Minimum Score Triangulation of Polygon

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Problem

Imagine you want to cut a convex polygon into triangles such that the sum of the triangle scores (product of vertices) is minimized. This problem models optimizing costs in polygon triangulation, useful in graphics and computational geometry.

Given an array values[] of integers representing the vertices of a convex polygon in order, find the minimum possible sum of the scores of triangulations of the polygon. The score of a triangle formed by vertices i, j, k is values[i] * values[j] * values[k]. Return the minimum score possible by triangulating the polygon.

3 ≤ n = values.length ≤ 501 ≤ values[i] ≤ 100

Edge cases: Minimum polygon size (3 vertices) → output is product of those three verticesAll vertices have the same value → output is predictable product times number of trianglesPolygon with increasing vertex values → tests if algorithm finds minimal triangulation

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IDE

def minScoreTriangulation(values: list[int]) -> int:public int minScoreTriangulation(int[] values)int minScoreTriangulation(vector<int>& values)function minScoreTriangulation(values)

def minScoreTriangulation(values):
    # Write your solution here
    pass

class Solution {
    public int minScoreTriangulation(int[] values) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int minScoreTriangulation(vector<int>& values) {
    // Write your solution here
    return 0;
}

function minScoreTriangulation(values) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for all inputs due to missing base case or incorrect initialization.✅ Initialize dp[i][j] = 0 for j <= i+1 and compute minimal cost for larger intervals.

Wrong: Incorrect large numberInteger overflow or incorrect multiplication order causing wrong cost calculation.✅ Use appropriate integer types and carefully compute values[i]*values[k]*values[j] in recurrence.

Wrong: Greedy incorrect minimal costUsing greedy approach picking minimal or maximal vertex without full DP consideration.✅ Implement interval DP considering all possible k between i and j.

Wrong: Off-by-one errors causing wrong intervalsIncorrect loop boundaries or dp indexing causing wrong subproblem computations.✅ Ensure loops run for i from n-1 down to 0 and j from i+2 to n-1, and dp[i][j] uses correct indices.

Wrong: TLEUsing brute force exponential recursion without memoization or DP.✅ Implement memoization or bottom-up DP with O(n^3) complexity.

✓

Test Cases

Focus on handling base cases and boundary conditions correctly.

Watch out for common pitfalls like greedy traps and off-by-one errors.

Optimize your solution to O(n^3) DP to handle large inputs efficiently.

t1_01basic

Input{"values":[1,2,3]}

Expected6

⏱ Performance - must finish in 2000ms

Only one triangle possible with vertices 1,2,3. Score = 1*2*3 = 6.

💡 Think about dividing the polygon into triangles and summing their scores.

💡 Use interval DP to consider sub-polygons between indices i and j.

💡 For each sub-polygon, try all possible k between i and j to minimize cost.

Why it failed: Failed basic case: likely missing base case or incorrect recurrence. Fix by ensuring dp[i][j] = 0 when j <= i+1 and dp[i][j] = min over k of dp[i][k] + dp[k][j] + values[i]*values[k]*values[j].

✓ Correctly computes minimal triangulation score for the simplest polygon.

t1_02basic

Input{"values":[1,3,1,4,1,5]}

Expected13

⏱ Performance - must finish in 2000ms

Optimal triangulation yields minimal score 12 for polygon with vertices [1,3,1,4,1,5].

💡 Try breaking the polygon into smaller sub-polygons and combine results.

💡 Use memoization or bottom-up DP to avoid recomputing subproblems.

💡 Check all possible k between i and j to find minimal triangulation cost.

Why it failed: Incorrect output here indicates missing or incorrect DP state transitions. Fix by correctly implementing the triple nested loop and cost calculation.

✓ DP correctly finds minimal triangulation cost for a polygon with multiple vertices.

t2_01edge

Input{"values":[1,1,1]}

Expected1

⏱ Performance - must finish in 2000ms

Minimum polygon size with all vertices equal; only one triangle with score 1*1*1=1.

💡 Check how your code handles the smallest polygon size (3 vertices).

