In the previous article, we learned the solution for the problem of “**Matrix chain multiplication”**. We had discussed the recursive and memoization solution. In this article, we will learn about the tabulation solution.

**Practice:**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

**Understanding the dp array**

In the memoization approach, we have taken a dp[N][N]. Let us revise it and understand what dp[i][j] means.

As we have to find A1.A2.A3.A4, we will return dp[1][4] as our final answer.

**Rules to write tabulation solution from the memoization solution**

- Write the base case.
- Write down the changing parameters in terms of loops.
- Write the recursive logic inside the loops.

**Base Case**

Let us discuss the base case first, in the memoization approach the base case was:

if (i==j) return 0

Now dp[i][i] means minimum operations to get the multiplication of Ai which means a single array which doesn’t mean anything in the context of this problem so we return 0. Therefore we write a loop to set the base case in our dp array as follows:

**Changing Parameters**

In tabulation, we follow a bottom-up approach that is we start from a smaller problem to the bigger problem, so here we start from (n-1)th matrix and move towards the first matrix i.e

A4_ -> A3_ -> A2_ -> A1_ .

These dashes are the matrix represented by the j pointer.

As j>i, we will start j from i+1 to N-1, thus the pattern of problem-solving becomes:

A4 -> A3.A4 -> A2.A3.A4 -> A1.A2.A3.A4.

This pattern can be achieved by writing the loops in the following way:

**Recursive Logic:**

Next, we just copy down the recursive logic and modify it as required.

**Code:**

## C++ Code

```
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum number of operations for matrix multiplication
int matrixMultiplication(vector<int>& arr, int N) {
// Create a DP table to store the minimum number of operations
vector<vector<int>> dp(N, vector<int>(N, -1));
// Initialize the diagonal elements of the DP table to 0
for (int i = 0; i < N; i++) {
dp[i][i] = 0;
}
// Loop for the length of the chain
for (int len = 2; len < N; len++) {
for (int i = 1; i < N - len + 1; i++) {
int j = i + len - 1;
dp[i][j] = INT_MAX;
// Try different partition points to find the minimum
for (int k = i; k < j; k++) {
int cost = dp[i][k] + dp[k + 1][j] + arr[i - 1] * arr[k] * arr[j];
dp[i][j] = min(dp[i][j], cost);
}
}
}
// The result is stored in dp[1][N-1]
return dp[1][N - 1];
}
int main() {
vector<int> arr = {10, 20, 30, 40, 50};
int n = arr.size();
cout << "The minimum number of operations for matrix multiplication is " << matrixMultiplication(arr, n) << endl;
return 0;
}
```

**Output:**

The minimum number of operations are 38000

**Time Complexity: O(N*N*N)**

Reason: There are N*N states and we explicitly run a loop inside the function which will run for N times, therefore at max ‘N*N*N’ new problems will be solved.

**Space Complexity: O(N*N)**

Reason: We are using a 2D array ( O(N*N)) space.

## Java Code

```
import java.util.*;
class TUF {
// Function to recursively calculate the minimum number of operations for matrix multiplication
static int f(int[] arr, int i, int j, int[][] dp) {
// Base condition
if (i == j) {
return 0;
}
// Check if the result is already computed
if (dp[i][j] != -1) {
return dp[i][j];
}
int minOperations = Integer.MAX_VALUE;
// Partitioning loop to find the optimal split point
for (int k = i; k <= j - 1; k++) {
int operations = f(arr, i, k, dp) + f(arr, k + 1, j, dp) + arr[i - 1] * arr[k] * arr[j];
minOperations = Math.min(minOperations, operations);
}
dp[i][j] = minOperations;
return minOperations;
}
// Function to find the minimum number of operations for matrix multiplication
static int matrixMultiplication(int[] arr, int N) {
int[][] dp = new int[N][N];
// Initialize the dp array with -1
for (int row[] : dp) {
Arrays.fill(row, -1);
}
// Initialize the diagonal with 0
for (int i = 1; i < N; i++) {
dp[i][i] = 0;
}
// Fill in the dp array using bottom-up approach
for (int i = N - 1; i >= 1; i--) {
for (int j = i + 1; j < N; j++) {
int minOperations = Integer.MAX_VALUE;
// Partitioning loop to find the optimal split point
for (int k = i; k <= j - 1; k++) {
int operations = dp[i][k] + dp[k + 1][j] + arr[i - 1] * arr[k] * arr[j];
minOperations = Math.min(minOperations, operations);
}
dp[i][j] = minOperations;
}
}
// The result is stored in dp[1][N-1]
return dp[1][N - 1];
}
public static void main(String[] args) {
int[] arr = {10, 20, 30, 40, 50};
int n = arr.length;
System.out.println("The minimum number of operations are " + matrixMultiplication(arr, n));
}
}
```

