0 1 Knapsack Problem in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array for the knapsack problem.

DSA Javascript

const dp = new Array(n + 1).fill().map(() => new Array([1] + 1).fill(0));

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

AW

Bn

C0

Dweights

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number of items 'n' instead of the capacity 'W' for the second dimension.

Initializing with zeros incorrectly.

We create a 2D array dp with dimensions (n+1) x (W+1), where W is the knapsack capacity.

2fill in blank

medium

Complete the code to check if the current item's weight fits in the knapsack capacity.

DSA Javascript

if (weights[i - 1] [1] w) {

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

A>

B<=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which means the item is heavier than capacity.

Using '==' which is too strict.

We check if the weight of the current item is less than or equal to the current capacity to decide if we can include it.

3fill in blank

hard

Fix the error in updating the dp table when including the current item.

DSA Javascript

dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - [1]] + values[i - 1]);

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

Aweights[i]

Bvalues[i - 1]

Cvalues[i]

Dweights[i - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using weights[i] which is out of bounds.

Using values instead of weights.

We subtract the weight of the current item (weights[i - 1]) from the capacity to find the remaining capacity.

4fill in blank

hard

Fill both blanks to complete the nested loops iterating over items and capacities.

DSA Javascript

for (let i = 1; i [1] n + 1; i++) {
  for (let w = 1; w [2] W + 1; w++) {

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' which causes index out of range.

Using '>' or '>=' which causes no iterations.

We loop from 1 up to but not including n+1 and W+1, so we use '<' for both loops.

5fill in blank

hard

Fill both blanks to return the final maximum value after filling the dp table.

DSA Javascript

return dp[[1]][[2]];

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

An

BW

C0

Dweights.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 which is the base case, not the answer.

Using weights.length instead of n.

The answer is at dp[n][W], representing max value with all items and full capacity.

Practice

1. What does the 0 1 Knapsack Problem aim to maximize given a set of items with weights and values?

easy

A. The total value of items without exceeding the weight limit

B. The total weight of items without exceeding the value limit

C. The number of items selected regardless of weight

D. The minimum weight of items selected

Solution

1. Step 1: Understand the problem goal

   The 0 1 Knapsack Problem tries to select items to maximize total value without exceeding the weight capacity.

2. Step 2: Identify the constraint

   The total weight of selected items must not exceed the knapsack's weight limit.

3. Final Answer:

   The total value of items without exceeding the weight limit -> Option A

4. Quick Check:

   Maximize value with weight limit = B [OK]

Hint: Remember: maximize value, not weight or count [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Thinking the problem maximizes number of items
• Ignoring the weight constraint

2. Which of the following is the correct way to initialize the DP array for the 0 1 Knapsack Problem in JavaScript?

easy

A. const dp = new Array(n + 1).fill(new Array(W + 1).fill(0));

B. const dp = Array(n + 1).fill(Array(W + 1).fill(0));

C. const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));

D. const dp = Array(n).fill(Array(W).fill(0));

Solution

1. Step 1: Understand DP array initialization

   We need a 2D array where each row is a separate array to avoid reference issues.

2. Step 2: Choose correct initialization

   Using Array.from with a function creates independent inner arrays, preventing shared references.

3. Final Answer:

   const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0)); -> Option C

4. Quick Check:

   Independent inner arrays = A [OK]

Hint: Use Array.from with function to avoid shared references [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Initializing with wrong dimensions
• Forgetting to fill inner arrays

3. What is the output of the following code snippet for weights = [1, 3, 4], values = [15, 50, 60], and W = 5?

function knapsack(weights, values, W) {
  const n = weights.length;
  const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));
  for (let i = 1; i <= n; i++) {
    for (let w = 1; w <= W; w++) {
      if (weights[i - 1] <= w) {
        dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] + values[i - 1]);
      } else {
        dp[i][w] = dp[i - 1][w];
      }
    }
  }
  return dp[n][W];
}

console.log(knapsack([1, 3, 4], [15, 50, 60], 5));

medium

A. 65

B. 80

C. 60

D. 75

Solution

1. Step 1: Trace DP table filling

   Check items one by one and weights up to 5:
   - Item 1 (weight 1, value 15): dp[1][1..5] = 15
   - Item 2 (weight 3, value 50): dp[2][3] = max(15, 0+50)=50, dp[2][4]=max(15,15+50)=65, dp[2][5]=max(15,15+50)=65
   - Item 3 (weight 4, value 60): dp[3][4]=max(65,0+60)=65, dp[3][5]=max(65,15+60)=75

2. Step 2: Find final answer

   The maximum value for weight 5 is dp[3][5] = 75.

3. Final Answer:

   75 -> Option D

4. Quick Check:

   Max value at W=5 = 75 [OK]

Hint: Trace DP table row by row for each item [OK]

Common Mistakes:

• Adding values without checking weight limit
• Confusing indices in dp array
• Forgetting to compare including vs excluding item

4. Identify the bug in this 0 1 Knapsack implementation and select the fix:

function knapsack(weights, values, W) {
  const n = weights.length;
  const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));
  for (let i = 1; i <= n; i++) {
    for (let w = 1; w <= W; w++) {
      if (weights[i] <= w) {
        dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - weights[i]] + values[i]);
      } else {
        dp[i][w] = dp[i - 1][w];
      }
    }
  }
  return dp[n][W];
}

medium

A. Change weights[i] and values[i] to weights[i - 1] and values[i - 1]

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

C. Initialize dp with zeros using fill(1)

D. Change inner loop to run from w = 0 to W - 1

Solution

1. Step 1: Identify index error in array access

   Arrays in JavaScript are zero-indexed, but loop starts at 1, so weights[i] accesses out-of-bound or wrong element.

2. Step 2: Correct index usage

   Use weights[i - 1] and values[i - 1] to access correct item corresponding to dp row i.

3. Final Answer:

   Change weights[i] and values[i] to weights[i - 1] and values[i - 1] -> Option A

4. Quick Check:

   Fix index offset = D [OK]

Hint: Remember dp row i corresponds to item i-1 [OK]

Common Mistakes:

• Using wrong indices causing undefined values
• Changing loop start incorrectly
• Incorrect dp initialization

5. Given items with weights = [2, 3, 5, 7], values = [10, 5, 15, 7], and knapsack capacity W = 10, which set of items maximizes value without exceeding weight? Select the correct maximum value.
Note: You can only take each item once.

hard

A. 25

B. 30

C. 27

D. 22

Solution

1. Step 1: Consider all item combinations within weight limit 10

   Possible combinations:
   - Items 1,2,3: weights 2+3+5=10, values 10+5+15=30 (exceeds capacity? No, equals 10)
   - Items 1,3,4: weights 2+5+7=14 (exceeds 10)
   - Items 2,3,4: weights 3+5+7=15 (exceeds 10)
   - Items 1,4: weights 2+7=9, values 10+7=17
   - Items 3 alone: weight 5, value 15
   - Items 1,2,4: weights 2+3+7=12 (exceeds 10)

2. Step 2: Find max value without exceeding weight

   Best valid combination is items 1,2,3 with total weight 10 and value 30. But since weight is exactly 10, it's valid.
   Check if any other combination gives more value:
   Items 1,3: weight 7, value 25
   Items 3,4: weight 12 (exceeds)
   So max value is 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Max value with W=10 = 30 [OK]

Hint: Try all combos within weight limit to find max value [OK]

Common Mistakes:

• Ignoring exact weight limit equality
• Adding values of items exceeding capacity
• Missing best combination