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# Matrix Chain Multiplication | DP-8

Given the dimension of a sequence of matrices in an array arr[], where the dimension of the ith matrix is (arr[i-1] * arr[i]), the task is to find the most efficient way to multiply these matrices together such that the total number of element multiplications is minimum.

Examples:

Input: arr[] = {40, 20, 30, 10, 30}
Output: 26000
Explanation:There are 4 matrices of dimensions 40×20, 20×30, 30×10, 10×30.
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.
The minimum is 20*30*10 + 40*20*10 + 40*10*30

Input: arr[] = {1, 2, 3, 4, 3}
Output: 30
Explanation: There are 4 matrices of dimensions 1×2, 2×3, 3×4, 4×3.
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.
The minimum number is 1*2*3 + 1*3*4 + 1*4*3 = 30

Input: arr[] = {10, 20, 30}
Output: 6000
Explanation: There are only two matrices of dimensions 10×20 and 20×30.
So there  is only one way to multiply the matrices, cost of which is 10*20*30

Recommended Practice

## Matrix Chain Multiplication using Recursion:

We can solve the problem using recursion based on the following facts and observations:

Two matrices of size m*n and n*p when multiplied, they generate a matrix of size m*p and the number of multiplications performed are m*n*p.

Now, for a given chain of N matrices, the first partition can be done in N-1 ways. For example, sequence of matrices A, B, C and D can be grouped as (A)(BCD), (AB)(CD) or (ABC)(D) in these 3 ways.

So a range [i, j] can be broken into two groups like {[i, i+1], [i+1, j]}, {[i, i+2], [i+2, j]}, . . . , {[i, j-1], [j-1, j]}.

• Each of the groups can be further partitioned into smaller groups and we can find the total required multiplications by solving for each of the groups.
• The minimum number of multiplications among all the first partitions is the required answer.

Follow the steps mentioned below to implement the approach:

• Create a recursive function that takes i and j as parameters that determines the range of a group.
• Iterate from k = i to j to partition the given range into two groups.
• Call the recursive function for these groups.
• Return the minimum value among all the partitions as the required minimum number of multiplications to multiply all the matrices of this group.
• The minimum value returned for the range 0 to N-1 is the required answer.

Below is the implementation of the above approach.

## C++

 `// C++ code to implement the ` `// matrix chain multiplication using recursion`   `#include ` `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` `mini = 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];`   `        ``mini = min(count, mini);` `    ``}`   `    ``// Return minimum count` `    ``return` `mini;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 4, 3 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// Function call` `    ``cout << ``"Minimum number of multiplications is "` `         ``<< MatrixChainOrder(arr, 1, N - 1);` `    ``return` `0;` `}`   `// This code is contributed by Shivi_Aggarwal`

## C

 `// C code to implement the ` `// matrix chain multiplication using recursion`   `#include ` `#include `   `// 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);`   `    ``// Function call` `    ``printf``(``"Minimum number of multiplications is %d "``,` `           ``MatrixChainOrder(arr, 1, N - 1));` `    ``getchar``();` `    ``return` `0;` `}`

## Java

 `// Java code to implement the` `// matrix chain multiplication using recursion` `import` `java.io.*;` `import` `java.util.*;` `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;`   `        ``// Function call` `        ``System.out.println(` `            ``"Minimum number of multiplications is "` `            ``+ MatrixChainOrder(arr, ``1``, N - ``1``));` `    ``}` `}` `/* This code is contributed by Rajat Mishra*/`

## Python3

 `# Python code to implement the` `# matrix chain multiplication using recursion`   `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` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``3``]` `    ``N ``=` `len``(arr)` `    `  `    ``# Function call` `    ``print``(``"Minimum number of multiplications is "``,` `      ``MatrixChainOrder(arr, ``1``, N``-``1``))`   `# This code is contributed by Aryan Garg`

## C#

 `// C# code to implement the ` `// matrix chain multiplication using recursion`   `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;`   `        ``// Function call` `        ``Console.Write(` `            ``"Minimum number of multiplications is "` `            ``+ MatrixChainOrder(arr, 1, N - 1));` `    ``}` `}`   `// This code is contributed by Sam007.`

## PHP

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## Javascript

 ``

Output

`Minimum number of multiplications is 30`

The time complexity of the solution is exponential
Auxiliary Space: O(1)

## Dynamic Programming Solution for Matrix Chain Multiplication using Memoization:

Below is the recursion tree for the 2nd example of the above recursive approach: If observed carefully you can find the following two properties:

1) Optimal Substructure: In the above case, we are breaking the bigger groups into smaller subgroups and solving them to finally find the minimum number of multiplications. Therefore, it can be said that the problem has optimal substructure property.

2) Overlapping Subproblems: We can see in the recursion tree that the same subproblems are called again and again and this problem has the Overlapping Subproblems property.

So Matrix Chain Multiplication problem has both properties of a dynamic programming problem. So recomputations of same subproblems can be avoided by constructing a temporary array dp[][] in a bottom up manner.

Follow the below steps to solve the problem:

• Build a matrix dp[][] of size N*N for memoization purposes.
• Use the same recursive call as done in the above approach:
• When we find a range (i, j) for which the value is already calculated, return the minimum value for that range (i.e., dp[i][j]).
• Otherwise, perform the recursive calls as mentioned earlier.
• The value stored at dp[N-1] is the required answer.

Below is the implementation of the above approach

## C++

 `// C++ program using memoization` `#include ` `using` `namespace` `std;` `int` `dp;`   `// 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);` `    ``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

 ``

Output

`Minimum number of multiplications is 18`

Time Complexity: O(N3 )
Auxiliary Space: O(N2) ignoring recursion stack space

## Dynamic Programming Solution for Matrix Chain Multiplication using Tabulation (Iterative Approach):

In iterative approach, we initially need to find the number of multiplications required to multiply two adjacent matrices. We can use these values to find the minimum multiplication required for matrices in a range of length 3 and further use those values for ranges with higher lengths.

Build on the answer in this manner till the range becomes [0, N-1].

Follow the steps mentioned below to implement the idea:

• Iterate from l = 2 to N-1 which denotes the length of the range:
• Iterate from i = 0 to N-1:
• Find the right end of the range (j) having l matrices.
• Iterate from k = i+1 to j which denotes the point of partition.
• Multiply the matrices in range (i, k) and (k, j).
• This will create two matrices with dimensions arr[i-1]*arr[k] and arr[k]*arr[j].
• The number of multiplications to be performed to multiply these two matrices (say X) are arr[i-1]*arr[k]*arr[j].
• The total number of multiplications is dp[i][k]+ dp[k+1][j] + X.
• The value stored at dp[N-1] is the required answer.

Below is the implementation of the above approach.

## C++

 `// See the Cormen book for details of the` `// following algorithm` `#include ` `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[n - 1];` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 4 };` `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``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 ` `#include `   `// 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[n - 1];` `}`   `// Driver  code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3, 4 };` `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``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` `import` `java.util.*;` `import` `java.io.*;` `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*/`

## Python3

 `# Dynamic Programming Python implementation of Matrix` `# Chain Multiplication. See the Cormen book for details` `# of the following algorithm` `import` `sys` `maxint``=``int``(``1e9``+``7``)` `# 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] ``=` `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

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## Javascript

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Output

`Minimum number of multiplications is 18`

Time Complexity: O(N3 )
Auxiliary Space: O(N2)

Matrix Chain Multiplication (A O(N^2) Solution)
Printing brackets in Matrix Chain Multiplication Problem