Printing brackets in Matrix Chain Multiplication Problem

Prerequisite : Dynamic Programming | Set 8 (Matrix Chain Multiplication)

Given a sequence of matrices, find the most efficient way to multiply these matrices together. The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications.

We have many options to multiply a chain of matrices because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B, C, and D, we would have: 

(ABC)D = (AB)(CD) = A(BCD) = ....

However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. For example, suppose A is a 10 × 30 matrix, B is a 30 × 5 matrix, and C is a 5 × 60 matrix. Then,  

(AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.

Clearly the first parenthesization requires less number of operations.

Given an array p[] which represents the chain of matrices such that the ith matrix Ai is of dimension p[i-1] x p[i]. We need to write a function MatrixChainOrder() that should return the minimum number of multiplications needed to multiply the chain.

Input:  p[] = {40, 20, 30, 10, 30}  
Output: Optimal parenthesization is  ((A(BC))D)
        Optimal cost of parenthesization is 26000
There are 4 matrices of dimensions 40x20, 20x30, 
30x10 and 10x30. Let the input 4 matrices be A, B, 
C and D.  The minimum number of  multiplications are 
obtained by putting parenthesis in following way
(A(BC))D --> 20*30*10 + 40*20*10 + 40*10*30

Input: p[] = {10, 20, 30, 40, 30} 
Output: Optimal parenthesization is (((AB)C)D)
        Optimal cost of parenthesization is 30000
There are 4 matrices of dimensions 10x20, 20x30, 
30x40 and 40x30. Let the input 4 matrices be A, B, 
C and D.  The minimum number of multiplications are 
obtained by putting parenthesis in following way
((AB)C)D --> 10*20*30 + 10*30*40 + 10*40*30

Input: p[] = {10, 20, 30}  
Output: Optimal parenthesization is (AB)
        Optimal cost of parenthesization is 6000
There are only two matrices of dimensions 10x20 
and 20x30. So there is only one way to multiply 
the matrices, cost of which is 10*20*30

This problem is mainly an extension of previous post. In the previous post, we have discussed algorithm for finding optimal cost only. Here we need print parenthesization also.

The idea is to store optimal break point for every subexpression (i, j) in a 2D array bracket[n][n]. Once we have bracket array us constructed, we can print parenthesization using below code. 

// Prints parenthesization in subexpression (i, j)
printParenthesis(i, j, bracket[n][n], name)
{
    // If only one matrix left in current segment
    if (i == j)
    {
        print name;
        name++;
        return;
    }

    print "(";

    // Recursively put brackets around subexpression
    // from i to bracket[i][j].
    printParenthesis(i, bracket[i][j], bracket, name);

    // Recursively put brackets around subexpression
    // from bracket[i][j] + 1 to j.
    printParenthesis(bracket[i][j]+1, j, bracket, name);

    print ")";
}

Below is the implementation of the above steps.

C++

// C++ program to print optimal parenthesization
// in matrix chain multiplication.
#include <bits/stdc++.h>
using namespace std;

// Function for printing the optimal
// parenthesization of a matrix chain product
void printParenthesis(int i, int j, int n, int* bracket,
                      char& name)
{
    // If only one matrix left in current segment
    if (i == j) {
        cout << name++;
        return;
    }

    cout << "(";

    // Recursively put brackets around subexpression
    // from i to bracket[i][j].
    // Note that "*((bracket+i*n)+j)" is similar to
    // bracket[i][j]
    printParenthesis(i, *((bracket + i * n) + j), n,
                     bracket, name);

    // Recursively put brackets around subexpression
    // from bracket[i][j] + 1 to j.
    printParenthesis(*((bracket + i * n) + j) + 1, j, n,
                     bracket, name);
    cout << ")";
}

// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
// Please refer below article for details of this
// function
// https://goo.gl/k6EYKj
void matrixChainOrder(int p[], int n)
{
    /* For simplicity of the program, one extra
       row and one extra column are allocated in
        m[][]. 0th row and 0th column of m[][]
        are not used */
    int m[n][n];

    // bracket[i][j] stores optimal break point in
    // subexpression from i to j.
    int bracket[n][n];

    /* m[i,j] = Minimum number of scalar multiplications
    needed to compute the matrix A[i]A[i+1]...A[j] =
    A[i..j] where dimension of A[i] is p[i-1] x p[i] */

    // cost is zero when multiplying one matrix.
    for (int i = 1; i < n; i++)
        m[i][i] = 0;

