Leetcode 1039: Minimum Score Triangulation of Polygon
Given an integer array representing the values at the vertices of a convex polygon, you need to triangulate the polygon in such a way that the sum of the products of the values at the vertices of all triangles is minimized. Each triangle must consist of three vertices of the polygon, and the number of triangles will always be n - 2, where n is the number of vertices in the polygon.
Problem
Approach
Steps
Complexity
Input: You are given an array 'values' where each element represents the value at the corresponding vertex of the polygon. The vertices are numbered in a clockwise direction.
Example: Input: values = [2, 5, 3]
Constraints:
• 3 <= n <= 50
• 1 <= values[i] <= 100
Output: Return the minimum possible score achievable by triangulating the polygon. The score is the sum of the products of the values of the vertices in each triangle.
Example: Output: 30
Constraints:
• The score should be minimized by appropriately triangulating the polygon.
Goal: The goal is to minimize the sum of the products of the values of vertices in the triangles formed by triangulating the polygon.
Steps:
• 1. Use dynamic programming (DP) to break the polygon into subproblems by selecting each pair of adjacent vertices and recursively solving for smaller polygons.
• 2. The key idea is to calculate the minimum score by selecting the right set of diagonals to form triangles.
Goal: The number of vertices is between 3 and 50, and the value at each vertex is between 1 and 100. The vertices are unique and numbered in a clockwise order.
Steps:
• 3 <= n <= 50
• 1 <= values[i] <= 100
Assumptions:
• The input polygon is convex and the values at the vertices are distinct.
• Input: Input: values = [1, 2, 3]
• Explanation: This polygon is already a triangle, and its score is simply the product of its three vertices: 1 * 2 * 3 = 6.
• Input: Input: values = [3, 7, 4, 5]
• Explanation: There are two possible triangulations for this polygon: the first has a score of 245 and the second has a score of 144. The minimum score of 144 is the correct result.
Approach: The approach uses dynamic programming to find the optimal triangulation of the polygon that minimizes the score. The DP table stores the minimum triangulation score for each subpolygon.
Observations:
• The problem involves dividing the polygon into triangles by drawing diagonals, and calculating the sum of the products of the values at the vertices of each triangle.
• A dynamic programming approach is suitable here because the problem involves finding the optimal way to triangulate the polygon. We can break the problem down into smaller subproblems.
Steps:
• 1. Define a DP table where dp[i][j] represents the minimum triangulation score for the polygon formed by vertices i to j.
• 2. Iterate over all pairs of vertices and for each pair, try different possible diagonals (subproblems) to divide the polygon into smaller subpolygons.
• 3. The score of a triangle formed by three vertices i, j, and k is the product of the values at these vertices, so we need to calculate the score recursively for smaller subproblems.
Empty Inputs:
• There will always be at least 3 vertices, so no need to handle empty inputs.
Large Inputs:
• For polygons with many vertices (up to 50), the solution should still work efficiently, considering the time complexity.
Special Values:
• Ensure that the DP approach handles cases with larger values (up to 100) efficiently.
Constraints:
• The number of vertices will always be between 3 and 50, ensuring that the problem size is manageable.
int minScoreTriangulation(vector<int>& vals) {
int dp[50][50] = {};
int n = vals.size();
for(int i = n - 1; i >= 0; i--)
for(int j = i + 1; j < n; j++)
for(int k = i + 1; k < j; k++)
dp[i][j] = min(dp[i][j] == 0? INT_MAX : dp[i][j], dp[i][k] + vals[i] * vals[k] * vals[j] + dp[k][j]);
return dp[0][n - 1];
}
1 : Function Definition
int minScoreTriangulation(vector<int>& vals) {
Define the function 'minScoreTriangulation' that calculates the minimum score for triangulating a polygon represented by 'vals'.
2 : DP Array Initialization
int dp[50][50] = {};
Initialize a 2D dynamic programming array 'dp' of size 50x50 to store the minimum triangulation scores for different subproblems.
3 : Array Size Calculation
int n = vals.size();
Determine the size of the input array 'vals', which represents the vertices of the polygon.
4 : Outer Loop
for(int i = n - 1; i >= 0; i--)
Start a loop from the last vertex to the first, iterating over possible starting points for triangulation.
5 : Middle Loop
for(int j = i + 1; j < n; j++)
Start an inner loop from 'i + 1' to 'n', iterating over possible end points for triangulation.
6 : Inner Loop
for(int k = i + 1; k < j; k++)
Start the innermost loop to iterate over all points between 'i' and 'j', representing potential intermediate points for the triangulation.
7 : DP Update
dp[i][j] = min(dp[i][j] == 0? INT_MAX : dp[i][j], dp[i][k] + vals[i] * vals[k] * vals[j] + dp[k][j]);
Update the dynamic programming table by calculating the minimum triangulation score for the subproblem defined by vertices 'i' and 'j'.
8 : Return Statement
return dp[0][n - 1];
Return the minimum score triangulation for the entire polygon, which is stored in dp[0][n - 1].
Best Case: O(n^3)
Average Case: O(n^3)
Worst Case: O(n^3)
Description: The time complexity is O(n^3) because for each pair of vertices (i, j), we check all possible diagonals to calculate the subproblems.
Best Case: O(n^2)
Worst Case: O(n^2)
Description: The space complexity is O(n^2) due to the DP table that stores the results for all subproblems of the polygon.
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