Matrix Chain Multiplication in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the minimum cost to zero for a single matrix.

DSA C++

for (int i = 1; i <= n; i++) {
    dp[i][i] = [1];
}

Drag options to blanks, or click blank then click option'

A0

B1

CINT_MAX

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting the cost to 1 or INT_MAX instead of 0.

Not initializing the diagonal of the dp table.

For a single matrix, no multiplication is needed, so the cost is zero.

2fill in blank

medium

Complete the code to calculate the cost of multiplying two matrix chains split at k.

DSA C++

int cost = dp[i][k] + dp[k + 1][j] + [1] * [2] * [3];

Drag options to blanks, or click blank then click option'

Ap[i - 1]

Bp[k]

Cp[j]

Dp[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using p[i] instead of p[i-1].

Confusing indices for dimensions.

The cost formula multiplies the dimensions of the matrices involved: p[i-1] * p[k] * p[j]. Here, the first dimension is p[i-1].

3fill in blank

hard

Fix the error in the loop that sets the chain length for matrix multiplication.

DSA C++

for (int len = 2; len <= [1]; len++) {
    for (int i = 1; i <= n - len + 1; i++) {
        int j = i + len - 1;
        // compute dp[i][j]
    }
}

Drag options to blanks, or click blank then click option'

Alen

Bn

Cj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using len or i instead of n in the loop condition.

Setting incorrect upper bound causing out-of-range errors.

The chain length should go up to n, the total number of matrices.

4fill in blank

hard

Fill both blanks to update the minimum multiplication cost correctly.

DSA C++

for (int k = i; k < j; k++) {
    int cost = dp[i][k] + dp[k + 1][j] + p[i - 1] * p[k] * p[j];
    if (cost [1] dp[i][j]) {
        dp[i][j] = [2];
    }
}

Drag options to blanks, or click blank then click option'

A<

B>

Ccost

Ddp[i][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < in the if condition.

Assigning dp[i][j] to dp[i][j] instead of cost.

If the new cost is less than the current dp[i][j], update dp[i][j] with the new cost.

5fill in blank

hard

Fill all three blanks to complete the recursive memoized function for matrix chain multiplication.

DSA C++

int matrixChainMemo(int i, int j, vector<int>& p, vector<vector<int>>& dp) {
    if (i == j) return [1];
    if (dp[i][j] != -1) return dp[i][j];
    dp[i][j] = [2];
    for (int k = i; k < j; k++) {
        int cost = matrixChainMemo(i, k, p, dp) + matrixChainMemo(k + 1, j, p, dp) + p[i - 1] * p[k] * p[j];
        if (cost < dp[i][j]) dp[i][j] = [3];
    }
    return dp[i][j];
}

Drag options to blanks, or click blank then click option'

A0

BINT_MAX

Ccost

Ddp[i][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 in base case.

Initializing dp[i][j] with 0 instead of INT_MAX.

Assigning dp[i][j] to dp[i][j] instead of cost.

Base case returns 0 cost for one matrix. Initialize dp[i][j] with INT_MAX. Update dp[i][j] with cost if smaller.

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

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 computation.

2. Step 2: Identify the correct goal

   The goal is to minimize the number of scalar multiplications, not just multiply or count matrices.

3. Final Answer:

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

4. Quick Check:

   Goal = Minimum multiplications [OK]

Hint: Remember: minimize multiplication cost, not just multiply [OK]

Common Mistakes:

• Thinking the goal is to multiply matrices directly without optimization
• Confusing counting matrices with minimizing operations
• Assuming maximum multiplications 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 C++?

easy

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

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

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

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

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, the dimensions array has n+1 elements representing rows and columns of each matrix.

