        for j in range(n):
            if obstacleGrid[i][j] == 1:
                dp[j] = 0
            else:
                if j > 0:
                    dp[j] += dp[j-1]
    return dp[-1]

medium

A. Line 4: dp[0] = 1 # Initialization without checking start cell obstacle

B. Line 7: if obstacleGrid[i][j] == 1: dp[j] = 0

C. Line 10: if j > 0: dp[j] += dp[j-1]

D. Line 3: dp = [0]*n

Solution

1. Step 1: Check initialization of dp[0]

   dp[0] is set to 1 unconditionally, ignoring if start cell is blocked.

2. Step 2: Consequence of bug

   If start cell is an obstacle, dp[0] should be 0, otherwise paths count is incorrectly non-zero.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Proper initialization must consider obstacleGrid[0][0] [OK]

Hint: Start cell obstacle must set dp[0] to 0 [OK]

Common Mistakes:

• Ignoring start cell obstacle
• Not zeroing dp on obstacles
• Misusing dp[j-1] before update

4. What is the time complexity of the optimal interval DP solution for Zuma Game (Min Moves to Clear) with board length n and hand size m, considering the state space and transitions?

medium

A. O(n^4 * m^5)

B. O(n^2 * m^3)

C. O(n^3 * m^4)

D. O(5^(n+m))

Solution

1. Step 1: Identify DP state dimensions

   The DP state depends on intervals of the board (O(n^2)) and hand counts (up to 5 colors, each up to m), leading to O(n^3 * m^4) states due to interval splits and hand count combinations.

2. Step 2: Analyze transitions and recursion

   Each state considers splits and insertions, adding an extra factor of n and m, resulting in O(n^3 * m^4) time complexity.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Complexity matches known analysis for interval DP with hand states [OK]

Hint: Interval splits and hand states multiply complexity [OK]

Common Mistakes:

• Confusing n^2 intervals with n^3 due to splits
• Ignoring hand state dimension leading to underestimation
• Mistaking brute force complexity for DP complexity

5. Suppose the Stone Game is modified so that players can pick stones from either end multiple times (i.e., reuse piles after picking). Which of the following changes correctly adapts the DP solution to handle this variant?

hard

A. Use a bottom-up DP with a sliding window over piles, but do not change the recurrence relation

B. Keep the same interval DP but add a loop to consider picking the same pile multiple times in one turn

C. Change dp state to represent maximum difference with unlimited picks and use a 1D dp array updated iteratively

D. Switch to a greedy approach since DP no longer applies when reuse is allowed

Solution

1. Step 1: Understand the impact of reuse

   Allowing reuse means piles are infinite or replenished, so intervals no longer shrink and the problem resembles unbounded knapsack.

2. Step 2: Adapt DP state and recurrence

   We need a 1D dp array representing maximum difference for each possible pile count, updated iteratively considering unlimited picks, similar to unbounded knapsack.

3. Step 3: Reject incorrect options

   Interval DP with multiple picks per turn or unchanged recurrence won't handle reuse correctly; greedy fails due to opponent's optimal play.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Reuse -> unbounded knapsack style DP with 1D array [OK]

Hint: Reuse -> unbounded knapsack DP, not interval DP [OK]

Common Mistakes:

• Trying to reuse interval DP unchanged
• Assuming greedy works with reuse