    // L is chain length.
    for (int L = 2; L < n; L++)
    {
        for (int i = 1; i < n - L + 1; i++)
        {
            int j = i + L - 1;
            m[i][j] = INT_MAX;
            for (int k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                int q = m[i][k] + m[k + 1][j]
                        + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                {
                    m[i][j] = q;

                    // Each entry bracket[i,j]=k shows
                    // where to split the product arr
                    // i,i+1....j for the minimum cost.
                    bracket[i][j] = k;
                }
            }
        }
    }

    // The first matrix is printed as 'A', next as 'B',
    // and so on
    char name = 'A';

    cout << "Optimal Parenthesization is : ";
    printParenthesis(1, n - 1, n, (int*)bracket, name);
    cout << "\nOptimal Cost is : " << m[1][n - 1];
}

// Driver code
int main()
{
    int arr[] = { 40, 20, 30, 10, 30 };
    int n = sizeof(arr) / sizeof(arr[0]);
    matrixChainOrder(arr, n);
    return 0;
}

Java

// Java program to print optimal parenthesization
// in matrix chain multiplication.
import java.io.*;
import java.util.*;
class GFG {
    static char name;

    // Function for printing the optimal
    // parenthesization of a matrix chain product
    static void printParenthesis(int i, int j, int n,
                                 int[][] bracket)
    {

        // If only one matrix left in current segment
        if (i == j) {
            System.out.print(name++);
            return;
        }
        System.out.print("(");

        // Recursively put brackets around subexpression
        // from i to bracket[i][j].
        // Note that "*((bracket+i*n)+j)" is similar to
        // bracket[i][j]
        printParenthesis(i, bracket[i][j], n, bracket);

        // Recursively put brackets around subexpression
        // from bracket[i][j] + 1 to j.
        printParenthesis(bracket[i][j] + 1, j, n, bracket);
        System.out.print(")");
    }

    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    // Please refer below article for details of this
    // function
    // https://goo.gl/k6EYKj
    static void matrixChainOrder(int p[], int n)
    {
        /*
             * For simplicity of the program,
             one extra row and one extra column are
             * allocated in m[][]. 0th row and
             0th column of m[][] are not used
             */
        int[][] m = new int[n][n];

        // bracket[i][j] stores optimal break point in
        // subexpression from i to j.
        int[][] bracket = new int[n][n];

        /*
             * m[i,j] = Minimum number of scalar
             multiplications needed to compute the
             * matrix A[i]A[i+1]...A[j] = A[i..j] where
             dimension of A[i] is p[i-1] x p[i]
             */

        // cost is zero when multiplying one matrix.
        for (int i = 1; i < n; i++)
            m[i][i] = 0;

        // L is chain length.
        for (int L = 2; L < n; L++) {
            for (int i = 1; i < n - L + 1; i++) {
                int j = i + L - 1;
                m[i][j] = Integer.MAX_VALUE;
                for (int k = i; k <= j - 1; k++) {

                    // q = cost/scalar multiplications
                    int q = m[i][k] + m[k + 1][j]
                            + p[i - 1] * p[k] * p[j];
                    if (q < m[i][j]) {
                        m[i][j] = q;

                        // Each entry bracket[i,j]=k shows
                        // where to split the product arr
                        // i,i+1....j for the minimum cost.
                        bracket[i][j] = k;
                    }
                }
            }
        }

        // The first matrix is printed as 'A', next as 'B',
        // and so on
        name = 'A';
        System.out.print("Optimal Parenthesization is : ");
        printParenthesis(1, n - 1, n, bracket);
        System.out.print("\nOptimal Cost is : "
                         + m[1][n - 1]);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 40, 20, 30, 10, 30 };
        int n = arr.length;
        matrixChainOrder(arr, n);
    }
}

// This code is contributed by sanjeev2552

Python3

# Python3 program to print optimal parenthesization
# in matrix chain multiplication.
name = 0;

# Function for printing the optimal
# parenthesization of a matrix chain product
def printParenthesis(i , j, n, bracket):

    global name

    # If only one matrix left in current segment
    if (i == j):

        print(name, end = "");
        name = chr(ord(name) + 1)
        return;

    print("(", end = "");

    # Recursively put brackets around subexpression
    # from i to bracket[i][j].
    # Note that "*((bracket+i*n)+j)" is similar to
    # bracket[i][j]
    printParenthesis(i, bracket[i][j], n, bracket);

