We add 1 at both ends of the input array to simplify edge cases, resulting in nums = [1,3,1,5,8,1]. Then, we initialize a 6x6 DP table filled with zeros to store maximum coins for intervals.
💡 Augmenting the array with 1s at both ends allows uniform calculation of coins without worrying about boundaries. Initializing dp with zeros sets the base for bottom-up computation.
Line:nums = [1] + nums + [1]
n = len(nums)
dp = [[0] * n for _ in range(n)]
💡 The problem is transformed into finding max coins for intervals in the augmented array, and dp table will store these results.
fill_row
Start interval length loop with length = 2
We begin filling the DP table by considering intervals of length 2, which represent the smallest subproblems with one balloon between left and right boundaries.
💡 Starting with the smallest intervals ensures that when we compute larger intervals, the smaller subproblems are already solved.
Line:for length in range(2, n):
💡 The DP table is filled diagonally by increasing interval length.
fill_cells
Set left pointer for interval length 2 at 0
We set the left pointer to 0, defining the interval [0, 2] which covers balloons nums[1] only. We will try bursting balloon i=1 last in this interval.
💡 Each interval is defined by left and right pointers; here, we start with the first interval of length 2.
Line:for left in range(n - length):
right = left + length
💡 Intervals are processed from left to right for each length.
compare
Try bursting balloon i=1 last in interval [0,2]
We calculate coins gained by bursting balloon 1 last: nums[0]*nums[1]*nums[2] + dp[0][1] + dp[1][2] = 1*3*1 + 0 + 0 = 3. Update dp[0][2] to max(0,3)=3.
💡 Bursting balloon i last means coins from bursting i plus coins from subintervals left and right of i.
💡 Choosing the optimal last balloon maximizes coins collected.
fill_row
Set interval length to 4 and left pointer to 0
Increase interval length to 4 and set left pointer to 0, defining interval [0,4]. We will try bursting balloons i=1, i=2, and i=3 last in this interval.
💡 Larger intervals combine more balloons and rely on previously computed smaller intervals.
Line:for length in range(2, n):
for left in range(n - length):
right = left + length
💡 DP table is filled progressively for all intervals.
💡 All candidates have been considered; dp[0][5] holds the max coins.
reconstruct
Return final answer dp[0][n-1]
The maximum coins collected by bursting all balloons is stored in dp[0][5], which is 167. We return this as the final answer.
💡 The dp table's top-right corner holds the solution to the original problem.
Line:return dp[0][n - 1]
💡 The bottom-up DP approach successfully computes the maximum coins.
def maxCoins(nums):
nums = [1] + nums + [1] # STEP 1: Augment array
n = len(nums) # STEP 1
dp = [[0] * n for _ in range(n)] # STEP 1
for length in range(2, n): # STEP 2
for left in range(n - length): # STEP 3,5,7,10,14
right = left + length # STEP 3,5,7,10,14
for i in range(left + 1, right): # STEP 4,6,8,9,11-13,15-19
coins = nums[left] * nums[i] * nums[right] + dp[left][i] + dp[i][right] # STEP 4,6,8,9,11-13,15-19
dp[left][right] = max(dp[left][right], coins) # STEP 4,6,8,9,11-13,15-19
return dp[0][n - 1] # STEP 20
if __name__ == '__main__':
print(maxCoins([3,1,5,8])) # Output: 167
📊
Burst Balloons - Watch the Algorithm Execute, Step by Step
Watching this visualization helps you understand how the bottom-up DP approach breaks down the problem into smaller intervals and builds up the solution, which is difficult to grasp from code alone.
Step 1/20
·Active fill★Answer cell
setup
1
0
3
1
1
2
5
3
8
4
1
5
Result: 0
fill_row
1
0
3
1
length
1
2
5
3
8
4
1
5
Result: 0
fill_cells
left
1
0
3
1
right
1
2
5
3
8
4
1
5
Result: 0
compare
left
1
0
i
3
1
right
1
2
5
3
8
4
1
5
Result: 3
fill_cells
1
0
left
3
1
length
1
2
right
5
3
8
4
1
5
Result: 3
compare
1
0
left
3
1
i
1
2
right
5
3
8
4
1
5
Result: 15
fill_row
left
1
0
3
1
1
2
right
5
3
8
4
1
5
Result: 15
compare
left
1
0
i
3
1
1
2
right
5
3
8
4
1
5
Result: 30
compare
left
1
0
3
1
i
1
2
right
5
3
8
4
1
5
Result: 30
fill_row
left
1
0
3
1
1
2
5
3
right
8
4
1
5
Result: 30
compare
left
1
0
i
3
1
1
2
5
3
right
8
4
1
5
Result: 30
compare
left
1
0
3
1
i
1
2
5
3
right
8
4
1
5
Result: 30
compare
left
1
0
3
1
1
2
i
5
3
right
8
4
1
5
Result: 70
fill_row
left
1
0
3
1
1
2
5
3
8
4
right
1
5
Result: 70
compare
left
1
0
i
3
1
1
2
5
3
8
4
right
1
5
Result: 70
compare
left
1
0
3
1
i
1
2
5
3
8
4
right
1
5
Result: 70
compare
left
1
0
3
1
1
2
i
5
3
8
4
right
1
5
Result: 70
compare
left
1
0
3
1
1
2
5
3
i
8
4
right
1
5
Result: 70
compare
left
1
0
i
3
1
1
2
5
3
8
4
right
1
5
Result: 78
record
left
1
0
3
1
1
2
5
3
8
4
right
1
5
Result: 167
Key Takeaways
✓ The DP table is filled by increasing interval length, ensuring smaller subproblems are solved before larger ones.
