Matrix Chain Multiplication | DP-8

Difficulty Level : Hard
Last Updated : 07 Aug, 2021

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,

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

(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: 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: 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: 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

1) Optimal Substructure:
A simple solution is to place parenthesis at all possible places, calculate the cost for each placement and return the minimum value. In a chain of matrices of size n, we can place the first set of parenthesis in n-1 ways. For example, if the given chain is of 4 matrices. let the chain be ABCD, then there are 3 ways to place first set of parenthesis outer side: (A)(BCD), (AB)(CD) and (ABC)(D). So when we place a set of parenthesis, we divide the problem into subproblems of smaller size. Therefore, the problem has optimal substructure property and can be easily solved using recursion.
Minimum number of multiplication needed to multiply a chain of size n = Minimum of all n-1 placements (these placements create subproblems of smaller size)

2) Overlapping Subproblems
Following is a recursive implementation that simply follows the above optimal substructure property.

Below is the implementation of the above idea:

C++

/* A naive recursive implementation that simply
follows the above optimal substructure property */
#include <bits/stdc++.h>
using namespace std;
 
// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1..n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;
 
    // place parenthesis at different places
    // between first and last matrix, recursively
    // calculate count of multiplications for
    // each parenthesis placement and return the
    // minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, 1, n - 1);
}
 
// This code is contributed by Shivi_Aggarwal

C

/* A naive recursive implementation that simply
  follows the above optimal substructure property */
#include <limits.h>
#include <stdio.h>
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;
 
    // place parenthesis at different places between first
    // and last matrix, recursively calculate count of
    // multiplications for each parenthesis placement and
    // return the minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, 1, n - 1));
 
    getchar();
    return 0;
}

Java

/* A naive recursive implementation that simply follows
   the above optimal substructure property */
class MatrixChainMultiplication {
    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    static int MatrixChainOrder(int p[], int i, int j)
    {
        if (i == j)
            return 0;
 
        int min = Integer.MAX_VALUE;
 
        // place parenthesis at different places between
        // first and last matrix, recursively calculate
        // count of multiplications for each parenthesis
        // placement and return the minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];
 
            if (count < min)
                min = count;
        }
 
        // Return minimum count
        return min;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4, 3 };
        int n = arr.length;
 
        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, n - 1));
    }
}
/* This code is contributed by Rajat Mishra*/

Python3

# A naive recursive implementation that
# simply follows the above optimal
# substructure property
import sys
 
# Matrix A[i] has dimension p[i-1] x p[i]
# for i = 1..n
 
 
def MatrixChainOrder(p, i, j):
 
    if i == j:
        return 0
 
    _min = sys.maxsize
 
    # place parenthesis at different places
    # between first and last matrix,
    # recursively calculate count of
    # multiplications for each parenthesis
    # placement and return the minimum count
    for k in range(i, j):
 
        count = (MatrixChainOrder(p, i, k)
                 + MatrixChainOrder(p, k + 1, j)
                 + p[i-1] * p[k] * p[j])
 
        if count < _min:
            _min = count
 
    # Return minimum count
    return _min
 
 
# Driver code
arr = [1, 2, 3, 4, 3]
n = len(arr)
 
print("Minimum number of multiplications is ",
      MatrixChainOrder(arr, 1, n-1))
 
# This code is contributed by Aryan Garg

C#

/* C# code for naive recursive implementation
that simply follows the above optimal
substructure property */
using System;
 
class GFG {
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int i, int j)
    {
 
        if (i == j)
            return 0;
 
        int min = int.MaxValue;
 
        // place parenthesis at different places
        // between first and last matrix, recursively
        // calculate count of multiplications for each
        // parenthesis placement and return the
        // minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];
 
            if (count < min)
                min = count;
        }
 
        // Return minimum count
        return min;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4, 3 };
        int n = arr.Length;
 
        Console.Write(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, n - 1));
    }
}
 
// This code is contributed by Sam007.

PHP

<?php
// A naive recursive implementation
// that simply follows the above
// optimal substructure property
 
// Matrix Ai has dimension
// p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(&$p, $i, $j)
{
    if($i == $j)
        return 0;
    $min = PHP_INT_MAX;
 
    // place parenthesis at different places
    // between first and last matrix, recursively
    // calculate count of multiplications for
    // each parenthesis placement and return
    // the minimum count
    for ($k = $i; $k < $j; $k++)
    {
        $count = MatrixChainOrder($p, $i, $k) +
                 MatrixChainOrder($p, $k + 1, $j) +
                                  $p[$i - 1] *
                                  $p[$k] * $p[$j];
 
        if ($count < $min)
            $min = $count;
    }
 
    // Return minimum count
    return $min;
}
 
// Driver Code
$arr = array(1, 2, 3, 4, 3);
$n = sizeof($arr);
 
echo "Minimum number of multiplications is " .
      MatrixChainOrder($arr, 1, $n - 1);
 