    # Recursively put brackets around subexpression
    # from bracket[i][j] + 1 to j.
    printParenthesis(bracket[i][j] + 1, j, n, bracket);
    print(")", end = "");

# Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
# Please refer below article for details of this
# function
# https:#goo.gl/k6EYKj
def matrixChainOrder( p , n):

    global name

    '''
         * For simplicity of the program,
         one extra row and one extra column are
         * allocated in m. 0th row and
         0th column of m are not used
         '''
    m = [ [0 for _ in range(n)] for _ in range(n)]

    # bracket[i][j] stores optimal break point in
    # subexpression from i to j.
    bracket = [ [0 for _ in range(n)] for _ in range(n)]

    '''
         * m[i,j] = Minimum number of scalar
         multiplications needed to compute the
         * matrix A[i]A[i+1]...A[j] = A[i..j] where
         dimension of A[i] is p[i-1] x p[i]
         '''

    # cost is zero when multiplying one matrix.
    for  i in range(1, n):
        m[i][i] = 0;

    # L is chain length.
    for L in range(2, n):

        for i in range(1, n - L + 1):
            j = i + L - 1;
            m[i][j] = 10 ** 8;
            for k in range(i, j):

                # q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
                if (q < m[i][j]) :

                    m[i][j] = q;

                # Each entry bracket[i,j]=k shows
                # where to split the product arr
                # i,i+1....j for the minimum cost.
                bracket[i][j] = k;

    # The first matrix is printed as 'A', next as 'B',
    # and so on
    name = 'A';
    print("Optimal Parenthesization is : ");
    printParenthesis(1, n - 1, n, bracket);
    print("\nOptimal Cost is :", m[1][n - 1]);

# Driver code
arr = [ 40, 20, 30, 10, 30 ];
n = len(arr);
matrixChainOrder(arr, n);

# This code is contributed by phasing17

C#

// C# program to print optimal parenthesization
// in matrix chain multiplication.
using System;

class GFG{

static char name;

// Function for printing the optimal
// parenthesization of a matrix chain product
static void printParenthesis(int i, int j,
                             int n, int[,] bracket)
{

    // If only one matrix left in current segment
    if (i == j)
    {
        Console.Write(name++);
        return;
    }
    Console.Write("(");

    // Recursively put brackets around subexpression
    // from i to bracket[i,j].
    // Note that "*((bracket+i*n)+j)" is similar to
    // bracket[i,j]
    printParenthesis(i, bracket[i, j], n, bracket);

    // Recursively put brackets around subexpression
    // from bracket[i,j] + 1 to j.
    printParenthesis(bracket[i, j] + 1, j, n, bracket);
    Console.Write(")");
}

// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
// Please refer below article for details of this
// function
// https://goo.gl/k6EYKj
static void matrixChainOrder(int []p, int n)
{

    /*
    * For simplicity of the program,
    one extra row and one extra column are
    * allocated in m[,]. 0th row and
    0th column of m[,] are not used
    */
    int[,] m = new int[n, n];

    // bracket[i,j] stores optimal break point in
    // subexpression from i to j.
    int[,] bracket = new int[n, n];

    /*
    * m[i,j] = Minimum number of scalar
    multiplications needed to compute the
    * matrix A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i]
    */

    // cost is zero when multiplying one matrix.
    for(int i = 1; i < n; i++)
        m[i, i] = 0;

    // L is chain length.
    for(int L = 2; L < n; L++)
    {
        for(int i = 1; i < n - L + 1; i++)
        {
            int j = i + L - 1;
            m[i, j] = int.MaxValue;
            for (int k = i; k <= j - 1; k++)
            {

                // q = cost/scalar multiplications
                int q = m[i, k] + m[k + 1, j] +
                       p[i - 1] * p[k] * p[j];

                if (q < m[i, j])
                {
                    m[i, j] = q;

                    // Each entry bracket[i,j]=k shows
                    // where to split the product arr
                    // i,i+1....j for the minimum cost.
                    bracket[i, j] = k;
                }
            }
        }
    }

    // The first matrix is printed as 'A', next as 'B',
    // and so on
    name = 'A';
    Console.Write("Optimal Parenthesization is : ");
    printParenthesis(1, n - 1, n, bracket);
    Console.Write("\nOptimal Cost is : " + m[1, n - 1]);
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 40, 20, 30, 10, 30 };
    int n = arr.Length;

    matrixChainOrder(arr, n);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// javascript program to print optimal parenthesization
// in matrix chain multiplication.

