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

Difficulty: interview

Starter code:

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        

Solutions:
class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        N = len(A)
        dp = [[0]*N for _ in range(N)]
        
        for i in range(N-2, -1, -1):
            for j in range(i+2, N):
                dp[i][j] = min(dp[i][k]+dp[k][j]+A[i]*A[j]*A[k] for k in range(i+1, j))
                
        return dp[0][-1]
                    
                


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        memo = [[0] * len(A) for _ in range(len(A))]
        for i in reversed(range(len(A))):
            for j in range(i + 2, len(A)):
                memo[i][j] = min(memo[i][k] + A[i] * A[k] * A[j] + memo[k][j] for k in range(i + 1, j))
        return memo[0][-1]

from functools import lru_cache

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        # If we pick a side of our polygon, it can form n - 2 triangles. Each such triangle forms 2 sub-polygons. We can analyze n - 2 triangles, and get the minimum score for sub-polygons using the recursion.

        # n = len(A)
        # dp = [[0] * n for _ in range(n)]
        # for d in range(2, n):
        #     for i in range(n - d):
        #         j = i + d
        #         dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        # return dp[0][n - 1]


        @lru_cache(None)
        def dp(i, j):
            return min([dp(i, k) + dp(k, j) + A[i] * A[k] * A[j] for k in range(i + 1, j)] or [0])
        return dp(0, len(A) - 1)


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        @lru_cache(maxsize=None)
        def helper(start, end):
            if start + 1 == end:
                return 0
            
            result = 0
            for k in range(start + 1, end):
                result = min(float('inf') if result == 0 else result,
                            helper(start, k) +
                            A[start] * A[k] * A[end] +
                            helper(k, end))
            return result
        return helper(0, len(A) - 1)

class Solution:
    def minScoreTriangulation(self, a: List[int]) -> int:
        n = len(a)
        dp = [[float('inf')]*n for _ in range(n)]
        for l in range(2, n):
            for i in range(n-l):
                j = i + l                
                for k in range(i+1, j):
                    dp[i][j] = min(dp[i][j], (dp[i][k] if k >= i + 2 else 0) + a[i]*a[k]*a[j]  + (dp[k][j] if j >= k + 2 else 0))
        return dp[0][n-1]

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[float('inf')]*n for _ in range(n)]
        for l in range(2, n):
            for i in range(n-l):
                j = i + l                
                for k in range(i+1, j):
                    dp[i][j] = min(dp[i][j], (dp[i][k] if k >= i + 2 else 0) + A[i]*A[k]*A[j]  + (dp[k][j] if j >= k + 2 else 0))
        return dp[0][n-1]

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0] * n for i in range(n)]
        for d in range(2, n):
            for i in range(n - d):
                j = i + d
                dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        return dp[0][n - 1]

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        @lru_cache(None)
        def dfs(i, j):
            if j < i + 2:
                return 0
            result = 10 ** 9
            for k in range(i + 1, j):
                result = min(result, A[i] * A[k] * A[j] + dfs(i, k) + dfs(k, j))
            return result
        n = len(A)
        return dfs(0, n - 1)

class Solution:
#     https://www.youtube.com/watch?v=eKkXU3uu2zk
    def minScoreTriangulation(self, a: List[int]) -> int:
        n = len(a)
        dp = [[(float('inf') if i-j>2 else 0)  for i in range(len(a))]for j in range(len(a)) ]
        for i in range(2, len(dp)):
            dp[i-2][i]=a[i-2]*a[i-1]*a[i]

        for l in range(3, len(dp)):
            for i in range(0, n-l):
                j= l+i
                if j>n:
                    break
                for k in range(i+1,j):
                    dp[i][j]=min(dp[i][j], a[k]*a[i]*a[j] + dp[i][k]+dp[k][j])
                    
        return dp[0][-1]
                    
                


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0] * n for i in range(n)]
        
        for d in range(2, n):
            for i in range(n - d):
                j = i + d
                dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        return dp[0][n - 1]