2. Step 2: Apply to given matrices

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

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Using fewer elements than needed in the dimensions array
• Including extra unrelated numbers
• Confusing matrix dimensions order

3. What is the output of the following C++ code snippet for Matrix Chain Multiplication cost calculation?

int dims[] = {5, 10, 3, 12, 5};
int n = 4;
int m[5][5] = {0};
for (int L = 2; L <= n; L++) {
  for (int i = 1; i <= n - L + 1; i++) {
    int j = i + L - 1;
    m[i][j] = 1000000;
    for (int k = i; k < j; k++) {
      int cost = m[i][k] + m[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < m[i][j]) m[i][j] = cost;
    }
  }
}
std::cout << m[1][n] << std::endl;

medium

A. 270

B. 150

C. 201

D. 405

Solution

1. Step 1: Understand matrix dimensions and indices

   Dimensions: 5x10, 10x3, 3x12, 12x5; n=4 matrices; m[i][j] stores min cost from i to j.

2. Step 2: Calculate minimum multiplication cost

   Length 2: m[1][2]=150, m[2][3]=360, m[3][4]=180
   Length 3: m[1][3]=min(0+360+600,150+0+180)=330; m[2][4]=min(0+180+150,360+0+600)=330
   Length 4: m[1][4]=min(0+330+250,150+180+75,330+0+300)=405

3. Final Answer:

   405 -> Option D

4. Quick Check:

   Minimum cost = 405 [OK]

Hint: Trace splits and pick minimum cost [OK]

Common Mistakes:

• Mixing up indices i, j, k
• Incorrect cost formula application
• Forgetting to initialize m[i][j] to max

4. Identify the error in the following C++ code snippet for Matrix Chain Multiplication and choose the correct fix:

int dims[] = {40, 20, 30, 10, 30};
int n = 4;
int m[5][5];
for (int i = 1; i <= n; i++)
  m[i][i] = 0;
for (int L = 2; L <= n; L++) {
  for (int i = 1; i <= n - L + 1; i++) {
    int j = i + L - 1;
    m[i][j] = 0;
    for (int k = i; k < j; k++) {
      int cost = m[i][k] + m[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < m[i][j])
        m[i][j] = cost;
    }
  }
}

medium

A. Change dims indexing to dims[i-1]*dims[k]*dims[j]

B. Initialize m[i][j] to INT_MAX instead of 0 before inner loop

C. Start i loop from 0 instead of 1

D. Remove the if condition inside the inner loop

Solution

1. Step 1: Identify initialization mistake

   m[i][j] is set to 0 before checking for minimum cost, which causes incorrect comparisons.

2. Step 2: Correct initialization

   m[i][j] should be initialized to a very large number (INT_MAX) to find minimum cost correctly.

3. Final Answer:

   Initialize m[i][j] to INT_MAX instead of 0 before inner loop -> Option B

4. Quick Check:

   Use INT_MAX to find minimum cost [OK]

Hint: Initialize costs to max before minimizing [OK]

Common Mistakes:

• Setting m[i][j] to 0 instead of INT_MAX
• Incorrect dims indexing causing wrong cost
• Ignoring initialization leading to wrong min

5. Given matrices with dimensions 10x15, 15x20, 20x25, and 25x30, what is the minimum number of scalar multiplications needed to multiply the entire chain? Use the Matrix Chain Multiplication approach.

hard

A. 15500

B. 15000

C. 13000

D. 11250

Solution

1. Step 1: List matrix dimensions and possible splits

   Matrices: A1(10x15), A2(15x20), A3(20x25), A4(25x30). dims=[10,15,20,25,30]. Calculate costs for chain length 2 to 4.

2. Step 2: Calculate minimum costs for each subchain

   Length 2: m[1][2]=10*15*20=3000; m[2][3]=15*20*25=7500; m[3][4]=20*25*30=15000
   Length 3: m[1][3]=min(0+7500+10*15*25=11250, 3000+0+10*20*25=8000)=8000;
   m[2][4]=min(0+15000+15*20*30=24000, 7500+0+15*25*30=18750)=18750
   Length 4: m[1][4]=min(0+18750+10*15*30=23250, 3000+15000+10*20*30=24000, 8000+0+10*25*30=15500)=15500

3. Final Answer:

   15500 -> Option A

4. Quick Check:

   Minimum scalar multiplications = 15500 [OK]

Hint: Calculate costs for smaller chains first, then combine [OK]

Common Mistakes:

• Incorrectly calculating subproblems costs
• Mixing up matrix dimensions in cost formula
• Not checking all split points k