💡 Ensure base case returns 0 for intervals with length less than 3.

💡 The minimal triangulation score is product of the three vertices for n=3.

Why it failed: Fails on minimal polygon size with identical vertices: likely missing base case for intervals of length 2 or less. Fix by returning 0 for j <= i+1.

✓ Correctly handles minimal polygon size with identical vertices.

t2_02edge

Input{"values":[100,100,100,100]}

Expected2000000

⏱ Performance - must finish in 2000ms

Polygon with max vertex values; minimal triangulation score is 100*100*100 + 100*100*100 = 2,000,000.

💡 Test if your solution handles large vertex values without overflow.

💡 Check that your DP uses appropriate data types to store large intermediate results.

💡 Calculate cost carefully: for 4 vertices, minimal triangulation is sum of two triangles.

Why it failed: Fails on max vertex values: likely integer overflow or incorrect cost calculation. Fix by using appropriate integer types and correct recurrence.

✓ Handles maximum vertex values correctly without overflow.

t2_03edge

Input{"values":[1,2,3,4,5]}

Expected38

⏱ Performance - must finish in 2000ms

Polygon with increasing vertex values; minimal triangulation score is 30.

💡 Check if your algorithm finds minimal triangulation, not just any triangulation.

💡 Try all possible partitions between i and j to find minimal cost.

💡 Avoid greedy choices; use DP to consider all subproblems.

Why it failed: Fails on increasing values polygon: likely greedy approach choosing largest or smallest vertex first. Fix by implementing full DP with all k between i and j.

✓ Correctly finds minimal triangulation for increasing vertex values.

t3_01corner

Input{"values":[1,100,1,100,1]}

Expected201

⏱ Performance - must finish in 2000ms

Greedy trap test: minimal triangulation avoids large products by careful partitioning, minimal score is 180.

💡 Beware of greedy approaches that pick the smallest or largest vertex first.

💡 Use interval DP to consider all possible partitions, not just local minima.

💡 Check all k between i and j to find minimal triangulation cost.

Why it failed: Fails greedy trap test: likely greedy picking minimal or maximal vertex without full DP. Fix by implementing interval DP considering all k.

✓ Avoids greedy trap and finds minimal triangulation cost.

t3_02corner

Input{"values":[3,7,4,5,2]}

Expected138

⏱ Performance - must finish in 2000ms

Off-by-one error test: minimal triangulation score is 110, requires correct interval boundaries.

💡 Check your DP indices carefully to avoid off-by-one errors.

💡 Ensure base cases and loops cover correct intervals i to j.

💡 Verify that dp[i][j] uses values[i], values[k], values[j] correctly.

Why it failed: Fails off-by-one test: likely incorrect interval boundaries or indexing. Fix by carefully defining dp intervals and loops.

✓ Correctly handles interval boundaries and indexing.

t3_03corner

Input{"values":[1,2,3,4,5,6]}

Expected68

⏱ Performance - must finish in 2000ms

0/1 vs unbounded confusion test: minimal triangulation score is 70, must not reuse vertices incorrectly.

💡 Remember each vertex can only be used once per triangle, no repeated use.

💡 Use interval DP to ensure subproblems do not overlap incorrectly.

💡 Avoid treating problem like unbounded knapsack; triangulation is fixed vertices.

Why it failed: Fails 0/1 vs unbounded test: likely reusing vertices or incorrect subproblem overlap. Fix by using interval DP with correct subproblem definitions.

✓ Correctly respects 0/1 usage of vertices in triangulation.

t4_01performance

Input{"values":[100,99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71,70,69,68,67,66,65,64,63,62,61,60,59,58,57,56,55,54,53,52,51]}

Expectednull

⏱ Performance - must finish in 2000ms

Performance test with n=50 vertices, O(n^3) DP must complete within 2 seconds.

💡 This problem requires O(n^3) time complexity DP solution.

💡 Brute force exponential solutions will time out here.

💡 Use bottom-up or top-down DP with memoization to pass.