// This code is contributed by ita_c
?>

Javascript

<script>
 
/* A naive recursive implementation that simply follows
   the above optimal substructure property */
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(p , i , j)
{
    if (i == j)
        return 0;
 
    var min = Number.MAX_VALUE;
 
    // place parenthesis at different places between
    // first and last matrix, recursively calculate
    // count of multiplications for each parenthesis
    // placement and return the minimum count
    var k=0;
    for (k = i; k < j; k++)
    {
        var count = MatrixChainOrder(p, i, k)
                    + MatrixChainOrder(p, k + 1, j)
                    + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver code
var arr = [ 1, 2, 3, 4, 3 ];
var n = arr.length;
 
document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, 1, n - 1));
 
// This code contributed by shikhasingrajput
 
</script>

Output

Minimum number of multiplications is 30

The time complexity of the above naive recursive approach is exponential. It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for a matrix chain of size 4. The function MatrixChainOrder(p, 3, 4) is called two times. We can see that there are many subproblems being called more than once.

Since same subproblems are called again, this problem has Overlapping Subproblems property. So Matrix Chain Multiplication problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array m[][] in bottom up manner.

Dynamic Programming Solution
Following is the implementation of the Matrix Chain Multiplication problem using Dynamic Programming (Tabulation vs Memoization)

Using Memoization –

C++

// C++ program using memoization
#include <bits/stdc++.h>
using namespace std;
int dp[100][100];
 
// Function for matrix chain multiplication
int matrixChainMemoised(int* p, int i, int j)
{
    if (i == j)
    {
        return 0;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
    dp[i][j] = INT_MAX;
    for (int k = i; k < j; k++)
    {
        dp[i][j] = min(
            dp[i][j], matrixChainMemoised(p, i, k)
                     + matrixChainMemoised(p, k + 1, j)
                       + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
int MatrixChainOrder(int* p, int n)
{
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    memset(dp, -1, sizeof dp);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, n);
}
 
// This code is contributed by Sumit_Yadav

Java

// Java program using memoization
import java.io.*;
import java.util.*;
class GFG
{
 
  static int[][] dp = new int[100][100];
 
  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j) 
    {
      return 0;
    }
    if (dp[i][j] != -1) 
    {
      return dp[i][j];
    }
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) 
    {
      dp[i][j] = Math.min(
        dp[i][j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
  }
 
  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    int arr[] = { 1, 2, 3, 4 };
    int n= arr.length;
 
    for (int[] row : dp)
      Arrays.fill(row, -1);
 
    System.out.println("Minimum number of multiplications is " + MatrixChainOrder(arr, n));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python program using memoization
import sys
dp = [[-1 for i in range(100)] for j in range(100)]
 
# Function for matrix chain multiplication
def matrixChainMemoised(p, i, j):
    if(i == j):
        return 0
     
    if(dp[i][j] != -1):
        return dp[i][j]
     
    dp[i][j] = sys.maxsize
     
    for k in range(i,j):
        dp[i][j] = min(dp[i][j], matrixChainMemoised(p, i, k) + matrixChainMemoised(p, k + 1, j)+ p[i - 1] * p[k] * p[j])
     
    return dp[i][j]
 
def MatrixChainOrder(p,n):
    i = 1
    j = n - 1   
    return matrixChainMemoised(p, i, j)
 
# Driver Code
arr = [1, 2, 3, 4]
n = len(arr)
print("Minimum number of multiplications is",MatrixChainOrder(arr, n))
 
# This code is contributed by rag2127

C#

// C# program using memoization
using System;
class GFG
{
     
    static int[,] dp = new int[100, 100];
  
  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j) 
    {
      return 0;
    }
    if (dp[i, j] != -1) 
    {
      return dp[i, j];
    }
    dp[i, j] = Int32.MaxValue;
    for (int k = i; k < j; k++) 
    {
      dp[i, j] = Math.Min(
        dp[i, j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i,j];
  }
  
  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }
   
  // Driver code
  static void Main()
  {
    int[] arr = { 1, 2, 3, 4 };
    int n = arr.Length;
     
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
        {
            dp[i, j] = -1;
        }
    }
  
    Console.WriteLine("Minimum number of multiplications is " +
                      MatrixChainOrder(arr, n));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

<script>
 
// Javascript program using memoization
let dp = new Array(100);
for(var i = 0; i < dp.length; i++)
{
    dp[i] = new Array(2);
}
 
