Minimum Score Triangulation of Polygon

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Steps

setup

Initialize DP Table with zeros

Create a 3x3 dp table filled with zeros because intervals of length less than 3 cannot form a triangle and thus have zero cost.

💡 Intervals too small to triangulate have zero cost, setting base cases for DP.

Line:

dp = [[0] * n for _ in range(n)]  # STEP 1

💡 Intervals with length less than 3 have zero triangulation cost.

fill_row

Start filling intervals of length 3

Set length = 3 to start considering the smallest polygon intervals that can form a triangle.

💡 Intervals of length 3 are the smallest that can form a triangle, so we start computing their minimal triangulation cost.

Line:

for length in range(3, n + 1):  # STEP 2

💡 We will fill dp cells for intervals of length 3 first.

fill_cells

Identify interval [0,2] to fill

For length=3, the only interval is from i=0 to j=2. We prepare to compute dp[0][2].

💡 We focus on the interval [0,2] which forms the smallest polygon triangle.

Line:

for i in range(n - length + 1):  # STEP 3
    j = i + length - 1  # STEP 3

💡 Intervals are processed in order of increasing length; here, length=3.

fill_cells

Initialize dp[0][2] to infinity

Set dp[0][2] to infinity to find the minimal triangulation cost by trying all possible splits.

💡 Setting dp[0][2] to infinity prepares to find the minimal cost by comparing all possible triangulations.

Line:

dp[i][j] = float('inf')  # STEP 3

💡 We are preparing to find the minimal triangulation cost for interval [0,2].

compare

Try k=1 to split interval [0,2]

Calculate cost for triangulation by splitting at k=1: dp[0][1] + dp[1][2] + values[0]*values[1]*values[2].

💡 Trying all possible k splits finds the minimal triangulation cost by combining smaller subproblems.

Line:

cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j]  # STEP 4

💡 The only possible triangle for [0,2] has cost 6.

fill_row

Intervals of length 3 done, no other intervals to fill

Since n=3, only one interval of length 3 exists. The dp table is now fully computed.

💡 For this input, only one triangle exists, so the dp table is complete.

Line:

return dp[0][n - 1]  # STEP 6

💡 The minimal triangulation score for the entire polygon is dp[0][2] = 6.

reconstruct

Read final answer from dp[0][2]

Return dp[0][2] which holds the minimal triangulation score for the polygon with vertices 0 to 2.

💡 The answer is stored in dp[0][n-1], representing the entire polygon.

Line:

return dp[0][n - 1]  # STEP 6

💡 The minimal triangulation score for the polygon is 6.

final

Summary: DP table fully computed

The dp table shows minimal triangulation costs for all intervals; only dp[0][2] was updated for this input.

💡 This final state confirms the algorithm's correctness and the minimal triangulation cost.

💡 The dp table structure and values reflect the minimal triangulation solution.

insight

Reinforce understanding of DP dependencies

Recall that dp[i][j] depends on dp[i][k] and dp[k][j] for i < k < j, ensuring smaller intervals are solved first.

💡 This dependency order is why we fill the dp table by increasing interval length.

Line:

for length in range(3, n + 1):  # STEP 2
    for i in range(n - length + 1):  # STEP 3
        j = i + length - 1  # STEP 3

💡 Understanding dependencies helps grasp why the algorithm fills the dp table in this order.

final

Confirm minimal triangulation cost matches expected output

The computed minimal triangulation cost dp[0][2] = 6 matches the expected output for the input polygon.

💡 Matching expected output validates the correctness of the DP approach.

Line:

return dp[0][n - 1]  # STEP 6

💡 The algorithm correctly computes the minimal triangulation cost for the polygon.

def minScoreTriangulation(values):
    n = len(values)  # STEP 1
    dp = [[0] * n for _ in range(n)]  # STEP 1
    for length in range(3, n + 1):  # STEP 2
        for i in range(n - length + 1):  # STEP 3
            j = i + length - 1  # STEP 3
            dp[i][j] = float('inf')  # STEP 3
            for k in range(i + 1, j):  # STEP 4
                cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j]  # STEP 4
                if cost < dp[i][j]:  # STEP 4
                    dp[i][j] = cost  # STEP 4
    return dp[0][n - 1]  # STEP 6

📊

Minimum Score Triangulation of Polygon - Watch the Algorithm Execute, Step by Step

Watching the algorithm fill the dp table step-by-step reveals how overlapping subproblems are solved and combined, which is hard to grasp from code alone.

Item 0 - wt:0 val:0

i\w	0	1	2
i=0	0	0	0
i=1	0	0	0
i=2	0	0	0

initialized to 0

Item 0 - wt:0 val:0

i\w	0	1	2
i=0	0	0	0
i=1	0	0	0
i=2	0	0	0

dp initialized

Item 0 - wt:0 val:0

i\w	0	1	2
i=0	0	0	0
i=1	0	0	0
i=2	0	0	0

interval [0,2]

Item 0 - wt:1 val:2

i\w	0	1	2
i=0	0	0	∞
i=1	0	0	0
i=2	0	0	0

dp[0][2] = ∞

Item 1 - wt:2 val:3

i\w	0	1	2
i=0	0	0	6
i=1	0	0	0
i=2	0	0	0

dp[0][2] = 6

Item 0 - wt:1 val:2

i\w	0	1	2
i=0	0	0	6
i=1	0	0	0
i=2	0	0	0

final answer

Item 0 - wt:1 val:2

i\w	0	1	2
i=0	0	0	6
i=1	0	0	0
i=2	0	0	0

answer = 6

Item 0 - wt:1 val:2

i\w	0	1	2
i=0	0	0	6
i=1	0	0	0
i=2	0	0	0

final answer

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Key Takeaways

✓ The DP table is filled by increasing interval length, ensuring all smaller subproblems are solved before larger ones.