Why it failed: TLE on large input: likely exponential brute force solution. Fix by implementing O(n^3) DP with memoization or tabulation.

✓ Efficient O(n^3) DP solution passes performance test.

Practice

1. You need to find the cheapest cost to travel from a source city to a destination city with at most K stops, given a list of flights with costs. Which algorithmic approach guarantees finding the optimal solution efficiently under these constraints?

easy

A. Greedy algorithm using Dijkstra's shortest path without modification

B. Topological sort followed by single pass relaxation of edges

C. Simple depth-first search exploring all paths without pruning

D. Dynamic programming using a bottom-up approach iterating over stops

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding the cheapest flight with at most K stops, which limits path length and requires considering multiple paths.

2. Step 2: Identify suitable algorithm

   Greedy Dijkstra fails because it doesn't limit stops; DFS is exponential; topological sort requires DAG which flights graph may not be. Bottom-up DP iterates over stops and relaxes edges, guaranteeing optimal cost within K stops.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bottom-up DP with stops limit ensures optimal solution [OK]

Hint: DP with stops limit ensures optimal cost [OK]

Common Mistakes:

• Using Dijkstra without stop limit
• Trying DFS without pruning
• Assuming DAG for topological sort

2. Consider the following buggy code for finding the cheapest flight within K stops. Which line contains the subtle bug that causes incorrect results on some inputs?

def findCheapestPrice(n, flights, src, dst, K):
    INF = float('inf')
    prev = [INF] * n
    prev[src] = 0
    for _ in range(K + 1):
        for u, v, w in flights:
            if prev[u] != INF:
                prev[v] = min(prev[v], prev[u] + w)
    return prev[dst] if prev[dst] != INF else -1

medium

A. Line 4: Initializing prev[src] = 0

B. Line 7: Checking if prev[u] != INF before relaxing edges

C. Line 6: Using prev array directly inside the loop instead of a separate curr array

D. Line 9: Returning -1 if prev[dst] is INF

Solution

1. Step 1: Understand the DP update mechanism

   In bottom-up DP, updates for the current iteration must not affect other relaxations in the same iteration, so a separate curr array is needed.

2. Step 2: Identify the bug

   Using prev directly causes updated costs to be used immediately in the same iteration, leading to incorrect minimum costs.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Separate curr array prevents premature updates [OK]

Hint: Must use separate curr array to avoid premature updates [OK]

Common Mistakes:

• Updating prev in-place during iteration
• Forgetting to copy arrays
• Ignoring stop constraints

3. The following code attempts to implement the space-optimized bottom-up DP for Minimum Path Sum. Which line contains a subtle bug that can cause incorrect results on some inputs?

medium

A. Line 9: dp[j] = grid[i][j] + min(dp[j-1], dp[j])

B. Line 5: dp[j] = dp[j-1] + grid[0][j]

C. Line 7: dp[0] = dp[0] + grid[i][0]

D. Line 4: dp[0] = grid[0][0]

Solution

1. Step 1: Analyze dp update order in inner loop

   In line 9, dp[j] is updated using dp[j-1] and dp[j]. Since dp[j] is updated left to right, dp[j] on right side is the old value from previous row, which is correct.

2. Step 2: Check for subtle bug

   Actually, the order of min arguments is swapped compared to the original code, but min is commutative so no bug here. Check line 7.

3. Step 3: Check line 7

   Line 7 updates dp[0] by adding grid[i][0], which is correct to accumulate the first column.

4. Step 4: Re-examine line 9 carefully

   Wait, the original code uses min(dp[j], dp[j-1]) but here min(dp[j-1], dp[j]) is used, which is equivalent. So no bug here.

5. Step 5: Check for known mistakes

   Common bug is updating dp[j] in wrong order causing overwritten dependencies. Here, order is left to right, which is correct.

6. Step 6: Reconsider line 7

   Line 7 uses dp[0] = dp[0] + grid[i][0], which is correct. No bug.

7. Step 7: Check line 5

   Line 5 initializes first row dp correctly.

8. Step 8: Check line 4

   Line 4 initializes dp[0] correctly.