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        N = len(A)
        dp = [
            [0 for j in range(N)] for i in range(N)
        ]
        
        for l in range(3, N + 1):
            for i in range(N - l + 1):
                j = i + l - 1
                for k in range(i + 1, j):
                    val = dp[i][k] + dp[k][j] + A[i] * A[k] * A[j]
                    if not dp[i][j] or dp[i][j] > val:
                        dp[i][j] = val
        return dp[0][N - 1]


class Solution:
    def minScoreTriangulation(self, a: List[int]) -> int:
        n = len(a)
        dp = [[(float('inf') if i-j>2 else 0)  for i in range(len(a))]for j in range(len(a)) ]
        for i in range(2, len(dp)):
            dp[i-2][i]=a[i-2]*a[i-1]*a[i]

        for l in range(3, len(dp)):
            for i in range(0, n-l):
                j= l+i
                if j>n:
                    break
                for k in range(i+1,j):
                    dp[i][j]=min(dp[i][j], a[k]*a[i]*a[j] + dp[i][k]+dp[k][j])
                    
        return dp[0][-1]
                    
                


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        def dfs(nums, i, j):
            if j-i <= 1:
                return 0
            
            if dp[i][j] != 0:
                return dp[i][j]
            
            ans = sys.maxsize
            for k in range(i+1, j):    # burst ith element
                ans = min(ans, dfs(nums, i, k) + dfs(nums, k, j) + nums[i]*nums[k]*nums[j])
                
            dp[i][j] = ans
            return ans
        
        n = len(A)
        dp = [[0] * n for _ in range(n)]
        total = dfs(A, 0, len(A)-1)
        return total

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0 for i in range(n)]for j in range(n)]
        for length in range(2,n):
            for i in range(n-length):
                j = i+length
                dp[i][j] = math.inf
                for k in range(i+1,j):
                    dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]+(A[i]*A[k]*A[j]))
        return sum(A) if dp[0][n-1]==math.inf else dp[0][n-1]

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)
        
                                                                                              
        
        
        
        


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        
        table=[[-1 for i in range(len(A))] for j in range(len(A))]
    
        def recursion(x=0,y=len(A)-1,A:list=A):
            if y-x<=1:
                return 0
            
            if table[x][y]!=-1:
                return table[x][y]
            else:
                m=float('inf')
                for i in range(x+1,y):
                    m=min(m,recursion(x,i,A)+recursion(i,y,A)+A[x]*A[y]*A[i])
                table[x][y]=m
                return m
        
        return recursion()


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)

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        
        def get(l, r):
            if r - l == 1:
                return 0
            if (l, r) in d:
                return d[(l, r)]
            res = min([A[l] * A[r] * A[i] + get(l, i) + get(i, r) for i in range(l + 1, r)])
            d[(l, r)] = res
            return res
        
        d = {}
        return get(0, len(A) - 1)


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        if len(A) == 3:
            return A[0]*A[1]*A[2]
            
        n = len(A)
        dp = [[float('inf') for j in range(n)] for i in range(n)]
        def helper(i, j):
            if j-i<=1:
                return 0
            
            if dp[i][j] != float('inf'):
                return dp[i][j]
            
            res = float('inf')
            for k in range(i+1, j):
                res = min(res, 
                          helper(i, k) + A[i]*A[k]*A[j] + helper(k, j))
                
            dp[i][j] = res
            return dp[i][j] 
        
        return helper(0, n-1)

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        N = len(A)
        if N == 3: return A[0]*A[1]*A[2]
        
        def solver(s,t, A):
            if t-s+1 < 3: return 0
            
            if dp[s][t] != float('inf'): return dp[s][t]
            

            for i in range(s+1, t):
                dp[s][t] = min(dp[s][t], solver(s,i, A) + A[s]*A[i]*A[t] + solver(i,t, A))
        
            return dp[s][t]
        
        dp = []
        for i in range(N):
            dp.append([float('inf')]*N)
        
        
        return solver(0,N-1,A)
        


