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# Printing brackets in Matrix Chain Multiplication Problem

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 ``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

 ``

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 ``#include ``#include ``#include ``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)

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