  var name=0;

  // Function for printing the optimal
  // parenthesization of a matrix chain product
  function printParenthesis(i , j, n, bracket)
  {

    // If only one matrix left in current segment
    if (i == j)
    {
      document.write(name++);
      return;
    }
    document.write("(");

    // Recursively put brackets around subexpression
    // from i to bracket[i][j].
    // Note that "*((bracket+i*n)+j)" is similar to
    // bracket[i][j]
    printParenthesis(i, bracket[i][j], n, bracket);

    // Recursively put brackets around subexpression
    // from bracket[i][j] + 1 to j.
    printParenthesis(bracket[i][j] + 1, j, n, bracket);
    document.write(")");
  }

  // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
  // Please refer below article for details of this
  // function
  // https://goo.gl/k6EYKj
  function matrixChainOrder( p , n)
  {
    /*
         * For simplicity of the program,
         one extra row and one extra column are
         * allocated in m. 0th row and
         0th column of m are not used
         */
    var m = Array(n).fill(0).map(x => Array(n).fill(0));

    // bracket[i][j] stores optimal break point in
    // subexpression from i to j.
    var bracket = Array(n).fill(0).map(x => Array(n).fill(0));

    /*
         * m[i,j] = Minimum number of scalar
         multiplications needed to compute the
         * matrix A[i]A[i+1]...A[j] = A[i..j] where
         dimension of A[i] is p[i-1] x p[i]
         */

    // cost is zero when multiplying one matrix.
    for (var i = 1; i < n; i++)
      m[i][i] = 0;

    // L is chain length.
    for (var L = 2; L < n; L++)
    {
      for (var i = 1; i < n - L + 1; i++)
      {
        var j = i + L - 1;
        m[i][j] = Number.MAX_VALUE;
        for (var k = i; k <= j - 1; k++)
        {

          // q = cost/scalar multiplications
          var q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j];
          if (q < m[i][j])
          {
            m[i][j] = q;

            // Each entry bracket[i,j]=k shows
            // where to split the product arr
            // i,i+1....j for the minimum cost.
            bracket[i][j] = k;
          }
        }
      }
    }

    // The first matrix is printed as 'A', next as 'B',
    // and so on
    name = 'A';
    document.write("Optimal Parenthesization is : ");
    printParenthesis(1, n - 1, n, bracket);
    document.write("\nOptimal Cost is : " + m[1][n - 1]);
  }

  // Driver code
  var arr = [ 40, 20, 30, 10, 30 ];
  var n = arr.length;
  matrixChainOrder(arr, n);

// This code is contributed by 29AjayKumar
</script>

Output

Optimal Parenthesization is : ((A(BC))D)
Optimal Cost is : 26000

Time Complexity: O(n³) 
Auxiliary Space: O(n²)

Another Approach:

This solution try to solve the problem using Recursion using permutations.

Let's take example:  {40, 20, 30, 10, 30}
n = 5

Let’s divide that into a Matrix

[ [40, 20], [20, 30], [30, 10], [10, 30] ]

[ A , B , C , D ]

it contains 4 matrices i.e. (n - 1)

We have 3 combinations to multiply  i.e.  (n-2)

AB    or    BC    or     CD

Algorithm:

1) Given array of matrices with length M, Loop through  M – 1 times

2) Merge consecutive matrices in each loop

for (int i = 0; i < M - 1; i++) {
   int cost =  (matrices[i][0] * 
                 matrices[i][1] * matrices[i+1][1]);
   
   // STEP - 3
   // STEP - 4
}

3) Merge the current two matrices into one, and remove merged matrices list from list.

If  A, B merged, then A, B must be removed from the List

and NEW matrix list will be like
newMatrices = [  AB,  C ,  D ]

We have now 3 matrices, in any loop
Loop#1:  [ AB,  C,   D ]
Loop#2:  [ A,   BC,  D ]
Loop#3   [ A,   B,   CD ]

4) Repeat: Go to STEP – 1  with  newMatrices as input M — recursion

5) Stop recursion, when we get 2 matrices in the list.

Workflow

Matrices are reduced in following way, 

and cost’s must be retained and summed-up during recursion with previous values of each parent step.

[ A, B , C, D ]

[(AB), C, D ]
 [ ((AB)C), D ]--> [ (((AB)C)D) ] 
 - return & sum-up total cost of this step.
 [ (AB),  (CD)] --> [ ((AB)(CD)) ] 
 - return .. ditto..

