Unbounded Knapsack Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with zeros.

DSA C

int dp[[1] + 1];
for (int i = 0; i <= W; i++) {
    dp[i] = 0;
}

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

AW

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp array size as W instead of W + 1

Not initializing dp array properly

The dp array size should be W + 1 to cover all weights from 0 to W.

2fill in blank

medium

Complete the code to update dp[i] inside the loop for unbounded knapsack.

DSA C

for (int i = 0; i <= W; i++) {
    for (int j = 0; j < n; j++) {
        if (wt[j] <= i) {
            dp[i] = [1](dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}

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Aabs

Bmin

Cmax

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max

Using sum or abs which are incorrect here

We use max to keep the maximum value achievable for weight i.

3fill in blank

hard

Fix the error in the loop condition to avoid out-of-bound access.

DSA C

for (int j = 0; j < [1]; j++) {
    // loop body
}

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AW

Bn

Cdp

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using W instead of n in inner loop

Using array names as loop limits

The inner loop must run over the number of items n, not weight W or arrays dp or val.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA C

int unboundedKnapsack(int [1], int [2], int val[], int wt[]) {
    // function body
    return dp[W];
}

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AW

Bn

Cweight

Ditems

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping order of n and W

Using unclear parameter names

The function takes number of items n and max weight W as parameters.

5fill in blank

hard

Fill all three blanks to complete the main loop and dp update for unbounded knapsack.

DSA C

for (int i = [1]; i <= [2]; i++) {
    for (int j = 0; j < [3]; j++) {
        if (wt[j] <= i) {
            dp[i] = max(dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}

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

A0

BW

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting i from 1 instead of 0

Using wrong loop limits for j

The outer loop runs from 0 to W (max weight), inner loop runs over n items.

Practice

1. What is the main difference between the 0/1 Knapsack Problem and the Unbounded Knapsack Problem?

easy

A. In Unbounded Knapsack, you can use unlimited copies of each item.

B. In Unbounded Knapsack, items cannot be repeated.

C. Unbounded Knapsack only works with fractional items.

D. Unbounded Knapsack requires items to be sorted by weight.

Solution

1. Step 1: Understand item usage in 0/1 Knapsack

   In 0/1 Knapsack, each item can be chosen at most once.

2. Step 2: Understand item usage in Unbounded Knapsack

   In Unbounded Knapsack, items can be chosen unlimited times.

3. Final Answer:

   In Unbounded Knapsack, you can use unlimited copies of each item. -> Option A

4. Quick Check:

   Unbounded means unlimited usage [OK]

Hint: Remember: Unbounded means unlimited item usage [OK]

Common Mistakes:

• Confusing 0/1 and unbounded usage
• Thinking items must be sorted
• Assuming fractional items allowed

2. Which of the following is the correct syntax to initialize the DP array dp for an Unbounded Knapsack problem with capacity W in C?

easy

A. int dp[W+1] = {0};

B. int dp[W]; memset(dp, 0, sizeof(dp));

C. int dp[W+1]; for(int i=0; i<=W; i++) dp[i] = 0;

D. int dp[W+1]; dp = {0};

Solution

1. Step 1: Understand array declaration in C

   Array size must be W+1 to include capacity 0 to W.

2. Step 2: Initialize all elements to zero

   Using a for loop to set dp[i] = 0 is correct and safe.

3. Final Answer:

   int dp[W+1]; for(int i=0; i<=W; i++) dp[i] = 0; -> Option C

4. Quick Check:

   Proper array size and initialization [OK]

Hint: Use loop to initialize arrays in C safely [OK]

Common Mistakes:

• Using dp = {0} after declaration (invalid)
• Declaring dp with size W instead of W+1
• Using memset incorrectly on non-char arrays

3. What will be the output of the following C code snippet for Unbounded Knapsack with weights = {1, 3}, values = {10, 40}, and capacity = 5?

