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. The following code snippet is a near-correct implementation of the Cherry Pickup DP solution. Identify the line containing the subtle bug that can cause incorrect cherry counting or invalid states.

for step in range(1, 2 * (n - 1) + 1):
    for r1 in range(max(0, step - (n - 1)), min(n, step + 1)):
        c1 = step - r1
        if c1 < 0 or c1 >= n:
            continue
        for r2 in range(max(0, step - (n - 1)), min(n, step + 1)):
            c2 = step - r2
            if c2 < 0 or c2 >= n:
                continue
            if grid[r1][c1] == -1 or grid[r2][c2] == -1:
                continue
            candidates = []
            if r1 > 0 and r2 > 0:
                candidates.append(dp[r1 - 1][c1][r2 - 1])
            if r1 > 0 and c2 > 0:
                candidates.append(dp[r1 - 1][c1][r2])
            if c1 > 0 and r2 > 0:
                candidates.append(dp[r1][c1 - 1][r2 - 1])
            if c1 > 0 and c2 > 0:
                candidates.append(dp[r1][c1 - 1][r2])
            best_prev = max(candidates) if candidates else -float('inf')
            if best_prev == -float('inf'):
                continue
            val = best_prev + grid[r1][c1]
            if r1 != r2:
                val += grid[r2][c2]
            dp[r1][c1][r2] = max(dp[r1][c1][r2], val)

medium

A. Lines appending candidates from dp states without verifying step synchronization

B. Line adding grid[r2][c2] to val when r1 != r2

C. Line checking if grid[r1][c1] == -1 or grid[r2][c2] == -1

D. Line initializing dp with -float('inf') for all states

Solution

1. Step 1: Examine candidate state transitions

   The candidates appended correspond to previous dp states, but the code does not ensure both players have taken the same number of steps, leading to invalid states.

2. Step 2: Identify the bug

   Not verifying that the previous states correspond to the same step count causes incorrect cherry counting or index errors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Ensuring step synchronization is critical for correctness [OK]

Hint: Check if previous states align on step count [OK]

Common Mistakes:

• Double counting cherries when players overlap
• Ignoring thorn cells early
• Using 4D dp instead of 3D

2. 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

1. Step 1: Understand dp state meaning

   dp[i][j] should represent the maximum difference in stones the current player can achieve over the opponent.

2. 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.

3. Final Answer:

   Option A -> Option A

4. 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

3. What is the time and space complexity of the optimal space-optimized bottom-up DP solution for the Unique Paths II problem on an m x n grid with obstacles?

medium

A. Time: O(m*n), Space: O(m*n)

B. Time: O(m+n), Space: O(m+n)

C. Time: O(m*n), Space: O(n)

D. Time: O(2^(m+n)), Space: O(m*n)

Solution

1. Step 1: Analyze time complexity

   The algorithm iterates over each cell once in nested loops: O(m*n).

2. Step 2: Analyze space complexity

   Uses a single 1D dp array of length n, so space is O(n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested loops over m and n, dp array size n [OK]

Hint: Nested loops over m and n, dp array size n [OK]

Common Mistakes:

• Confusing recursion stack space with dp array
• Assuming full 2D dp array used
• Mistaking linear time for exponential

4. Suppose the problem is modified so that the matrix can contain '1's and '0's, but you want to find the largest square containing only '1's where you are allowed to flip at most one '0' inside the square to '1'. Which of the following approaches correctly adapts the DP solution to handle this variant efficiently?

hard

A. Maintain two DP arrays: one for squares with zero flips and one for squares with one flip allowed, updating both simultaneously

B. Use the original DP but treat all '0's as '1's to allow flipping implicitly

C. Run the original DP multiple times, each time flipping a different '0' to '1' and taking the max result

D. Use a greedy approach to find the largest square ignoring zeros, then check if flipping one zero inside is possible

Solution

1. Step 1: Understand the variant

   Allowing one flip means tracking squares formed with zero or one zero flipped inside.

2. Step 2: Adapt DP states

   Maintain two DP states per cell: max square size without flips and with one flip used, updating both based on neighbors.

3. Step 3: Reject naive approaches

   Original DP ignores flips; multiple runs are inefficient; greedy ignores DP dependencies.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two DP arrays track flip states efficiently [OK]

Hint: Track DP states with and without flips [OK]

Common Mistakes:

• Ignoring flip state in DP
• Trying multiple brute force runs
• Using greedy without DP

5. Suppose the Strange Printer problem is modified so that the printer can reuse previously printed characters anywhere in the string without reprinting them. Which modification to the DP approach correctly handles this variant?

hard

A. Modify DP to consider non-contiguous intervals and allow merging non-adjacent segments

B. Add a global visited set to track printed characters and skip them

C. No change needed; original DP already accounts for reuse

D. Use a greedy approach printing each unique character once

Solution

1. Step 1: Understand reuse variant

   Reusing printed characters anywhere means intervals are no longer strictly contiguous; DP must consider merging non-adjacent segments.

2. Step 2: Identify correct DP modification

   Original DP assumes contiguous intervals. To handle reuse, DP must be extended to track sets of printed characters or allow merging non-contiguous intervals, which requires more complex state representation.

3. Step 3: Evaluate options

   No change (C) ignores new reuse property. Global visited set (B) is insufficient for interval merging. Greedy (D) fails on complex merges. Only (A) correctly adapts DP for non-contiguous merges.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Reuse breaks contiguous assumption; DP must handle non-adjacent merges [OK]

Hint: Reuse breaks contiguous intervals; DP must track complex merges [OK]

Common Mistakes:

• Assuming original DP handles reuse without changes
• Trying greedy approach