 [ A, (BC), D ]
 [ (A(BC)), D ]--> [ ((A(BC))D) ] 
  - return
 [ A, ((BC)D) ]--> [ (A((BC)D)) ] 
  - return
    
 [ A, B, (CD) ]
 [ A, (B(CD)) ]--> [ (A(B(CD))) ] 
  - return
 [ (AB), (CD) ]--> [ ((AB)(CD)) ] 
  - return .. ditto..

on return i.e. at final step of each recursion, check if  this value smaller than of any other.

Below is JAVA,c# and Javascript implementation of above steps.

C++

#include <algorithm>
#include <climits>
#include <iostream>
#include <string>
using namespace std;

// FinalCost class stores the final label and cost of the
// optimal solution
class FinalCost {
public:
    string label = "";
    int cost = INT_MAX;
};

class MatrixMultiplyCost {
public:
    // Recursive function that finds the optimal cost and
    // label

    void optimalCost(int** matrices, string* labels,
                     int prevCost, FinalCost& finalCost,
                     int len)
    {
        // Base case: If there are no or only one matrix,
        // the cost is 0 and there is no need for a label
        if (len < 2) {
            finalCost.cost = 0;
            return;
        }

        // Base case: If there are only two matrices, the
        // cost is the product of their dimensions and the
        // label is the concatenation of their labels
        else if (len == 2) {
            int cost = prevCost
                       + (matrices[0][0] * matrices[0][1]
                          * matrices[1][1]);
            if (cost < finalCost.cost) {
                finalCost.cost = cost;
                finalCost.label
                    = "(" + labels[0] + labels[1] + ")";
            }
            return;
        }

        // Iterate through all possible matrix combinations
        for (int i = 0; i < len - 1; i++) {
            int j;

            // Create new matrices and labels after merging
            // two matrices
            int** newMatrix = new int*[len - 1];
            string* newLabels = new string[len - 1];
            int subIndex = 0;

            // Calculate the cost of merging matrices[i] and
            // matrices[i+1]
            int cost = (matrices[i][0] * matrices[i][1]
                        * matrices[i + 1][1]);

            // Copy over the matrices and labels before the
            // merge
            for (j = 0; j < i; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            // Add the merged matrix and label
            newMatrix[subIndex] = new int[2];
            newMatrix[subIndex][0] = matrices[i][0];
            newMatrix[subIndex][1] = matrices[i + 1][1];
            newLabels[subIndex++]
                = "(" + labels[i] + labels[i + 1] + ")";

            // Copy over the matrices and labels after the
            // merge
            for (j = i + 2; j < len; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            // Recursively call the function with the new
            // matrices and labels
            optimalCost(newMatrix, newLabels,
                        prevCost + cost, finalCost,
                        len - 1);
        }
    }

    FinalCost findOptionalCost(int* arr, int len)
    {

        // Create matrices and labels from the input array
        int** matrices = new int*[len - 1];
        string* labels = new string[len- 1];
        for (int i = 0; i < len - 1; i++) {
            matrices[i] = new int[2];
            matrices[i][0] = arr[i];
            matrices[i][1] = arr[i + 1];
            labels[i] = char(
                65 + i); // Assign labels as A, B, C, etc.
        }
        FinalCost finalCost;

        // Call the recursive function to find the optimal
        // cost and label
        optimalCost(matrices, labels, 0, finalCost,
                    len - 1);
        return finalCost;
    }
};
 void printMatrix(int ** matrices, int len) {
        cout << "matrices = " << endl << "[";
        for (int i = 0; i < len; i++) {
            cout << "[" << matrices[i][0] << " " << matrices[i][1] << "]" << " ";
        }
        cout << "]" << endl;
    }

int main() {
    MatrixMultiplyCost calc;

    int arr[] = {40, 20, 30, 10, 30};
    int len = sizeof(arr) / sizeof(arr[0]);
    int **matrices = new int*[len - 1];
    string *labels = new string[len - 1];

    for (int i = 0; i < len - 1; i++) {
        matrices[i] = new int[2];
        matrices[i][0] = arr[i];
        matrices[i][1] = arr[i + 1];
        labels[i] = char(65 + i);
    }

    printMatrix(matrices, len-1);

    FinalCost cost = calc.findOptionalCost(arr, len);
    cout << "Final labels: \n" << cost.label << endl;
    cout << "Final Cost:\n" << cost.cost << endl;

    return 0;
}

// This code is contributed by lokeshpotta20.