class Solution:
    def minScoreTriangulation_v2(self, A: List[int]) -> int:
      length = len(A)
      DP = [[float('inf')] * length for _ in range(len(A))]
      for e in range(2, length):
        for i in range(length - e):
          j = i + e
          for k in range(i+1, j):
            left_point = DP[i][k] if k >= i + 2 else 0
            right_point = DP[k][j] if j >= k + 2 else 0
            DP[i][j] = min(DP[i][j], left_point + A[i] * A[j] * A[k] + right_point)
      return DP[0][len(A)-1]

    def minScoreTriangulation(self, A: List[int]) -> int:
      length = len(A)
      memo = [[float('inf')] * length for _ in range(len(A))]
        
      def memo_dp(start, end):
        if end - start < 2:
          return 0
        if end - start == 2:
          memo[start][end] = A[start] * A[start+1] * A[end]
          return memo[start][end]
        if memo[start][end] != float('inf'):
          return memo[start][end]
        
        val = memo[start][end]
        for i in range(start+1, end):
            val = min(val, A[start] * A[i] * A[end] + memo_dp(start, i) + memo_dp(i, end))
        memo[start][end] = val
        return memo[start][end]
        
      memo_dp(0, length-1)
      return memo[0][length-1]

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        def dfs(nums, i, j):
            if j-i <= 1:
                return 0
            
            key = tuple([i, j])
            if key in dp:
                return dp[key]
            
            ans = sys.maxsize
            for k in range(i+1, j):    # burst ith element
                ans = min(ans, dfs(nums, i, k) + dfs(nums, k, j) + nums[i]*nums[k]*nums[j])
                
            dp[key] = ans
            return ans
        
        dp = {}
        total = dfs(A, 0, len(A)-1)
        return total

def tri_score(a1,a2,a3):
    return a1*a2*a3

class Solution:
    def __init__(self):
        self.memory = []
    def minScoreTriangulation(self, A) -> int:
        n=len(A)
        self.memory=[[None for _ in range(n)] for _ in range(n)]
        return self.minScoreTriangulationPart(A,0,len(A)-1)

    def minScoreTriangulationPart(self, A, start: int, end: int) -> int:
        if end-start<2:
            return 0
        elif self.memory[start][end]:
            return self.memory[start][end]
        elif end-start==2:
            res = tri_score(A[start],A[start+1],A[start+2])
            self.memory[start][end] = res
            return res
        res = min(tri_score(A[start],A[end],A[k])+self.minScoreTriangulationPart(A,start,k)+self.minScoreTriangulationPart(A,k,end) for k in range(start+1,end))
        self.memory[start][end] = res
        return res

def tri_score(a1,a2,a3):
    return a1*a2*a3

class Solution:
    def __init__(self):
        self.memory = []
    def minScoreTriangulation(self, A) -> int:
        n=len(A)
        self.memory=[[None for _ in range(j+1)] for j in range(n)]
        return self.minScoreTriangulationPart(A,0,len(A)-1)

    def minScoreTriangulationPart(self, A, start: int, end: int) -> int:
        if end-start<2:
            return 0
        elif self.memory[end][start]:
            return self.memory[end][start]
        elif end-start==2:
            res = tri_score(A[start],A[start+1],A[start+2])
            self.memory[end][start] = res
            return res
        res = min(tri_score(A[start],A[end],A[k])+self.minScoreTriangulationPart(A,start,k)+self.minScoreTriangulationPart(A,k,end) for k in range(start+1,end))
        self.memory[end][start] = res
        return res

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)


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        dp = {}
        def recursive_helper(left, right):
            if left > right:
                return 0
            # B/C for when invalidation
            
            pair = (left, right)
            if pair in dp:
                return dp[pair]
            dp[pair] = float('inf')
            for k in range(left+1, right):
                dp[pair] = min(dp[pair], recursive_helper(left, k) + recursive_helper(k, right) + A[left] * A[k] * A[right])
            
            if dp[pair] == float('inf'):
                dp[pair] = 0
            # Fix k, and try to traingulate on left side and right side
            
            #dp[pair] = min([recursive_helper(left, k) + recursive_helper(k, right) + A[left] * A[k] * A[right] for k in range(left + 1, right)] or [0])
            
            return dp[pair]
        
        return recursive_helper(0, len(A) - 1)
        