// Function for matrix chain multiplication
function matrixChainMemoised(p, i, j)
{
    if (i == j) 
    {
        return 0;
    }
    if (dp[i][j] != -1) 
    {
        return dp[i][j];
    }
     
    dp[i][j] = Number.MAX_VALUE;
    for(let k = i; k < j; k++) 
    {
        dp[i][j] = Math.min(
            dp[i][j], matrixChainMemoised(p, i, k) +
                      matrixChainMemoised(p, k + 1, j) +
                      p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
 
function MatrixChainOrder(p, n)
{
    let i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver code
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;
 
for(var i = 0; i < dp.length; i++)
{
    for(var j = 0; j < dp.length; j++)
    {
        dp[i][j] = -1;
    }
}
 
document.write("Minimum number of multiplications is " +
               MatrixChainOrder(arr, n));
                
// This code is contributed by target_2
 
</script>

Output

Minimum number of multiplications is 18

Using Tabulation –

C++

// See the Cormen book for details of the
// following algorithm
#include <bits/stdc++.h>
using namespace std;
 
// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1..n
int 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];
 
    int i, j, k, L, q;
 
    /* 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 = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // 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;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, size);
 
    getchar();
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// See the Cormen book for details of the following
// algorithm
#include <limits.h>
#include <stdio.h>
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int 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];
 
    int i, j, k, L, q;
 
    /* 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 = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++) {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // 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;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver  code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, size));
 
    getchar();
    return 0;
}

Java

// Dynamic Programming Java implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm
class MatrixChainMultiplication
{
 
    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    static int 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];
 
        int i, j, k, L, q;
 
        /* 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 = 1; i < n; i++)
            m[i][i] = 0;
 
        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i][j] = Integer.MAX_VALUE;
                for (k = i; k <= j - 1; k++)
                {
                    // 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;
                }
            }
        }
 
        return m[1][n - 1];
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4 };
        int size = arr.length;
 
        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, size));
    }
}
/* This code is contributed by Rajat Mishra*/

Python

# Dynamic Programming Python implementation of Matrix
# Chain Multiplication. See the Cormen book for details
# of the following algorithm
import sys
 
# Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
 
 
def 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
    m = [[0 for x in range(n)] for x 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] = sys.maxint
            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
 
    return m[1][n-1]
 
 
# Driver code
arr = [1, 2, 3, 4]
size = len(arr)
 
print("Minimum number of multiplications is " +
      str(MatrixChainOrder(arr, size)))
# This Code is contributed by Bhavya Jain

C#

// Dynamic Programming C# implementation of
// Matrix Chain Multiplication.
// See the Cormen book for details of the
// following algorithm
using System;
 
class GFG
{
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int 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];
 
        int i, j, k, L, q;
 
        /* 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 = 1; i < n; i++)
            m[i, i] = 0;
 
        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i, j] = int.MaxValue;
                for (k = i; k <= j - 1; k++)
                {
                    // 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;
                }
            }
        }
 
        return m[1, n - 1];
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4 };
        int size = arr.Length;
 
        Console.Write("Minimum number of "
                      + "multiplications is "
                      + MatrixChainOrder(arr, size));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// Dynamic Programming Python implementation
// of Matrix Chain Multiplication.
 
// See the Cormen book for details of the
// following algorithm Matrix Ai has
// dimension p[i-1] x p[i] for i = 1..n
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 */
    $m[][] = array($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 ($i = 1; $i < $n; $i++)
        $m[$i][$i] = 0;
 
    // L is chain length.
    for ($L = 2; $L < $n; $L++)
    {
        for ($i = 1; $i < $n - $L + 1; $i++)
        {
            $j = $i + $L - 1;
            if($j == $n)
                continue;
            $m[$i][$j] = PHP_INT_MAX;
            for ($k = $i; $k <= $j - 1; $k++)
            {
                // 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;
            }
        }
    }
 
    return $m[1][$n-1];
}
 
// Driver Code
$arr = array(1, 2, 3, 4);
$size = sizeof($arr);
 
echo"Minimum number of multiplications is ".
              MatrixChainOrder($arr, $size);
 
// This code is contributed by Mukul Singh
?>

Javascript

<script>
 
// Dynamic Programming javascript implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm
 
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
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));
 
    var i, j, k, L, q;
 
    /* 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 = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            if (j == n)
                continue;
            m[i][j] = Number.MAX_VALUE;
            for (k = i; k <= j - 1; k++)
            {
                // 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;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver code
var arr = [ 1, 2, 3, 4 ];
var size = arr.length;
 
document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, size));
 
// This code contributed by Princi Singh
 
</script>