int weights[] = {1, 3};
int values[] = {10, 40};
int W = 5;
int dp[6] = {0};
for (int i = 0; i <= W; i++) {
  for (int j = 0; j < 2; j++) {
    if (weights[j] <= i) {
      int val = dp[i - weights[j]] + values[j];
      if (val > dp[i]) dp[i] = val;
    }
  }
}
printf("%d\n", dp[W]);

medium

A. 50

B. 80

C. 70

D. 60

Solution

1. Step 1: Calculate dp values step-by-step

   dp[0]=0; dp[1]=max(dp[0]+10)=10; dp[2]=max(dp[1]+10)=20; dp[3]=max(dp[0]+40, dp[2]+10)=max(40,30)=40; dp[4]=max(dp[1]+40, dp[3]+10)=max(50,50)=50; dp[5]=max(dp[2]+40, dp[4]+10)=max(60,60)=60.

2. Step 2: Final dp[5] value is 60

   This is the maximum value achievable with capacity 5.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   dp[5] = 60 [OK]

Hint: Build dp array from 0 to W, update max value [OK]

Common Mistakes:

• Not considering multiple uses of items
• Mixing up indices in dp array
• Forgetting to check if weight fits capacity

4. Identify the error in the following Unbounded Knapsack code snippet and choose the correct fix:

int dp[W+1];
for (int i = 0; i < W; i++) {
  dp[i] = 0;
}
for (int i = 1; i <= W; i++) {
  for (int j = 0; j < n; j++) {
    if (weights[j] <= i) {
      dp[i] = max(dp[i], dp[i - weights[j]] + values[j]);
    }
  }
}

medium

A. Change first loop to for (int i = 0; i <= W; i++) to initialize dp correctly.

B. Change second loop to start from i = 0 instead of i = 1.

C. Remove the first loop; dp array is automatically zero-initialized.

D. Change max function to min function for correct calculation.

Solution

1. Step 1: Check dp array initialization loop

   The dp array size is W+1, but loop runs only till i < W, missing dp[W].

2. Step 2: Fix initialization loop range

   Change loop to i <= W to initialize all dp elements including dp[W].

3. Final Answer:

   Change first loop to for (int i = 0; i <= W; i++) to initialize dp correctly. -> Option A

4. Quick Check:

   Initialize full dp array size [OK]

Hint: Initialize dp array fully from 0 to W inclusive [OK]

Common Mistakes:

• Initializing dp array partially
• Starting dp loop from 0 instead of 1
• Using min instead of max function

5. Given items with weights = {2, 3, 5}, values = {50, 70, 100}, and capacity = 8, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 250

B. 200

C. 220

D. 270

Solution

1. Step 1: Calculate dp array values for capacity 0 to 8

   dp[0]=0; dp[1]=0 (no item fits); dp[2]=max(dp[0]+50)=50; dp[3]=max(dp[0]+70)=70; dp[4]=max(dp[2]+50, dp[1]+70)=max(100,0)=100; dp[5]=max(dp[0]+100, dp[3]+50, dp[2]+70)=max(100,120,120)=120; dp[6]=max(dp[4]+50, dp[3]+70, dp[1]+100)=max(150,140,0)=150; dp[7]=max(dp[5]+50, dp[4]+70, dp[2]+100)=max(170,170,150)=170; dp[8]=max(dp[6]+50, dp[5]+70, dp[3]+100)=max(200,190,170)=200.

2. Step 2: Re-check combinations for better value

   Try dp[8] as dp[3]+dp[5] or dp[2]+dp[6], but dp[8] = 200 is max from above. However, check dp[8] = dp[3]+dp[5] = 70 + 120 = 190, dp[8] = dp[5]+dp[3] = 120 + 70 = 190. Try dp[8] = dp[8-5]+100 = dp[3]+100 = 70 + 100 = 170. So 200 is max so far.

3. Step 3: Consider multiple 3s and 5s

   Try 3*3=9 > 8, so max is 200.

4. Final Answer:

   200 -> Option B

5. Quick Check:

   Maximum value dp[8] = 200 [OK]

Hint: Build dp array and check all item combinations for max value [OK]

Common Mistakes:

• Ignoring multiple item usage
• Not checking all item weights for each capacity
• Assuming only one item type per capacity