Update matrix_chain_order calculation with more details and test. (TheAlgorithms#12759)

Author: mindaugl

dynamic_programming/matrix_chain_order.py

@@ -5,13 +5,19 @@
 Implementation of Matrix Chain Multiplication
 Time Complexity: O(n^3)
 Space Complexity: O(n^2)
+
+Reference: https://en.wikipedia.org/wiki/Matrix_chain_multiplication
 """
 
 
-def matrix_chain_order(array):
+def matrix_chain_order(array: list[int]) -> tuple[list[list[int]], list[list[int]]]:
+    """
+    >>> matrix_chain_order([10, 30, 5])
+    ([[0, 0, 0], [0, 0, 1500], [0, 0, 0]], [[0, 0, 0], [0, 0, 1], [0, 0, 0]])
+    """
     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)]
+    matrix = [[0 for _ in range(n)] for _ in range(n)]
+    sol = [[0 for _ in range(n)] for _ in range(n)]
 
     for chain_length in range(2, n):
         for a in range(1, n - chain_length + 1):
@@ -28,26 +34,33 @@ def matrix_chain_order(array):
     return matrix, sol
 
 
-# Print order of matrix with Ai as Matrix
-def print_optiomal_solution(optimal_solution, i, j):
+def print_optimal_solution(optimal_solution: list[list[int]], i: int, j: int):
+    """
+    Print order of matrix with Ai as Matrix.
+    """
+
     if i == j:
         print("A" + str(i), end=" ")
     else:
         print("(", end=" ")
-        print_optiomal_solution(optimal_solution, i, optimal_solution[i][j])
-        print_optiomal_solution(optimal_solution, optimal_solution[i][j] + 1, j)
+        print_optimal_solution(optimal_solution, i, optimal_solution[i][j])
+        print_optimal_solution(optimal_solution, optimal_solution[i][j] + 1, j)
         print(")", end=" ")
 
 
 def main():
+    """
+    Size of matrix created from array [30, 35, 15, 5, 10, 20, 25] will be:
+    30*35 35*15 15*5 5*10 10*20 20*25
+    """
+
     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, optimal_solution = matrix_chain_order(array)
 
     print("No. of Operation required: " + str(matrix[1][n - 1]))
-    print_optiomal_solution(optimal_solution, 1, n - 1)
+    print_optimal_solution(optimal_solution, 1, n - 1)
 
 
 if __name__ == "__main__":