**Output:**

The minimum number of operations are 38000

**Time Complexity: O(N*N*N)**

Reason: There are N*N states and we explicitly run a loop inside the function which will run for N times, therefore at max ‘N*N*N’ new problems will be solved.

**Space Complexity: O(N*N)**

Reason: We are using a 2D array ( O(N*N)) space.

## Python Code

```
def matrix_multiplication(arr):
N = len(arr)
# Initialize a 2D dp list with -1 values
dp = [[-1 for _ in range(N)] for _ in range(N)]
# Initialize the diagonal elements of the dp table to 0
for i in range(N):
dp[i][i] = 0
# Loop through the dp table to calculate the minimum number of operations
for i in range(N - 1, 0, -1):
for j in range(i + 1, N):
mini = float('inf')
# Partitioning loop
for k in range(i, j):
ans = dp[i][k] + dp[k + 1][j] + arr[i - 1] * arr[k] * arr[j]
mini = min(mini, ans)
dp[i][j] = mini
# The result is stored in the top-right corner of the dp table
return dp[1][N - 1]
if __name__ == "__main__":
arr = [10, 20, 30, 40, 50]
print("The minimum number of operations is:", matrix_multiplication(arr))
```

**Output:**

The minimum number of operations are 38000

**Time Complexity: O(N*N*N)**

Reason: There are N*N states and we explicitly run a loop inside the function which will run for N times, therefore at max ‘N*N*N’ new problems will be solved.

**Space Complexity: O(N*N)**

Reason: We are using a 2D array ( O(N*N)) space.

## JavaScript Code

```
function matrixMultiplication(arr) {
const N = arr.length;
// Create a 2D DP array to store the minimum number of operations
const dp = new Array(N).fill(null).map(() => new Array(N).fill(-1));
// Initialize the base case where i == j
for (let i = 0; i < N; i++) {
dp[i][i] = 0;
}
// Loop through the matrix chain sizes starting from the smallest chains
for (let chainSize = 2; chainSize < N; chainSize++) {
for (let i = 1; i < N - chainSize + 1; i++) {
const j = i + chainSize - 1; // Ending index of the chain
dp[i][j] = Infinity; // Initialize dp[i][j] to a large value
// Partitioning loop to find the minimum number of operations
for (let k = i; k < j; k++) {
const operations = dp[i][k] + dp[k + 1][j] + arr[i - 1] * arr[k] * arr[j];
dp[i][j] = Math.min(dp[i][j], operations);
}
}
}
// The result will be stored in dp[1][N-1], which represents the minimum operations for the entire chain
return dp[1][N - 1];
}
// Main function
function main() {
const arr = [10, 20, 30, 40, 50];
const result = matrixMultiplication(arr);
console.log("The minimum number of operations is:", result);
}
// Call the main function
main();
```

**Output:**

The minimum number of operations are 38000

**Time Complexity: O(N*N*N)**

**Space Complexity: O(N*N)**

Reason: We are using a 2D array ( O(N*N)) space.

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