Java

import java.util.Arrays;

public class MatrixMultiplyCost {

    static class FinalCost
    {
        public String label = "";
        public int cost = Integer.MAX_VALUE;
    }

    private void optimalCost(int[][] matrices,
                             String[] labels, int prevCost,
                             FinalCost finalCost)
    {
        int len = matrices.length;

        if (len < 2)
        {
            finalCost.cost = 0;
            return;
        }
        else if (len == 2)
        {
            int cost = prevCost
                       + (matrices[0][0] *
                          matrices[0][1] *
                          matrices[1][1]);

            // This is where minimal cost has been caught
            // for whole program
            if (cost < finalCost.cost)
            {
                finalCost.cost = cost;
                finalCost.label
                    = "(" + labels[0]
                    + labels[1] + ")";
            }
            return;
        }

        // recursive Reduce
        for (int i = 0; i < len - 1; i++)
        {
            int j;
            int[][] newMatrix = new int[len - 1][2];
            String[] newLabels = new String[len - 1];
            int subIndex = 0;

            // STEP-1:
            //   - Merge two matrices's into one - in each
            //   loop, you move merge position
            //        - if i = 0 THEN  (AB) C D ...
            //        - if i = 1 THEN  A (BC) D ...
            //        - if i = 2 THEN  A B (CD) ...
            //   - and find the cost of this two matrices
            //   multiplication
            int cost = (matrices[i][0] * matrices[i][1]
                        * matrices[i + 1][1]);

            // STEP - 2:
            //    - Build new matrices after merge
            //    - Keep track of the merged labels too
            for (j = 0; j < i; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            newMatrix[subIndex][0] = matrices[i][0];
            newMatrix[subIndex][1] = matrices[i + 1][1];
            newLabels[subIndex++]
                = "(" + labels[i] + labels[i + 1] + ")";

            for (j = i + 2; j < len; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            optimalCost(newMatrix, newLabels,
                        prevCost + cost, finalCost);
        }
    }

    public FinalCost findOptionalCost(int[] arr)
    {
        // STEP -1 : Prepare and convert inout as Matrix
        int[][] matrices = new int[arr.length - 1][2];
        String[] labels = new String[arr.length - 1];

        for (int i = 0; i < arr.length - 1; i++) {
            matrices[i][0] = arr[i];
            matrices[i][1] = arr[i + 1];
            labels[i] = Character.toString((char)(65 + i));
        }
        printMatrix(matrices);

        FinalCost finalCost = new FinalCost();
        optimalCost(matrices, labels, 0, finalCost);

        return finalCost;
    }

    /**
     * Driver Code
     */
    public static void main(String[] args)
    {
        MatrixMultiplyCost calc = new MatrixMultiplyCost();

        // ======= *** TEST CASES **** ============

        int[] arr = { 40, 20, 30, 10, 30 };
        FinalCost cost = calc.findOptionalCost(arr);
        System.out.println("Final labels: \n" + cost.label);
        System.out.println("Final Cost:\n" + cost.cost
                           + "\n");
    }

    /**
     * Ignore this method
     * - THIS IS for DISPLAY purpose only
     */
    private static void printMatrix(int[][] matrices)
    {
        System.out.print("matrices = \n[");
        for (int[] row : matrices) {
            System.out.print(Arrays.toString(row) + " ");
        }
        System.out.println("]");
    }
}

// This code is contributed by suvera

Python3

# Python3 code to implement the approach

class FinalCost:
    def __init__(self):
        self.label = ""
        self.cost = float("inf")

def optimalCost(matrices, labels, prevCost, finalCost):
    length = len(matrices)
    if length < 2:
        finalCost.cost = 0
    elif length == 2:
        cost = prevCost + matrices[0][0] * matrices[0][1] * matrices[1][1]
        # This is where minimal cost has been caught
        # for whole program
        if cost < finalCost.cost:
            finalCost.cost = cost
            finalCost.label = "(" + labels[0] + labels[1] + ")"
    else:
        # recursive Reduce
        for i in range(length - 1):
            newMatrix = [[0] * 2 for i in range(length - 1)]
            newLabels = [0] * (length - 1)
            subIndex = 0

