matrix chain multiplication (or matrix Chain order problem, MCOP) is an optimization problem that can be solved use dynamic programming. " check each possible parenthesization (brute force) would necessitate a run-time that is exponential in the number of matrix, which is very slow and impractical for large n. A quicker solution to this problem can be achieved by breaking up the problem into a set of related subproblems.

COMING SOON!

```
import sys
"""
Dynamic Programming
Implementation of Matrix Chain Multiplication
Time Complexity: O(n^3)
Space Complexity: O(n^2)
"""
def MatrixChainOrder(array):
N = len(array)
Matrix = [[0 for x in range(N)] for x in range(N)]
Sol = [[0 for x in range(N)] for x in range(N)]
for ChainLength in range(2, N):
for a in range(1, N - ChainLength + 1):
b = a + ChainLength - 1
Matrix[a][b] = sys.maxsize
for c in range(a, b):
cost = (
Matrix[a][c] + Matrix[c + 1][b] + array[a - 1] * array[c] * array[b]
)
if cost < Matrix[a][b]:
Matrix[a][b] = cost
Sol[a][b] = c
return Matrix, Sol
# Print order of matrix with Ai as Matrix
def PrintOptimalSolution(OptimalSolution, i, j):
if i == j:
print("A" + str(i), end=" ")
else:
print("(", end=" ")
PrintOptimalSolution(OptimalSolution, i, OptimalSolution[i][j])
PrintOptimalSolution(OptimalSolution, OptimalSolution[i][j] + 1, j)
print(")", end=" ")
def main():
array = [30, 35, 15, 5, 10, 20, 25]
n = len(array)
# Size of matrix created from above array will be
# 30*35 35*15 15*5 5*10 10*20 20*25
Matrix, OptimalSolution = MatrixChainOrder(array)
print("No. of Operation required: " + str(Matrix[1][n - 1]))
PrintOptimalSolution(OptimalSolution, 1, n - 1)
if __name__ == "__main__":
main()
```