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 2D binary matrix filled with 0's and 1's. The task is to find the largest square containing only 1's and return its area. Which algorithmic approach guarantees an optimal solution with polynomial time complexity?

easy

A. Greedy approach that expands squares from each cell until a zero is found

B. Sorting rows and columns to find the largest continuous block of 1's

C. Depth-First Search exploring all possible squares recursively without memoization

D. Dynamic Programming that builds solutions for smaller squares to find larger ones

Solution

1. Step 1: Understand problem constraints

   The problem requires finding the largest square of 1's, which depends on overlapping subproblems of smaller squares.

2. Step 2: Recognize DP pattern

   Dynamic Programming efficiently computes maximal squares by using previously computed smaller squares, avoiding redundant checks.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP uses subproblem solutions to build larger squares [OK]

Hint: Largest square -> DP using smaller squares [OK]

Common Mistakes:

• Thinking greedy expansion always works
• Confusing DFS with DP
• Assuming sorting helps find squares

2. Consider the following Python code for computing unique paths in a grid:

def uniquePaths(m, n):
    dp = [1] * n
    for i in range(1, m):
        for j in range(1, n):
            dp[j] += dp[j - 1]
    return dp[-1]

print(uniquePaths(3, 3))

What is the value of dp after the outer loop completes its second iteration (i=2)?

easy

A. [1, 3, 6]

B. [1, 2, 3]

C. [1, 3, 5]

D. [1, 4, 10]

Solution

1. Step 1: Initialize dp array

   Initially, dp = [1, 1, 1] for n=3.

2. Step 2: Trace iterations

   First outer iteration (i=1): - j=1: dp[1] = dp[1] + dp[0] = 1 + 1 = 2 - j=2: dp[2] = dp[2] + dp[1] = 1 + 2 = 3 After i=1, dp = [1, 2, 3] Second outer iteration (i=2): - j=1: dp[1] = dp[1] + dp[0] = 2 + 1 = 3 - j=2: dp[2] = dp[2] + dp[1] = 3 + 3 = 6 After i=2, dp = [1, 3, 6]

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   dp array accumulates path counts correctly -> [1,3,6] [OK]

Hint: Track dp updates carefully per iteration [OK]

Common Mistakes:

• Off-by-one errors in loop indices
• Mis-updating dp[j] with wrong dp[j-1] value

3. What is the time complexity of the bottom-up dynamic programming solution for the Dungeon Game problem on an m x n grid, and why might some candidates mistakenly think it is higher?

medium

A. O(2^{m+n}) because all paths are explored recursively

B. O(m + n) since only one path is considered at a time

C. O(m * n * max(|dungeon[i][j]|)) due to health value range affecting computations

D. O(m * n) because each cell is computed once using constant time operations

Solution

1. Step 1: Identify loops

   The bottom-up DP uses nested loops over m rows and n columns, each cell computed once.

2. Step 2: Analyze per-cell work

   Each dp[i][j] calculation is O(1), involving min and max operations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP table size m*n and constant work per cell [OK]

Hint: DP fills m*n table once, no recursion overhead [OK]

Common Mistakes:

• Confusing brute force recursion with DP
• Thinking health values affect complexity

4. Suppose the Cherry Pickup problem is modified so that players can revisit cells multiple times (i.e., cycles allowed), and cherries are replenished after being picked (can be collected multiple times). Which of the following algorithmic changes correctly adapts the solution to this variant?

hard

A. Use the same 3D DP with memoization but remove the check that prevents revisiting cells.

B. Apply a greedy approach that always moves players to the next cell with the highest cherry count.

C. Use a 4D DP tracking both players' full positions and number of visits per cell to avoid double counting.

D. Switch to a shortest path algorithm like Dijkstra on a state graph representing both players' positions and steps, allowing cycles.

Solution

1. Step 1: Understand the impact of allowing revisits and replenished cherries

   Cycles and replenished cherries mean the problem is no longer acyclic and DP assumptions break down.

2. Step 2: Identify suitable algorithm

   Modeling the problem as a shortest path on a state graph with both players' positions and steps allows handling cycles and repeated cherry collection correctly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DP fails with cycles; shortest path algorithms handle repeated states [OK]

Hint: Cycles break DP; use graph shortest path [OK]

Common Mistakes:

• Removing DP constraints without changing algorithm
• Adding dimensions to DP without handling cycles
• Using greedy which ignores global optimality

5. Suppose the problem is modified so that you can move right, down, or diagonally down-right, still avoiding obstacles. Which of the following changes to the space-optimized DP approach correctly accounts for the new diagonal move?

hard

A. Modify dp[j] += dp[j-1] to dp[j] += dp[j-1] + dp[j] without extra storage.

B. Use a 2D dp array where dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1], updating in row-major order.

C. Add dp[j-1] and dp[j] plus dp[j-1] from previous row, requiring a 2D dp array to track diagonal paths.

D. Keep 1D dp array but add dp[j-1] twice to account for diagonal moves.

Solution

1. Step 1: Understand diagonal dependency

   Diagonal move depends on dp[i-1][j-1], which cannot be tracked with only 1D dp array updated in-place.

2. Step 2: Correct approach

   Use 2D dp array to store counts for all cells, updating dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1], ensuring all dependencies are available.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Diagonal requires previous row and previous column info simultaneously [OK]

Hint: Diagonal moves require 2D dp to track previous row and column [OK]

Common Mistakes:

• Trying to reuse 1D dp without extra storage
• Double counting paths
• Ignoring diagonal dependency