            # STEP-1:
            #   - Merge two matrices's into one - in each
            #   loop, you move merge position
            #        - if i = 0 THEN  (AB) C D ...
            #        - if i = 1 THEN  A (BC) D ...
            #        - if i = 2 THEN  A B (CD) ...
            #   - and find the cost of this two matrices
            #   multiplication
            cost = matrices[i][0] * matrices[i][1] * matrices[i + 1][1]

            # STEP - 2:
            #    - Build new matrices after merge
            #    - Keep track of the merged labels too
            for j in range(i):
                newMatrix[subIndex] = matrices[j]
                newLabels[subIndex] = labels[j]
                subIndex += 1

            newMatrix[subIndex][0] = matrices[i][0];
            newMatrix[subIndex][1] = matrices[i + 1][1];
            newLabels[subIndex] = "(" + str(labels[i]) + str(labels[i + 1]) + ")";
            subIndex+= 1

            for j in range(i + 2, length):
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex] = labels[j];
                subIndex+= 1
            optimalCost(newMatrix, newLabels, prevCost + cost, finalCost);


def findOptionalCost(arr):
    # STEP -1 : Prepare and convert inout as Matrix
    matrices = [[0] * 2 for i in range(len(arr) - 1)]
    labels = [0] * (len(arr) - 1)

    for i in range(len(arr) - 1):
        matrices[i][0] = arr[i]
        matrices[i][1] = arr[i + 1]
        labels[i] = chr(65 + i)

    print("matrices =", matrices)


    finalCost = FinalCost()
    optimalCost(matrices, labels, 0, finalCost)

    return finalCost

# Driver Code

# ======= *** TEST CASES **** ============

arr = [40, 20, 30, 10, 30]


cost = findOptionalCost(arr)
print("Final labels:" + cost.label)
print("Final Cost:" + str(cost.cost))



# This code is contributed by phasing17

C#

using System;
using System.Collections.Generic;

public class Cost
{
    public string label = "";
    public int cost =Int32.MaxValue;
}

public class MatrixMultiplyCost {

    private void optimalCost(int[][] matrices,
                             string[] labels, int prevCost,
                             Cost Cost)
    {
        int len = matrices.Length;

        if (len < 2)
        {
            Cost.cost = 0;
            return;
        }
        else if (len == 2)
        {
            int cost = prevCost
                       + (matrices[0][0] *
                          matrices[0][1] *
                          matrices[1][1]);

            // This is where minimal cost has been caught
            // for whole program
            if (cost < Cost.cost)
            {
                Cost.cost = cost;
                Cost.label
                    = "(" + labels[0]
                    + labels[1] + ")";
            }
            return;
        }

        // recursive Reduce
        for (int i = 0; i < len - 1; i++)
        {
            int j;
            int[][] newMatrix = new int[len - 1][];

            for (int x = 0; x < len - 1; x++)
                newMatrix[x] = new int[2];

            string[] newLabels = new string[len - 1];
            int subIndex = 0;

            // STEP-1:
            //   - Merge two matrices's into one - in each
            //   loop, you move merge position
            //        - if i = 0 THEN  (AB) C D ...
            //        - if i = 1 THEN  A (BC) D ...
            //        - if i = 2 THEN  A B (CD) ...
            //   - and find the cost of this two matrices
            //   multiplication
            int cost = (matrices[i][0] * matrices[i][1]
                        * matrices[i + 1][1]);

            // STEP - 2:
            //    - Build new matrices after merge
            //    - Keep track of the merged labels too
            for (j = 0; j < i; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            newMatrix[subIndex][0] = matrices[i][0];
            newMatrix[subIndex][1] = matrices[i + 1][1];
            newLabels[subIndex++]
                = "(" + labels[i] + labels[i + 1] + ")";

            for (j = i + 2; j < len; j++) {
                newMatrix[subIndex] = matrices[j];
                newLabels[subIndex++] = labels[j];
            }

            optimalCost(newMatrix, newLabels,
                        prevCost + cost, Cost);
        }
    }

    public Cost findOptionalCost(int[] arr)
    {
        // STEP -1 : Prepare and convert inout as Matrix
        int[][] matrices = new int[arr.Length - 1][];
        string[] labels = new string[arr.Length - 1];

        for (int i = 0; i < arr.Length - 1; i++) {
            matrices[i] = new int[2];
            matrices[i][0] = arr[i];
            matrices[i][1] = arr[i + 1];
            labels[i] = Convert.ToString((char)(65 + i));
        }
        printMatrix(matrices);