        # COmpelxity is O(n^3), Space is O(n^2)


class Solution:
    
    
    def minScoreTriangulation(self, A: List[int]) -> int:
        
        #dp = [[0 for i in range(50)] for j in range(50)]
        dp = dict()
        def helper(i, j):
            if (i, j) in dp:
                return dp[(i, j)]
            
            if j-i <= 1:
                dp[(i, j)] = 0
                return dp[(i, j)]
            
            minval = float('inf')
            for k in range(i+1, j):
                #print(i, k, j, A[i]*A[j]*A[k])
                val = helper(i, k) + helper(k, j) + A[i]*A[j]*A[k]
                minval = min(minval, val)
            dp[(i, j)] = minval
            #print(i, j, minval)
            return dp[(i, j)]
        
        return helper(0, len(A)-1)
    
    def minScoreTriangulation3(self, A: List[int]) -> int:
        
        if len(A) < 3:
            return 0
        
        if len(A) == 3:
            return A[0]*A[1]*A[2]
        
        def multi_array(B, j):
            #print(B[j%len(B)],B[(j+1)%len(B)],B[(j+2)%len(B)])
            return B[j%len(B)]*B[(j+1)%len(B)]*B[(j+2)%len(B)]
        
        # 
        sum1, sum2 = 0, 0
        A_copy = A.copy()
        sum1 += multi_array(A, 0)
        A_copy.pop(1)
        sum1 += self.minScoreTriangulation(A_copy)
        
        sum1 += multi_array(A, len(A)-1)
        A.pop(0)
        sum2 += self.minScoreTriangulation(A_copy)
        
        return min(sum1, sum2)
    
    def minScoreTriangulation2(self, A: List[int]) -> int:
        
        if len(A) < 3:
            return 0
        
        if len(A) == 3:
            return A[0]*A[1]*A[2]
        
        def multi_array(B, j):
            print(B[j%len(B)],B[(j+1)%len(B)],B[(j+2)%len(B)])
            return B[j%len(B)]*B[(j+1)%len(B)]*B[(j+2)%len(B)]
        
        # 
        sum1, sum2 = 0, 0
        next_A = []
        for i in range(len(A)//2):
            sum1 += multi_array(A, i*2)
            print(i, sum1, multi_array(A, i*2))
            next_A.append(A[(i*2)%len(A)])
           
        #print(i, i*2, )
        if (i*2+2)%len(A) != 0:
            next_A.append(A[(i*2+2)%len(A)])
            
        print(next_A, sum1)
        sum1 += self.minScoreTriangulation(next_A)
            
        print(sum1, '\
')
        val = A.pop(0)
        A.append(val)
        next_A = []
        for i in range(len(A)//2):
            
            sum2 += multi_array(A, i*2)
            print(i, sum2)
            next_A.append(A[(i*2)%len(A)])
        if (i*2+2)%len(A) != 0:
            next_A.append(A[(i*2+2)%len(A)])
        print(next_A, sum2)
        sum2 += self.minScoreTriangulation(next_A)
        
        return min(sum1, sum2)

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        
        
        def helper(i, j):
            key = (i, j)
            if key in d:
                return d[key]
            if j - i < 2:
                return 0
            ans = float('inf')
            
            for k in range(i+1, j):
                
                lans = helper(i, k)
                rans = helper(k , j)
                ans = min(ans, lans + rans + A[i]*A[j]*A[k])
            d[key] = ans
            return d[key]
        d = dict()
        return helper(0, len(A) - 1)
            


from functools import lru_cache

class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0] * n for _ in range(n)]
        for d in range(2, n):
            for i in range(n - d):
                j = i + d
                dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        return dp[0][n - 1]


class Solution:
    def minScoreTriangulation(self, A: List[int]) -> int:
        n = len(A)
        dp = [[0] * n for i in range(n)]
        for d in range(2, n):
            for i in range(n - d):
                j = i + d
                dp[i][j] = min(dp[i][k] + dp[k][j] + A[i] * A[j] * A[k] for k in range(i + 1, j))
        return dp[0][n - 1]