How do you solve Minimum Score Triangulation of Polygon on LeetCode?

Minimum Score Triangulation of Polygon (LeetCode #1039) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n²) space complexity. Key insight: Dynamic programming helps break down the problem into smaller, manageable subproblems.

What is the brute force solution for Minimum Score Triangulation of Polygon?

The brute force approach for Minimum Score Triangulation of Polygon is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible ways to triangulate the polygon and calculating the score for each triangulation. This method guarantees finding the minimum score but is inefficient due to the exponential number of triangulations. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Minimum Score Triangulation of Polygon?

Minimum Score Triangulation of Polygon uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Minimum Score Triangulation of Polygon?

Explain your thought process clearly while solving the problem. Discuss the trade-offs between brute force and optimal solutions. Practice dynamic programming problems to become familiar with the approach.

What are common mistakes when solving Minimum Score Triangulation of Polygon?

Not considering all possible triangles during triangulation. Failing to properly initialize the DP table or mismanaging indices.

#1039

Minimum Score Triangulation of Polygon

Medium

ArrayDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n²)

💡

Intuition

Time O(n³)Space O(n²)

The optimal solution uses dynamic programming to build up solutions to subproblems, storing results to avoid redundant calculations. This significantly reduces the number of computations needed compared to brute force.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the minimum score to triangulate the polygon from vertex i to vertex j.
2Step 2: Iterate over the lengths of the segments and fill the DP table using previously computed values.
3Step 3: Return dp[0][n-1] which contains the minimum score for the entire polygon.

solution.py14 lines

1# Full working Python code
2
3def minScoreTriangulation(values):
4    n = len(values)
5    dp = [[0] * n for _ in range(n)]
6    for length in range(2, n):
7        for i in range(n - length):
8            j = i + length
9            dp[i][j] = float('inf')
10            for k in range(i + 1, j):
11                score = values[i] * values[k] * values[j] + dp[i][k] + dp[k][j]
12                dp[i][j] = min(dp[i][j], score)
13    return dp[0][n - 1]
14

ℹ

Complexity note: The time complexity is O(n³) due to the nested loops iterating over the vertices and the pairs of vertices being considered for triangulation. The space complexity is O(n²) because we maintain a DP table of size n x n.

1Dynamic programming helps break down the problem into smaller, manageable subproblems.
2Storing intermediate results prevents redundant calculations, making the solution efficient.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.