        Cost Cost = new Cost();
        optimalCost(matrices, labels, 0, Cost);

        return Cost;
    }

    /**
     * Driver Code
     */
    public static void Main(string[] args)
    {
        MatrixMultiplyCost calc = new MatrixMultiplyCost();

        // ======= *** TEST CASES **** ============

        int[] arr = { 40, 20, 30, 10, 30 };
        Cost cost = calc.findOptionalCost(arr);
        Console.WriteLine(" labels: \n" + cost.label);
        Console.WriteLine(" Cost:\n" + cost.cost
                           + "\n");
    }

    /**
     * Ignore this method
     * - THIS IS for DISPLAY purpose only
     */
    private static void printMatrix(int[][] matrices)
    {
        Console.Write("matrices = \n[");
        foreach (int[] row in matrices) {
            Console.Write("[ " + string.Join(" ", row) + " " + "], ");
        }
        Console.WriteLine("]");
    }
}

// This code is contributed by phasing17

Javascript

class FinalCost {
        constructor() {
          this.label = "";
          this.cost = Number.MAX_VALUE;
        }
      }

      function optimalCost(matrices, labels, prevCost, finalCost) {
        var len = matrices.length;
        if (len < 2) {
          finalCost.cost = 0;
          return;
        } else if (len == 2) {
          var Cost =
            prevCost + matrices[0][0] * matrices[0][1] * matrices[1][1];

          // This is where minimal cost has been caught
          // for whole program
          if (Cost < finalCost.cost) {
            finalCost.cost = Cost;
            finalCost.label = "(" + labels[0] + labels[1] + ")";
          }
          return;
        }

        // recursive Reduce
        for (var i = 0; i < len - 1; i++) {
          var j;
          let newMatrix = Array.from(Array(len - 1), () => new Array(2));
          let newLabels = new Array(len - 1);
          subIndex = 0;

          // STEP-1:
          //   - Merge two matrices's into one - in each
          //   loop, you move merge position
          //        - if i = 0 THEN  (AB) C D ...
          //        - if i = 1 THEN  A (BC) D ...
          //        - if i = 2 THEN  A B (CD) ...
          //   - and find the cost of this two matrices
          //   multiplication
          Cost = matrices[i][0] * matrices[i][1] * matrices[i + 1][1];

          // STEP - 2:
          //    - Build new matrices after merge
          //    - Keep track of the merged labels too
          for (j = 0; j < i; j++) {
            newMatrix[subIndex] = matrices[j];
            newLabels[subIndex++] = labels[j];
          }

          newMatrix[subIndex][0] = matrices[i][0];
          newMatrix[subIndex][1] = matrices[i + 1][1];
          newLabels[subIndex++] = "(" + labels[i] + labels[i + 1] + ")";

          for (j = i + 2; j < len; j++) {
            newMatrix[subIndex] = matrices[j];
            newLabels[subIndex++] = labels[j];
          }

          optimalCost(newMatrix, newLabels, prevCost + Cost, finalCost);
        }
      }

      function findOptionalCost(arr) {
        // STEP -1 : Prepare and convert inout as Matrix
        let matrices = Array.from(Array(arr.length - 1), () => new Array(2));
        let labels = new Array(arr.length - 1);

        for (var i = 0; i < arr.length - 1; i++) {
          matrices[i][0] = arr[i];
          matrices[i][1] = arr[i + 1];
          labels[i] = String.fromCharCode(65 + i);
        }
        printMatrix(matrices);

        let finalCost = new FinalCost();
        optimalCost(matrices, labels, 0, finalCost);

        return finalCost;
      }

      /**
       * Driver Code
       */

      // ======= *** TEST CASES **** ============

      var arr = [40, 20, 30, 10, 30];
      cost = findOptionalCost(arr);
      console.log("Final labels:" + cost.label);
      console.log("Final Cost:" + cost.cost);

      /**
       * Ignore this method
       * - THIS IS for DISPLAY purpose only
       */
      function printMatrix(matrices) {
        console.log("matrices = ");
        for (var k = 0; k < matrices.length; k++) {
          console.log(matrices[k]);
        }
      }

      // This code is contributed by satwiksuman.

Output

matrices = 
[[40 20] [20 30] [30 10] [10 30] ]
Final labels: 
((A(BC))D)
Final Cost:
26000

Time Complexity : O(n²)

Auxiliary Space:O(n²)

This article is contributed by Yasin Zafar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.