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. Given the following code snippet for MedianFinder, after adding the numbers 1, 5, and 3 in that order, what is the value returned by findMedian()?
easy
A. 4.0
B. 5.0
C. 1.0
D. 3.0
Solution
Step 1: Insert 1
small is empty, so push -1 to small. small_size=1, large_size=0.
Step 2: Insert 5
5 > -small[0] (which is 1), so push 5 to large. small_size=1, large_size=1. Heaps balanced.
Step 3: Insert 3
3 <= -small[0] (1)? No, so push 3 to large. large_size=2, small_size=1. Now large_size > small_size, so move smallest from large (3) to small (push -3). small_size=2, large_size=1.
Step 4: Find median
small_size > large_size, median is -small[0] = 3.0.
Final Answer:
Option D -> Option D
Quick Check:
Median after [1,5,3] is 3.0 [OK]
Hint: Median is top of larger heap when sizes differ [OK]
Common Mistakes:
Not balancing heaps after insertion
Returning average when sizes differ
Forgetting to negate values in max heap
2. 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
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.
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.
Final Answer:
Option B -> Option B
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
3. Examine the following modified code for Largest Rectangle in Histogram. Which line contains a subtle bug that can cause incorrect maximum area calculation?
def largestRectangleArea(heights):
stack = []
max_area = 0
heights.append(0) # Sentinel to flush stack at end
for i, h in enumerate(heights):
while stack and heights[stack[-1]] >= h:
height = heights[stack.pop()]
left = stack[-1] if stack else -1
width = i - left - 1
max_area = max(max_area, height * width)
stack.append(i)
heights.pop() # Restore original list
return max_area
medium
A. Line with 'while stack and heights[stack[-1]] >= h:'
B. Line with 'height = heights[stack.pop()]'
C. Line with 'width = i - left - 1'
D. Line with 'heights.append(0)'
Solution
Step 1: Identify condition in while loop
The condition uses '>=' which causes bars of equal height to be popped prematurely, merging areas incorrectly.
Step 2: Correct condition
Changing '>=' to '>' ensures bars of equal height remain on stack, preserving correct width calculations.
Final Answer:
Option A -> Option A
Quick Check:
Using '>=' pops equal height bars too early, causing wrong max area [OK]
Hint: Use '>' not '>=' to avoid premature popping [OK]
Common Mistakes:
Using '>=' instead of '>' in while condition
Miscomputing width boundaries
Forgetting sentinel bar
4. What is the time complexity of the optimal solution for Word Search II that uses backtracking with a Trie and in-place board modification, assuming the board size is NxM, the number of words is W, and the maximum length of a word is L?
medium
A. O(W * N * M * 4^L) because each word is searched independently
B. O(W * L) for building Trie plus O(N * M * 4^L) for backtracking
C. O(N * M * L * W) because Trie traversal multiplies word length and count
D. O(N * M * 4^L) due to backtracking pruning with Trie prefix checks
Solution
Step 1: Identify Trie construction cost
Building the Trie takes O(W * L) time, inserting each word character by character.
Step 2: Analyze backtracking complexity
Backtracking explores up to 4^L paths from each cell, so O(N * M * 4^L) dominates search time.
Final Answer:
Option B -> Option B
Quick Check:
Trie build plus backtracking combined yields O(W*L + N*M*4^L) [OK]
Hint: Trie build separate from backtracking cost [OK]
Common Mistakes:
Confusing independent word search with Trie pruning
Ignoring Trie build cost
Multiplying all parameters incorrectly
5. Suppose the problem is modified so that the path can reuse nodes multiple times (cycles allowed). Which of the following statements about adapting the maximum path sum algorithm is correct?
hard
A. The existing recursive approach works without changes because it already considers all paths.
B. The problem reduces to finding the maximum sum cycle in a graph, which can be solved by the same recursion.
C. We can simply remove the max(..., 0) checks to allow negative gains and reuse nodes.
D. We must detect and avoid cycles, possibly using memoization or graph algorithms, since the tree is no longer acyclic.
Solution
Step 1: Recognize that allowing node reuse introduces cycles, so the structure is no longer a tree.
The original recursion assumes acyclic structure and no repeated nodes.
Step 2: Understand that to handle cycles, the algorithm must detect and avoid infinite recursion, typically via memoization or graph cycle detection.
Standard tree recursion breaks down without cycle checks.
Step 3: Note that removing max(..., 0) or reusing nodes without cycle detection leads to infinite loops or incorrect sums.
Maximum sum cycle detection is a different problem requiring graph algorithms.
Final Answer:
Option D -> Option D
Quick Check:
Cycle detection and memoization are necessary for graphs with cycles [OK]
Hint: Cycles require cycle detection and memoization [OK]
Common Mistakes:
Assuming tree recursion works unchanged
Removing zero cutoff to allow reuse
Confusing max path sum with max cycle sum
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