# 1039. Minimum Score Triangulation of Polygon
Given N, consider a convex N-sided polygon with vertices labelled A, A[i], ..., A[N-1] in clockwise order.
Suppose you triangulate the polygon into N-2 triangles. For each triangle, the value of that triangle is the product of the labels of the vertices, and the total score of the triangulation is the sum of these values over all N-2 triangles in the triangulation.
Return the smallest possible total score that you can achieve with some triangulation of the polygon.
Input: [1,2,3] Output: 6 Explanation: The polygon is already triangulated, and the score of the only triangle is 6.
Input: [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.
Input: [1,3,1,4,1,5] Output: 13 Explanation: The minimum score triangulation has score 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13.
Approach 1: DP. Let score[i][j] be the min score formed by polygon A[i] to A[j] (there's an edge from A[i] to A[j]).
Approach 2: memoized calls.
# Code (Python)
def minScoreTriangulation(self, A): # DP: score[i][j] -- min score formed by polygon A[i] to A[j] (there's an edge from A[i] to A[j]) score = [[float('inf') for _ in range(len(A))] for _ in range(len(A))] for distance in range(1, len(A)): for i in range(len(A) - distance): j = i + distance # helps handle the special case for k == i + 1 and j - 1 if distance == 1: score[i][j] = 0 elif distance == 2: # not neccessary for we have distance == 1 score[i][j] = A[i] * A[i+1] * A[i+2] else: # draw triangles with bottom i~j, and vertex k from i + 1 to j - 1 for k in range(i + 1, j): score[i][j] = min(score[i][j], score[i][k] + score[k][j] + A[i] * A[j] * A[k]) return score[len(A)-1]
def minScoreTriangulation(self, A): # cool higher order function for python3 @functools.lru_cache(None) # no max size for cache def score(i, j): # func args i and j should be hashable if i + 1 == j: return 0 else: return min((score(i, k) + score(k, j) + A[i] * A[j] * A[k] for k in range(i + 1, j))) return score(0, len(A) - 1)
# Code (C++)