Minimum Score Triangulation Of Polygon - Leetcode Solution

Problem Statement:

Given a convex polygon with n vertices, triangulate the polygon by using n - 2 triangles in such a way that the minimum score is achieved. The score is the sum of the product of the lengths of the sides of each triangle.

Solution:

This problem can be solved using dynamic programming. Let's define dp[i][j] as the minimum score needed to triangulate the polygon with vertices from i to j. The base case is when j - i + 1 == 3 which means there are only three vertices and we can simply connect them to form a triangle and the score would be the product of the three sides.

For the general case, we need to find the minimum score needed to triangulate the polygon using the diagonals between vertex i and vertex k, vertex k and vertex j, and vertex i and vertex j for all k such that i < k < j. We can calculate the score of the resulting triangles using the side lengths of these diagonals and add them to the minimum scores of triangulating the resulting two smaller polygons on each side of the diagonals. The final solution would be the minimum score of all possible triangulations.

Here is the algorithm in more detail:

// Define dp[i][j] as the minimum score for triangulating the polygon from i to j for (int len = 4; len <= n; len++) { // Loop through all possible lengths of polygon sides for (int i = 0; i <= n - len; i++) { // Loop through all starting positions int j = i + len - 1; // Calculate the end position for (int k = i + 1; k < j; k++) { // Loop through all possible diagonals int score = dp[i][k] + dp[k][j] + (polygon[k] - polygon[i]) * (polygon[j] - polygon[k]); // Calculate score of the resulting triangles dp[i][j] = min(dp[i][j], score); // Update the minimum score } } } return dp[0][n - 1]; // Return the minimum score for the entire polygon

Time Complexity: O(n^3)

Space Complexity: O(n^2)

This algorithm has a time complexity of O(n^3) since we are looping through all possible triangulations of the polygon. However, by memoizing the results, we can reduce the time complexity to O(n^2). The space complexity is also O(n^2) since we are using a two-dimensional array to store the results.

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