This order is crucial because each dp[i][j] depends on dp[i][k] and dp[k][j] for smaller intervals.

✓ Each dp cell represents the minimal triangulation cost for a polygon interval, combining subproblems and the cost of the triangle formed.

Visualizing the dp table helps understand how the problem is broken down into overlapping subproblems.

✓ For the smallest polygon (length 3), only one triangle exists, so the dp cell is directly computed without further splits.

This base case is the foundation for building solutions for larger polygons.

Practice

1. You are given a grid where some cells are blocked and others are free. You need to find the number of unique paths from the top-left corner to the bottom-right corner, moving only down or right, but you cannot pass through blocked cells. Which algorithmic approach guarantees an efficient and correct solution for this problem?

easy

A. Dynamic Programming that builds solutions using previously computed subproblems while skipping blocked cells

B. Pure brute force recursion exploring all paths without memoization

C. Greedy algorithm that always moves right if possible, else down

D. Dijkstra's shortest path algorithm treating grid cells as graph nodes

Solution

1. Step 1: Understand problem constraints

   The problem requires counting all unique paths avoiding obstacles, which involves overlapping subproblems and optimal substructure.

2. Step 2: Identify suitable algorithm

   Dynamic Programming efficiently computes the number of paths by reusing results and handling obstacles by zeroing paths through blocked cells.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   DP handles obstacles and overlapping subproblems correctly [OK]

Hint: DP handles obstacles and overlapping subproblems correctly [OK]

Common Mistakes:

• Thinking greedy can find all paths
• Using brute force without pruning
• Confusing shortest path with counting paths

2. What is the time complexity of the space-optimized bottom-up DP solution for the Cheapest Flights Within K Stops problem, given n cities, E flights, and maximum K stops?

medium

A. O(K * E) because each iteration relaxes all edges up to K+1 times

B. O(n^3) due to nested loops over cities and stops

C. O(E * log n) similar to Dijkstra's algorithm

D. O(n * K^2) due to dynamic programming over stops and cities

Solution

1. Step 1: Identify loops in the algorithm

   The outer loop runs K+1 times, and the inner loop iterates over all E flights to relax edges.

2. Step 2: Calculate total complexity

   Each iteration processes E edges, so total time is O(K * E). No nested loops over n^2 or log factors appear.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Algorithm iterates over edges K+1 times [OK]

Hint: Outer loop K+1 times, inner loop over E edges [OK]

Common Mistakes:

• Confusing E with n^2
• Assuming Dijkstra complexity
• Counting recursion stack space

3. The following code attempts to compute the minimum falling path sum using bottom-up DP. Identify the line containing the subtle bug that causes incorrect results on some inputs.

def minFallingPathSum(matrix):
    n = len(matrix)
    dp = matrix[0][:]

    for i in range(1, n):
        for j in range(n):
            best = dp[j]
            if j > 0:
                best = min(best, dp[j-1])
            if j < n - 1:
                best = min(best, dp[j+1])
            dp[j] = matrix[i][j] + best

    return min(dp)

medium

A. Line 7: Updating dp[j] inside the inner loop without backup

B. Line 9: Checking if j > 0 before accessing dp[j-1]

C. Line 3: dp initialization with matrix[0][:]

D. Line 12: Returning min(dp) after the loops

Solution

1. Step 1: Understand dp update logic

   dp array holds minimum sums for previous row. Updating dp[j] in place overwrites values needed for neighbors in the same iteration.

2. Step 2: Identify the bug

   Updating dp[j] inside the inner loop without using a separate array causes premature overwriting, leading to incorrect neighbor comparisons.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Using a separate new_dp array fixes the bug [OK]

Hint: Don't overwrite dp while still reading from it in same iteration [OK]

Common Mistakes:

• Updating dp in place without backup
• Ignoring boundary checks

4. 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)

5. What is the time complexity of the bottom-up DP solution for the Strange Printer problem after string compression, where m is the length of the compressed string?

medium

A. O(m^2)

B. O(m^3)

C. O(n^3) where n is original string length

D. O(m^2 * log m)

Solution

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

   There are three nested loops: length (1 to m), start index i (up to m), and partition index k (between i and j), each up to m.

2. Step 2: Calculate complexity

   Overall complexity is O(m * m * m) = O(m^3). Compression reduces n to m, so complexity depends on compressed length.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Three nested loops over compressed length m [OK]

Hint: Three nested loops over compressed string length cause cubic time [OK]

Common Mistakes:

• Confusing original length n with compressed length m
• Forgetting the inner loop over k