0 1 Knapsack Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the knapsack capacity variable.

DSA C

int [1] = 50; // Maximum weight capacity of the knapsack

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

Asize

Bweight

CmaxWeight

Dcapacity

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names that don't clearly represent knapsack capacity.

Using 'weight' which might confuse with item weight.

The variable capacity is commonly used to represent the maximum weight the knapsack can hold.

2fill in blank

medium

Complete the code to initialize the DP table with zeros.

DSA C

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

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

Aweight

Bn

Ccapacity

DmaxWeight

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'n' which is the number of items, not capacity.

Using undefined variables.

The inner loop runs up to the knapsack capacity, so we use capacity here.

3fill in blank

hard

Fix the error in the condition to check if the current item's weight fits in the knapsack.

DSA C

if (wt[i - 1] [1] w) {
    dp[i][w] = max(val[i - 1] + dp[i - 1][w - wt[i - 1]], dp[i - 1][w]);
} else {
    dp[i][w] = dp[i - 1][w];
}

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

A<=

B==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which is the opposite condition.

Using '==' which excludes items lighter than capacity.

The item can be included only if its weight is less than or equal to the current capacity w.

4fill in blank

hard

Fill both blanks to complete the max function and the loop variable.

DSA C

int max(int a, int b) {
    return (a [1] b) ? a : b;
}

for (int [2] = 1; [2] <= n; [2]++) {
    // loop body
}

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

A>

B<

Ci

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' in max function which returns the smaller value.

Using 'j' as loop variable when 'i' is standard here.

The max function returns the greater of two values using '>'. The loop variable is commonly 'i' for iterating items.

5fill in blank

hard

Fill all three blanks to complete the DP table update inside the nested loops.

DSA C

for (int i = 1; i <= n; i++) {
    for (int w = 1; w <= capacity; w++) {
        if (wt[i - 1] [1] w) {
            dp[i][w] = max(val[i - 1] + dp[i - 1][w [2] wt[i - 1]], dp[i - 1][w]);
        } else {
            dp[i][w] = dp[i - 1][w];
        }
    }
}

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

A<=

B-

C+

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' when calculating remaining capacity.

Using '>' in the condition which excludes fitting items.

The condition checks if the item weight is less than or equal to current capacity. The remaining capacity is calculated by subtracting the item's weight. The max function compares two values.

Practice

1. In the 0/1 Knapsack problem, why can't we take fractions of items like in the Fractional Knapsack?

easy

A. Because the DP table only stores integers

B. Because each item must be taken as a whole unit or not at all - fractional inclusion violates the problem definition

C. Because fractions make the recursion infinite

D. Because the weight array only accepts integer values

Solution

1. 0/1 Constraint

   The '0/1' in the name literally means each item has two states: not taken (0) or fully taken (1). There is no state for 0.5 or any fraction.

2. Why Greedy Fails

   Because you cannot take fractions, the greedy 'highest value/weight ratio first' strategy fails - taking a high-ratio item may block a better combination of whole items.

3. Why DP is Needed

   DP explores all combinations of 0/1 choices, which is required when greedy fails due to the binary selection constraint.

4. Final Answer:

   Each item is binary - take it whole or skip it. → Option B

Hint: 0/1 = binary choice per item: take it whole or skip entirely. [OK]

Common Mistakes:

• Confusing 0/1 Knapsack with Fractional Knapsack
• Thinking DP table data type restricts selection
• Assuming greedy works for 0/1 Knapsack

2. For 0/1 Knapsack with n=3 items and capacity W=5, what is the size of the standard DP table dp[n+1][W+1] in C?

easy

A. 3 × 5

B. 4 × 5

C. 3 × 6

D. 4 × 6

Solution

1. Table Dimensions

   The DP table is dp[n+1][W+1]. With n=3 items and W=5 capacity: dp[4][6].

2. Why +1

   Index 0 represents the base case: dp[0][w] = 0 (no items) and dp[i][0] = 0 (zero capacity). The +1 makes these base cases fit naturally without off-by-one adjustments.

3. In C

   int dp[4][6]; - 4 rows (0 to 3 items), 6 columns (capacity 0 to 5).

4. Final Answer:

   4 × 6 → Option D

Hint: dp[n+1][W+1] → always add 1 to both dimensions for the base case. [OK]

Common Mistakes:

• Using dp[n][W] and missing the base case row/column
• Adding +1 to only one dimension
• Confusing n items with n+1 rows

3. In the 0/1 Knapsack DP recurrence, when weight[i] > w (current item is heavier than current capacity), what is dp[i][w]?

medium

A. 0

B. dp[i-1][w] - same value as without this item

C. dp[i][w-1]

D. value[i] + dp[i-1][w - weight[i]]

Solution

1. Can't Include Overweight Item

   If weight[i] > w, item i cannot fit - we must skip it.

2. Inherit from Previous Row

   dp[i][w] = dp[i-1][w]: the best value with first i items equals the best with first i-1 items.

3. Full Recurrence

   if (weight[i] > w): dp[i][w] = dp[i-1][w]. else: dp[i][w] = max(dp[i-1][w], value[i] + dp[i-1][w-weight[i]])

4. Final Answer:

   dp[i-1][w] → Option B

Hint: Item too heavy → skip → copy from row above: dp[i-1][w]. [OK]

Common Mistakes:

• Setting dp[i][w]=0 instead of carrying forward the previous best
• Applying include formula when item doesn't fit
• Reading dp[i][w-1] instead of dp[i-1][w]

4. What is the time and space complexity of the standard bottom-up 0/1 Knapsack DP solution?

medium

A. Time O(n×W), Space O(n×W)

B. Time O(n×W), Space O(W)

C. Time O(2^n), Space O(n)

D. Time O(n log W), Space O(n×W)

Solution

1. Time Complexity

   Two nested loops: outer over n items, inner over W+1 capacities. Each cell O(1). Total: O(n×W).

2. Space Complexity

   Full table dp[n+1][W+1] → O(n×W) space.

3. Note

   Space can be reduced to O(W) with 1D rolling array, but standard 2D table is O(n×W).

4. Final Answer:

   Time O(n×W), Space O(n×W) → Option A

Hint: Standard 2D table = O(n×W) time AND space. 1D optimized = O(W) space only. [OK]

Common Mistakes:

• Confusing standard O(n×W) space with optimized O(W)
• Claiming O(2^n) which applies only to naive recursion
• Mixing up time and space complexities

5. In the space-optimized 1D 0/1 Knapsack in C, why must the inner capacity loop iterate from W down to weight[i], not from 0 up to W?

hard

A. To avoid negative array indices

B. To prevent using item i more than once - forward iteration would allow the same item to be counted multiple times in the same pass

C. Because C arrays are stored in column-major order

D. To match the order of the weight array

Solution

1. Problem With Left-to-Right

   With a 1D array, processing left to right means dp[w-weight[i]] has already been updated in this pass - item i gets counted again at smaller capacities.

2. Right-to-Left Prevents Reuse

   Iterating from W down to weight[i] ensures dp[w-weight[i]] still holds the pre-item-i value. Each item counted at most once.

3. 0/1 vs Unbounded

   Right-to-left = 0/1 Knapsack. Left-to-right = Unbounded Knapsack.

4. Final Answer:

   Prevents reusing item i in the same pass. → Option B

Hint: Right-to-left = 0/1 (no reuse). Left-to-right = Unbounded (reuse allowed). [OK]

Common Mistakes:

• Using left-to-right which implements Unbounded Knapsack accidentally
• Thinking loop direction is about array bounds not item reuse
• Not knowing right-to-left is the key 0/1 vs Unbounded distinction