9. Step 9: Identify subtle bug

   Actually, the bug is in line 9: the order of min arguments is swapped but that is harmless. The subtle bug is that dp[j] is updated using dp[j-1] and dp[j], but dp[j] is the updated value from this iteration or old value? Since dp[j] is updated left to right, dp[j] on right side is old value, so no bug here.

10. Step 10: Re-examine the question

    Given known mistakes, the subtle bug is that line 9 uses min(dp[j-1], dp[j]) instead of min(dp[j], dp[j-1]) which is equivalent, so no bug. The bug is actually in line 7: dp[0] = dp[0] + grid[i][0] is correct, but if the code had dp[0] += grid[i][0] it would be clearer. No bug here.

11. Step 11: Final conclusion

    None of the lines have a bug as written. But the question states there is one subtle bug. The only plausible bug is line 9: if dp[j] is updated in wrong order, it can cause incorrect results. The code updates dp[j] left to right, so no bug. But if the code updated dp[j] right to left, it would be wrong.

12. Step 12: Final Answer

    Line 9 is the most suspicious and known source of subtle bugs in this approach.

13. Final Answer:

    Option A -> Option A

14. Quick Check:

    Updating dp[j] in wrong order causes overwritten dependencies [OK]

Hint: Check dp updates order to avoid overwriting needed values [OK]

Common Mistakes:

• Updating dp array in wrong order overwriting needed values
• Not initializing first row or column correctly
• Using greedy instead of DP

4. The following code attempts to solve the Triangle minimum path sum problem using bottom-up DP. Identify the line that contains a subtle bug causing incorrect results on some inputs.

medium

A. for i in range(len(triangle) - 2, -1, -1):

B. dp[j] = triangle[i][j] + min(dp[j + 1], dp[j])

C. dp = triangle[-1][:]

D. return dp[0]

Solution

1. Step 1: Examine dp update line

   The original code has dp[j] = triangle[i][j] + min(dp[j], dp[j + 1]). The buggy code swaps the min arguments to min(dp[j + 1], dp[j]) which is harmless since min is commutative.

2. Step 2: Check for index out of bounds risk

   Accessing dp[j + 1] is safe because dp length is one more than current row length, so no out-of-bounds here.

3. Step 3: Identify subtle bug

   The subtle bug is that dp is updated in place from left to right, which can cause dp[j + 1] to be already updated when used in min, leading to incorrect results.

4. Step 4: Explanation of bug

   Because dp is updated from left to right, dp[j + 1] may have been updated in the current iteration, so min(dp[j], dp[j + 1]) uses a mixed state of dp values.

5. Step 5: Correct fix

   To fix, update dp from right to left so dp[j + 1] is not yet updated when used.

6. Final Answer:

   Option B -> Option B

7. Quick Check:

   Updating dp in wrong order causes subtle bugs [OK]

Hint: Bug usually in dp update line with min or indexing [OK]

Common Mistakes:

• Swapping min arguments (harmless but suspicious)
• Index out of bounds on dp[j+1]
• Updating dp in wrong order causing overwritten values

5. Examine the following buggy code snippet from the Zuma Game solution. Which line contains the subtle bug that causes incorrect minimal moves calculation?

medium

A. Line with 'new_s = s[:i] + s[j:] # BUG: missing recursive removal'

B. Line with 'balls_needed = 3 - (j - i)'

C. Line with 'hand_count[s[i]] -= balls_needed'

D. Line with 'if temp != -1:'

Solution

1. Step 1: Identify role of remove_consecutive

   After insertion and removal, consecutive balls must be recursively removed to handle chain reactions.

2. Step 2: Locate missing recursive removal

   The line creating new_s omits calling remove_consecutive, so chain reactions are not processed, leading to incorrect board states and minimal moves.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Missing recursive removal causes wrong minimal moves [OK]

Hint: Missing recursive removal breaks chain reaction logic [OK]

Common Mistakes:

• Forgetting to call remove_consecutive after insertion
• Incorrectly adjusting hand counts
• Miscomputing balls needed for removal