Matrix Chain Multiplication in DSA Typescript - Time & Space Complexity

Time Complexity: Matrix Chain Multiplication

O(n³)

Understanding Time Complexity

Matrix Chain Multiplication finds the best way to multiply a chain of matrices with minimum cost.

We want to know how the time needed grows as the number of matrices increases.

Scenario Under Consideration

Analyze the time complexity of this dynamic programming solution.

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const dp: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
  for (let length = 2; length <= n; length++) {
    for (let i = 0; i <= n - length; i++) {
      let j = i + length - 1;
      dp[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        const cost = dp[i][k] + dp[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (cost < dp[i][j]) dp[i][j] = cost;
      }
    }
  }
  return dp[0][n - 1];
}
    

This code finds the minimum multiplication cost for a chain of matrices using dynamic programming.

Identify Repeating Operations

Look at the loops that repeat work:

• Primary operation: The innermost loop tries all split points between matrices.
• How many times: For each pair of matrices, it tries all possible splits, repeating many times.

How Execution Grows With Input

As the number of matrices grows, the operations increase quickly because of three nested loops.

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

Pattern observation: Operations grow roughly by the cube of the input size.

Final Time Complexity

Time Complexity: O(n³)

This means if you double the number of matrices, the time needed grows about eight times.

Common Mistake

[X] Wrong: "The solution runs in linear time because it just loops over the matrices."

[OK] Correct: There are three nested loops, so the work grows much faster than just going through the matrices once.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over brute force and shows your grasp of nested loops and optimization.

Self-Check

"What if we used memoization with recursion instead of bottom-up DP? How would the time complexity change?"

Practice

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

easy

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

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

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

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

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 what is minimized

   The cost minimized is the number of scalar multiplications, not the number of matrices or direct multiplication.

3. Final Answer:

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

4. Quick Check:

   Goal = minimum scalar multiplications [OK]

Hint: Remember: minimize multiplication cost, not count matrices [OK]

Common Mistakes:

• Confusing multiplication count with matrix count
• Thinking the problem multiplies matrices directly
• Assuming maximum matrices multiplication is the goal

2. Which of the following is the correct TypeScript function signature for a recursive Matrix Chain Multiplication function that takes an array of matrix dimensions?

easy

A. function matrixChainOrder(dimensions: string): number

B. function matrixChainOrder(dimensions: number): number[]

C. function matrixChainOrder(dimensions: number[]): number

D. function matrixChainOrder(dimensions: number[][]): void

Solution

1. Step 1: Identify input type for matrix dimensions

   Matrix dimensions are given as an array of numbers representing sizes, so number[] is correct.

2. Step 2: Identify return type for minimum cost

   The function returns the minimum number of scalar multiplications, which is a number.

3. Final Answer:

   function matrixChainOrder(dimensions: number[]): number -> Option C

4. Quick Check:

   Input = number[], Output = number [OK]

Hint: Input is array of numbers, output is a single number [OK]

Common Mistakes:

• Using string instead of number array for dimensions
• Returning an array instead of a number
• Using 2D array or void return incorrectly

3. What is the output of the following TypeScript code snippet?

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const m: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i < n - L + 1; i++) {
      const j = i + L - 1;
      m[i][j] = Number.MAX_SAFE_INTEGER;
      for (let k = i; k < j; k++) {
        const q = m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}

console.log(matrixChainOrder([10, 20, 30, 40]));

medium

A. 18000

B. 24000

C. 30000

D. 60000

Solution

1. Step 1: Understand the input and matrix dimensions

   The input array [10, 20, 30, 40] represents matrices of dimensions 10x20, 20x30, and 30x40.

2. Step 2: Calculate minimum multiplication cost

   Possible parenthesizations:
   - (A1 x (A2 x A3)) = 20*30*40 + 10*20*40 = 24000 + 8000 = 32000
   - ((A1 x A2) x A3) = 10*20*30 + 10*30*40 = 6000 + 12000 = 18000 (minimum)

3. Final Answer:

   18000 -> Option A

4. Quick Check:

   Minimum multiplications = 18000 [OK]

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

Common Mistakes:

• Calculating cost for only one parenthesization
• Mixing matrix dimensions order
• Forgetting to add multiplication costs correctly

4. Identify the error in the following TypeScript code snippet for Matrix Chain Multiplication and select the fix:

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const m: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i < n - L + 1; i++) {
      const j = i + L - 1;
      m[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        const q = m[i][k] + m[k + 1][j] + p[i] * p[k] * p[j];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}

medium

A. Change p[i] * p[k] * p[j] to p[i] * p[k + 1] * p[j + 1]

B. Initialize m with Infinity instead of 0

C. Change loop for (let k = i; k <= j; k++) to k < j

D. Return m[0][n] instead of m[0][n - 1]

Solution

1. Step 1: Identify the multiplication cost formula error

   The cost formula uses incorrect indices: p[i] * p[k] * p[j] should use p[k + 1] and p[j + 1] to represent matrix dimensions correctly.

2. Step 2: Confirm other parts are correct

   Initialization with 0 is correct for diagonal, loop ranges are correct, and return index is correct.

3. Final Answer:

   Change p[i] * p[k] * p[j] to p[i] * p[k + 1] * p[j + 1] -> Option A

4. Quick Check:

   Correct cost formula indices fix error [OK]

Hint: Check matrix dimension indices carefully in cost formula [OK]

Common Mistakes:

• Using wrong indices for matrix dimensions in cost calculation
• Changing loop bounds unnecessarily
• Returning wrong matrix cell

5. Given matrices with dimensions [5, 10, 3, 12, 5, 50, 6], what is the minimum number of scalar multiplications needed to multiply the entire chain? Use the Matrix Chain Multiplication approach.

hard

A. 1510

B. 2010

C. 4050

D. 3500

Solution

1. Step 1: Understand matrix dimensions and chain length

   The array represents matrices: 5x10, 10x3, 3x12, 12x5, 5x50, 50x6 (6 matrices).

2. Step 2: Use dynamic programming to find minimum cost

   By calculating all splits and costs:
   - Minimum cost found is 2010 scalar multiplications.

3. Final Answer:

   2010 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 2010 [OK]

Hint: Use DP table to check all splits and pick minimum [OK]

Common Mistakes:

• Calculating only one split option
• Mixing up matrix dimensions order
• Forgetting to add multiplication costs for subproblems