You have a convex n-sided polygon where each vertex has an integer value. You are given an integer array values where values[i] is the value of the ith vertex in clockwise order.
Polygon triangulation is a process where you divide a polygon into a set of triangles and the vertices of each triangle must also be vertices of the original polygon. Note that no other shapes other than triangles are allowed in the division. This process will result in n - 2 triangles.
You will triangulate the polygon. For each triangle, the weight of that triangle is the product of the values at its vertices. The total score of the triangulation is the sum of these weights over all n - 2 triangles.
Return the minimum possible score that you can achieve with some triangulation of the polygon.
Example 1:
Input: values = [1,2,3]
Output: 6
Explanation: The polygon is already triangulated, and the score of the only triangle is 6.
Example 2:
Input: values = [3,7,4,5]
Output: 144
Explanation: There are two triangulations, with possible scores: 3*7*5 + 4*5*7 = 245, or 3*4*5 + 3*4*7 = 144.
The minimum score is 144.
Example 3:
Input: values = [1,3,1,4,1,5]
Output: 13
Explanation: The minimum score triangulation is 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13.
Constraints:
n == values.length3 <= n <= 501 <= values[i] <= 100This problem is a classic example of Dynamic Programming. The idea is to partition the polygon using dynamic programming where you maintain a dp table, dp[i][j] represents the minimum score to triangulate a polygon starting from vertex i to vertex j.
To solve the problem, we iterate through all possible partitions and calculate their respective scores, updating the dp table to reflect the minimum scores possible for triangulating various sub-polygons.
This C code uses a dynamic programming approach to solve the polygon triangulation problem. The dp table, dp[i][j], stores the minimum score for triangulating the sub-polygon from vertex i to vertex j. We iterate over possible lengths len of the sub-polygons, then calculate the minimum score considering the product of the chosen vertices.
C++
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Time Complexity: O(n^3), where n is the number of vertices in the polygon, due to the three nested loops.
Space Complexity: O(n^2), for storing the dp table.
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