This order is hard to see from code alone but is critical for correctness.
✓ Each dp[left][right] entry represents the maximum coins obtainable by bursting balloons strictly between left and right.
Understanding this interval definition clarifies why the augmented array has 1s at both ends.
✓ Choosing which balloon to burst last in an interval is the key decision that leads to the optimal solution.
The visualization shows how trying all candidates and taking the max builds the solution.
Practice
(1/5)
1. Consider the following code snippet for the Stone Game problem. What is the value of dp[0][3] after the completion of all loops when input piles = [5, 3, 4, 5]?
def stoneGame(piles):
n = len(piles)
dp = [[0]*n for _ in range(n)]
for i in range(n):
dp[i][i] = piles[i]
for length in range(2, n+1):
for i in range(n - length + 1):
j = i + length - 1
dp[i][j] = max(piles[i] - dp[i+1][j], piles[j] - dp[i][j-1])
return dp[0][n-1] > 0
easy
A. 2
B. 0
C. 3
D. 1
Solution
Step 1: Initialize dp for single piles
dp[i][i] = piles[i], so dp[0][0]=5, dp[1][1]=3, dp[2][2]=4, dp[3][3]=5.
Step 2: Compute dp for length=2 to 4
For length=2, dp[0][1] = max(5 - 3, 3 - 5) = max(2, -2) = 2; similarly compute others. For length=4, dp[0][3] = max(5 - dp[1][3], 5 - dp[0][2]) = max(5 - 1, 5 - 4) = max(4, 1) = 4. But since dp stores difference, the final dp[0][3] = 1 after all calculations.
Final Answer:
Option D -> Option D
Quick Check:
Manual trace confirms dp[0][3] = 1 [OK]
Hint: Trace dp table diagonally from single piles [OK]
Common Mistakes:
Off-by-one errors in indices
Confusing dp values with total stones
2. Consider the following bottom-up DP code for the Strange Printer problem. What is the final returned value when the input string is "aba"?
easy
A. 3
B. 4
C. 1
D. 2
Solution
Step 1: Compress input string "aba"
Compression does not change string since no consecutive duplicates: s = "aba", n=3.
Step 2: Trace dp table filling
Base cases: dp[0][0]=1, dp[1][1]=1, dp[2][2]=1.
For length=2:
- dp[0][1]: dp[0][0]+1=2, s[0]!=s[1], so dp[0][1]=2
- dp[1][2]: dp[1][1]+1=2, s[1]!=s[2], so dp[1][2]=2
For length=3:
- dp[0][2]: dp[0][1]+1=3
Check k=0: s[0]==s[2] ('a'=='a'), dp[0][0]+dp[1][1]=1+1=2 < 3, so dp[0][2]=2
Forgetting to merge intervals when characters match
3. What is the time complexity of the bottom-up dynamic programming solution for the minimum score triangulation of a polygon with n vertices?
medium
A. O(n^2)
B. O(n^4)
C. O(n^3)
D. O(n^3 * log n)
Solution
Step 1: Identify loops in the code
There are three nested loops: length (up to n), i (up to n), and k (up to n), each iterating up to n times.
Step 2: Calculate total complexity
Multiplying these loops gives O(n * n * n) = O(n^3). No extra log factor or higher power arises.
Final Answer:
Option C -> Option C
Quick Check:
Three nested loops over n vertices -> cubic time [OK]
Hint: Count nested loops carefully to avoid underestimating complexity [OK]
Common Mistakes:
Confusing recursion stack space
Assuming only two nested loops
Adding unnecessary log factors
4. The following code attempts to solve the Stone Game problem using bottom-up DP. Identify the line containing the subtle bug that causes incorrect results on some inputs.
def stoneGame(piles):
n = len(piles)
dp = [[0]*n for _ in range(n)]
for i in range(n):
dp[i][i] = piles[i]
for length in range(2, n+1):
for i in range(n - length + 1):
j = i + length - 1
dp[i][j] = max(piles[i] + dp[i+1][j], piles[j] + dp[i][j-1])
return dp[0][n-1] > 0
medium
A. Line computing dp[i][j] = max(piles[i] + dp[i+1][j], piles[j] + dp[i][j-1])
B. Line initializing dp[i][i] = piles[i]
C. Line defining the outer loop for length in range(2, n+1)
D. Line returning dp[0][n-1] > 0
Solution
Step 1: Understand dp state meaning
dp[i][j] should represent the maximum difference in stones the current player can achieve over the opponent.
Step 2: Identify incorrect recurrence
The buggy line adds piles[i] + dp[i+1][j], which ignores the opponent's optimal response. The correct formula subtracts dp[i+1][j] to account for opponent's best play.
Final Answer:
Option A -> Option A
Quick Check:
Correct recurrence uses subtraction, not addition [OK]
Hint: DP recurrence must subtract opponent's score [OK]
Common Mistakes:
Adding dp values instead of subtracting opponent's score
Forgetting base case initialization
5. 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
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.
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.
Final Answer:
Option C -> Option C
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
Recommended
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