LeetCode: Minimum Score Triangulation of Polygon

Posted on August 5, 2019July 26, 2020 by braindenny

Minimum Score Triangulation of Polygon

Similar Problems:

Given N, consider a convex N-sided polygon with vertices labelled A[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 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: [1,2,3]
Output: 6
Explanation: The polygon is already triangulated, and the score of the only triangle is 6.

Example 2:

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.

Example 3:

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.

Note:

3 <= A.length <= 50
1 <= A[i] <= 100

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/minimum-score-triangulation-of-polygon
// Basic Ideas: math
//
// Observation: for each triangle, one node can be removed
//   So dp(i) fall back to dp(i-1)
// Observation: There must be one triangle with A[0], A[N-1] + A[i]
//    The sum would be A[0...i], A[i...N-1] + A[0]*A[N-1]*A[i]
//    Loop all the possiblity of i, we can get the result.
//
//   dp(i, j): value from A[i...j]
//
// Notice: there may not exist a central point to divide the polygon
//
// Complexity: Time O(n), Space O(1)
func minScoreTriangulation(A []int) int {
    dp := make([][]int, len(A))
    for i, _ := range dp {
        dp[i] = make([]int, len(A))
    }
    // dp from diagonal line
    for i:=len(A)-1; i>=0; i-- {
        for j:=i; j<len(A); j++ {
            // A[i...j]
            for k:=i+1; k<j; k++ {
                if dp[i][j] == 0 {
                    // set initial value
                    dp[i][j] = 1<<31-1
                }
                v := dp[i][k] + A[i]*A[k]*A[j] + dp[k][j]
                if dp[i][j] > v {
                    dp[i][j] = v
                }
            }
        }
    }
    return dp[0][len(A)-1]
}

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