Difficulty : Medium
Related Topics : DynamicProgramming
Given
N
, consider a convexN
-sided polygon with vertices labelledA[0], 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 allN-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.
3 <= A.length <= 50
1 <= A[i] <= 100
- mine
- Java
- DP
Runtime: 3 ms, faster than 71.28%, Memory Usage: 37.6 MB, less than 28.99% of Java online submissions
// O(N^3)time // O(N^2)space public int minScoreTriangulation(int[] A) { int n = A.length; int[][] dp = new int[n][n]; for(int len = 2; len < n; len++){ for(int i = 0; i < n - len; i++){ int j = i + len; 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[k] * A[j]); } } } return dp[0][n - 1]; }
- DP
- Java
- the most votes
- DP
Runtime: 3 ms, faster than 71.28%, Memory Usage: 37.6 MB, less than 28.99% of Java online submissions
// O(N^3)time // O(N^2)space 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]; }