Algorithm
Problem Name: 1039. Minimum Score Triangulation of Polygon
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 (i.e., clockwise order).
You will triangulate the polygon into n - 2 triangles. For each triangle, the value of that triangle is the product of the values of its 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.
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 has score 1*1*3 + 1*1*4 + 1*1*5 + 1*1*1 = 13.
Constraints:
n == values.length3 <= n <= 501 <= values[i] <= 100
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int minScoreTriangulation(int[] A) {
int n = A.length;
int[][] dp = new int[n][n];
for (int d = 2; d < n; ++d) {
for (int i = 0; i + d < n; ++i) {
int j = i + d;
dp[i][j] = Integer.MAX_VALUE;
for (int k = i + 1; k < j; ++k) {
dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j] + A[i] * A[j] * A[k]);
}
}
}
return dp[0][n - 1];
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const minScoreTriangulation = function(A) {
if(A.length <= 2) return 0
if(A.length === 3) return A[0] * A[1] * A[2]
return chk(A, A.length)
};
function cost(points, i, j, k) {
let p1 = points[i],
p2 = points[j],
p3 = points[k]
return p1 * p2 * p3
}
function chk(points, n) {
if (n < 3) return 0
const table = Array.from({ length: n }, (> => new Array(n).fill(0))
for (let gap = 0; gap < n; gap++) {
for (let i = 0, j = gap; j < n; i++, j++) {
if (j < i + 2) table[i][j] = 0
else {
table[i][j] = Number.MAX_VALUE
for (let k = i + 1; k < j; k++) {
let val = table[i][k] + table[k][j] + cost(points, i, j, k)
if (table[i][j] > val) table[i][j] = val
}
}
}
}
return table[0][n - 1]
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def minScoreTriangulation(self, A: List[int]) -> int:
memo = {}
def dp(i, j):
if (i, j) not in memo:
memo[i, j] = min([dp(i, k) + dp(k, j) + A[i] * A[j] * A[k] for k in range(i + 1, j)] or [0])
return memo[i, j]
return dp(0, len(A) - 1)
Input
Output