Matrix Chain Multiplication in DSA Java - Time & Space Complexity

Time Complexity: Matrix Chain Multiplication

O(n³)

Understanding Time Complexity

We want to understand how the time needed to solve the Matrix Chain Multiplication problem grows as the number of matrices increases.

Specifically, how the number of calculations changes when we try to find the best way to multiply many matrices.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int matrixChainOrder(int[] dims, int n) {
    int[][] dp = new int[n][n];
    for (int length = 2; length <= n; length++) {
        for (int i = 0; i <= n - length; i++) {
            int j = i + length - 1;
            dp[i][j] = Integer.MAX_VALUE;
            for (int k = i; k < j; k++) {
                int cost = dp[i][k] + dp[k+1][j] + dims[i]*dims[k+1]*dims[j+1];
                if (cost < dp[i][j]) dp[i][j] = cost;
            }
        }
    }
    return dp[0][n-1];
}
    

This code finds the minimum number of scalar multiplications needed to multiply a chain of matrices.

Identify Repeating Operations

• Primary operation: The triple nested loops that check all ways to split the matrix chain.
• How many times: The outer loop runs about n times, the middle loop runs up to n times, and the inner loop runs up to n times, making the main operation repeat roughly n x n x n times.

How Execution Grows With Input

As the number of matrices (n) grows, the number of operations grows roughly by the cube of n.

Input Size (n)	Approx. Operations
10	About 1,000
100	About 1,000,000
1000	About 1,000,000,000

Pattern observation: When n doubles, the work increases about eight times (since 2³ = 8), showing cubic growth.

Final Time Complexity

Time Complexity: O(n³)

This means the time needed grows very quickly as the number of matrices increases, roughly proportional to the cube of the number of matrices.

Common Mistake

[X] Wrong: "The problem can be solved in linear time because we just multiply matrices one by one."

[OK] Correct: The order of multiplication greatly affects the cost, so we must check many combinations, which takes much more time than just multiplying in order.

Interview Connect

Understanding this time complexity helps you explain why dynamic programming is needed to avoid repeated work and how the problem scales with input size.

Self-Check

"What if we used a recursive solution without memoization? How would the time complexity change?"

Practice

1. What is the main goal of the Matrix Chain Multiplication problem?

easy

A. To find the minimum number of scalar multiplications needed to multiply a chain of matrices

B. To multiply all matrices in the chain directly without optimization

C. To find the maximum number of matrices that can be multiplied

D. To count the total number of matrices in the chain

Solution

1. Step 1: Understand the problem objective

   The Matrix Chain Multiplication problem aims to find the best way to multiply matrices to minimize cost.

2. Step 2: Identify the cost metric

   The cost is measured by the number of scalar multiplications needed, not just multiplying matrices directly.

3. Final Answer:

   To find the minimum number of scalar multiplications needed to multiply a chain of matrices -> Option A

4. Quick Check:

   Goal = minimum scalar multiplications [OK]

Hint: Focus on minimizing multiplication cost, not counting matrices [OK]

Common Mistakes:

• Thinking the goal is to multiply matrices directly without optimization
• Confusing number of matrices with multiplication cost
• Assuming maximum matrices multiplied is the goal

2. Which of the following is the correct way to declare the dimensions array for 4 matrices with dimensions 10x20, 20x30, 30x40, and 40x50 in Java?

easy

A. int[] dims = {10, 20, 20, 30, 30, 40, 40, 50};

B. int[] dims = {10, 20, 30, 40};

C. int[] dims = {10, 20, 30, 40, 50, 60};

D. int[] dims = {10, 20, 30, 40, 50};

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, the dimensions array has length n+1, storing rows of first matrix and columns of each matrix.

2. Step 2: Apply to given matrices

   Given matrices: 10x20, 20x30, 30x40, 40x50, so dims = {10, 20, 30, 40, 50}.

3. Final Answer:

   int[] dims = {10, 20, 30, 40, 50}; -> Option D

4. Quick Check:

   Length = 5 for 4 matrices [OK]

Hint: Dimensions array length = number of matrices + 1 [OK]

Common Mistakes:

• Using length equal to number of matrices instead of number+1
• Repeating dimensions unnecessarily
• Including extra unrelated dimensions

3. What will be the output of the following Java code snippet for matrix chain multiplication cost calculation?

int[] dims = {5, 10, 3};
int n = dims.length;
int[][] dp = new int[n][n];
for (int l = 2; l < n; l++) {
  for (int i = 1; i < n - l + 1; i++) {
    int j = i + l - 1;
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) {
      int cost = dp[i][k] + dp[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < dp[i][j]) dp[i][j] = cost;
    }
  }
}
System.out.println(dp[1][n-1]);

medium

A. 0

B. 75

C. 150

D. Syntax Error

Solution

1. Step 1: Understand the input matrices

   Dimensions: 5x10 and 10x3 matrices. Only one way to multiply.

2. Step 2: Calculate multiplication cost

   Cost = 5*10*3 = 150 scalar multiplications.

3. Final Answer:

   150 -> Option C

4. Quick Check:

   5*10*3 = 150 [OK]

Hint: Multiply dims[i-1]*dims[k]*dims[j] for cost [OK]

Common Mistakes:

• Confusing indices in cost calculation
• Assuming zero cost for single multiplication
• Syntax errors from incorrect loops

4. Identify the error in the following Java code snippet for matrix chain multiplication:

int[] dims = {2, 3, 4};
int n = dims.length;
int[][] dp = new int[n][n];
for (int l = 2; l <= n; l++) {
  for (int i = 1; i <= n - l + 1; i++) {
    int j = i + l - 1;
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) {
      int cost = dp[i][k] + dp[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < dp[i][j]) dp[i][j] = cost;
    }
  }
}
System.out.println(dp[1][n-1]);

medium

A. Loop condition l <= n causes ArrayIndexOutOfBoundsException

B. Incorrect cost calculation formula

C. dp array should be one-dimensional

D. Missing initialization of dp array

Solution

1. Step 1: Analyze loop bounds

   Outer loop runs l from 2 to n inclusive, but dp indices go up to n-1.

2. Step 2: Identify index error

   When l = n, j = i + l - 1 can exceed dp array size causing ArrayIndexOutOfBoundsException.

3. Final Answer:

   Loop condition l <= n causes ArrayIndexOutOfBoundsException -> Option A

4. Quick Check:

   Loop bound off by one causes index error [OK]

Hint: Check loop bounds carefully to avoid index errors [OK]

Common Mistakes:

• Using <= n instead of < n for loop limit
• Ignoring array size limits
• Misunderstanding dp array dimensions

5. Given matrices with dimensions {40, 20, 30, 10, 30}, what is the minimum number of scalar multiplications needed to multiply the entire chain?

hard

A. 30000

B. 26000

C. 18000

D. 35000

Solution

1. Step 1: Understand matrix dimensions

   Matrices: 40x20, 20x30, 30x10, 10x30.

2. Step 2: Calculate minimum cost using dynamic programming

   Possible parenthesizations and costs:
   - ((40x20)x30)x(10x30):
   Cost1 = 40*20*30=24000 + 40*30*10=12000 + 0 = 36000 (too high)
   - (40x(20x30))x(10x30):
   Cost2 = 20*30*10=6000 + 40*20*10=8000 + 40*10*30=12000 = 26000 (minimum)

3. Final Answer:

   26000 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 26000 [OK]

Hint: Try all splits and pick minimum cost [OK]

Common Mistakes:

• Calculating cost for only one parenthesization
• Mixing up matrix dimensions order
• Ignoring intermediate multiplication costs