0 1 Knapsack Problem in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table with zeros.

DSA Python

dp = [[[1] for _ in range(capacity + 1)] for _ in range(n + 1)]

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

A-1

BNone

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 1 or None instead of 0.

The DP table should be initialized with zeros because initially no items are taken, so the maximum value is zero.

2fill in blank

medium

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

DSA Python

if weights[i - 1] [1] w:

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A<=

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of <=.

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

3fill in blank

hard

Fix the error in the code to correctly update the DP table when including the current item.

DSA Python

dp[i][w] = max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] [1] values[i - 1])

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

A+

B-

C//

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

We add the current item's value to the best value without this item's weight to update the DP table.

4fill in blank

hard

Fill both blanks to complete the loop that builds the DP table.

DSA Python

for i in range(1, [1] + 1):
    for w in range(1, [2] + 1):

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

An

Bcapacity

Cweights

Dvalues

Attempts:

3 left

💡 Hint

Common Mistakes

Using weights or values instead of n or capacity.

The outer loop runs over items (n), and the inner loop runs over capacities (capacity) to fill the DP table.

5fill in blank

hard

Fill all three blanks to complete the function that returns the maximum value for the knapsack.

DSA Python

def knapsack([1], [2], [3]):
    n = len(values)
    dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]
    for i in range(1, n + 1):
        for w in range(1, capacity + 1):
            if weights[i - 1] <= w:
                dp[i][w] = 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][capacity]

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

Aweights

Bvalues

Ccapacity

Ditems

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'items' instead of weights or values.

The function takes weights, values, and capacity as inputs to solve the knapsack problem.

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 regardless of value

C. The number of items selected

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: Clarify constraints

   Each item can be chosen only once (0 or 1 times), so the goal is to maximize value under weight limit.

3. Final Answer:

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

4. Quick Check:

   Maximize value under weight limit = The total value of items without exceeding the weight limit [OK]

Hint: Focus on maximizing value, not weight or count [OK]

Common Mistakes:

• Confusing weight maximization with value maximization
• Thinking the problem counts items instead of values
• Ignoring the weight limit constraint

2. Which of the following is the correct Python syntax to initialize a 2D list dp of size (n+1) x (W+1) filled with zeros for the 0 1 Knapsack DP table?

easy

A. dp = [[0 for _ in range(W+1)] for _ in range(n+1)]

B. dp = [[0] * n for _ in range(W)]

C. dp = [0] * (n+1) * (W+1)

D. dp = [[0] * (W+1)] * (n+1)

Solution

1. Step 1: Understand 2D list initialization

   Using list comprehension with nested loops creates independent inner lists, avoiding shared references.

2. Step 2: Analyze each option

   dp = [[0 for _ in range(W+1)] for _ in range(n+1)] correctly uses nested list comprehension to create a (n+1) x (W+1) zero matrix.

3. Final Answer:

   dp = [[0 for _ in range(W+1)] for _ in range(n+1)] -> Option A

4. Quick Check:

   Nested list comprehension for 2D zero matrix = dp = [[0 for _ in range(W+1)] for _ in range(n+1)] [OK]

Hint: Use nested list comprehension for 2D lists [OK]

Common Mistakes:

• Using multiplication for inner lists causing shared references
• Creating 1D list instead of 2D
• Swapping dimensions incorrectly

3. What is the output of the following Python code snippet implementing 0 1 Knapsack DP for items with weights [1, 2, 3] and values [10, 20, 30] and capacity 5?

def knapsack(weights, values, W):
    n = len(weights)
    dp = [[0]*(W+1) for _ in range(n+1)]
    for i in range(1, n+1):
        for w in range(1, W+1):
            if weights[i-1] <= w:
                dp[i][w] = 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]

print(knapsack([1,2,3], [10,20,30], 5))

medium

A. 60

B. 40

C. 50

D. 30

Solution

1. Step 1: Trace DP table filling

   Items: (1,10), (2,20), (3,30), Capacity=5. Check combinations:
   - Take items 2 and 3: weight=5, value=50
   - Take items 1 and 3: weight=4, value=40
   - Take all items 1,2,3: weight=6 (exceeds capacity)
   - Take items 1 and 2: weight=3, value=30

2. Step 2: Find max value within capacity

   Best is items 2 and 3 with total value 50, but check if better:
   Actually, items 1,2,3 total weight 6 > 5, so not allowed.
   Items 1,2,3 can't all be taken.
   Check items 1,2,3 subsets:
   Items 2,3: 5 weight, 50 value
   Items 1,3: 4 weight, 40 value
   Items 1,2: 3 weight, 30 value
   So max value is 50.

3. Final Answer:

   50 -> Option C

4. Quick Check:

   Max value under weight 5 = 50 [OK]

Hint: Check all item subsets under capacity for max value [OK]

Common Mistakes:

• Assuming all items fit without checking weight
• Mixing up indices in DP table
• Forgetting to subtract weight when adding value

4. Find the error in this 0 1 Knapsack DP code snippet and select the fix:

def knapsack(weights, values, W):
    n = len(weights)
    dp = [[0]*(W+1) for _ in range(n+1)]
    for i in range(1, n+1):
        for w in range(1, W+1):
            if weights[i] <= w:
                dp[i][w] = 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 range(1, W+1) to range(W)

B. Change dp[i-1][w] to dp[i][w]

C. Initialize dp with ones instead of zeros

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

Solution

1. Step 1: Identify index error in array access

   Python lists are zero-indexed, but loop uses i from 1 to n, so weights[i] and values[i] cause IndexError or wrong access.

2. Step 2: Correct indexing for weights and values

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

3. Final Answer:

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

4. Quick Check:

   Fix indexing to avoid off-by-one errors = Change weights[i] to weights[i-1] and values[i] to values[i-1] [OK]

Hint: Use i-1 to index weights and values inside loops [OK]

Common Mistakes:

• Using i instead of i-1 for zero-based lists
• Changing dp indices incorrectly
• Altering loop ranges unnecessarily

5. Given items with weights [2, 3, 4, 5] and values [3, 4, 5, 6], and a knapsack capacity of 5, which items should be selected to maximize value? Also, what is the maximum value achievable?

hard

A. Select item with weight 5 only; max value 6

B. Select items with weights 2 and 3; max value 7

C. Select items with weights 3 and 4; max value 9

D. Select items with weights 2 and 4; max value 8

Solution

1. Step 1: List possible item combinations under capacity 5

   Check combinations:
   - Item 1 (2,3) + Item 2 (3,4): total weight 5, value 7
   - Item 3 (4,5) alone: weight 4, value 5
   - Item 4 (5,6) alone: weight 5, value 6
   - Item 2 (3,4) + Item 3 (4,5): weight 7 (exceeds capacity)
   - Item 1 (2,3) + Item 3 (4,5): weight 6 (exceeds capacity)

2. Step 2: Choose combination with max value

   Items 1 and 2 together fit exactly and give value 7, which is higher than any single item value under capacity.

3. Final Answer:

   Select items with weights 2 and 3; max value 7 -> Option B

4. Quick Check:

   Max value with capacity 5 = 7 by items 2 and 3 weights [OK]

Hint: Try all item pairs fitting capacity, pick max value [OK]

Common Mistakes:

• Choosing heavier single item over lighter combo
• Ignoring exact capacity limit
• Assuming max value is from single item