0 1 Knapsack Problem in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of 0 1 Knapsack DP Table

What is the final DP table (2D list) after running the 0 1 Knapsack algorithm on the given input?

DSA Python

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

weights = [1, 3, 4]
values = [15, 50, 60]
capacity = 5
result = knapsack(weights, values, capacity)
print(result)

A [[0, 0, 0, 0, 0, 0], [0, 15, 15, 15, 15, 15], [0, 15, 15, 50, 50, 65], [0, 15, 15, 50, 60, 75]]

B [[0, 0, 0, 0, 0, 0], [0, 15, 15, 15, 15, 15], [0, 15, 15, 50, 65, 65], [0, 15, 15, 50, 60, 80]]

C [[0, 0, 0, 0, 0, 0], [0, 15, 15, 15, 15, 15], [0, 15, 15, 50, 65, 65], [0, 15, 15, 50, 60, 60]]

D [[0, 0, 0, 0, 0, 0], [0, 15, 15, 15, 15, 15], [0, 15, 15, 50, 65, 65], [0, 15, 15, 50, 60, 75]]

💡 Hint

Trace the DP table row by row, considering if the current item fits or not.

The DP table stores max values for each item and capacity. For item 3 (weight 4, value 60), at capacity 5, max is max of excluding or including it. Including it adds 60 + dp of capacity 1 from previous row (15), total 75.

🧠 Conceptual

intermediate

Maximum Value in 0 1 Knapsack

Given weights = [2, 3, 5], values = [20, 30, 50], and capacity = 5, what is the maximum value achievable?

A 30

B 50

C 20

D 80

💡 Hint

Check which items fit without exceeding capacity and sum their values.

Only the item with weight 5 and value 50 fits exactly. Other combinations exceed capacity or have less value.

🔧 Debug

advanced

Identify the error in this 0 1 Knapsack code snippet

What error does this code produce when run? ```python def knapsack(weights, values, capacity): n = len(weights) dp = [[0] * (capacity + 1) for _ in range(n)] for i in range(n): for w in range(capacity + 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-1][capacity] weights = [1, 2] values = [10, 20] capacity = 3 print(knapsack(weights, values, capacity)) ```

A IndexError

B TypeError

C KeyError

D No error, outputs 30

💡 Hint

Check how dp is initialized and accessed at dp[i-1].

dp is initialized with n rows, but dp[i-1] for i=0 accesses dp[-1], which is valid in Python but logically incorrect here. Also, dp[0][w] tries to access dp[-1][w], which is out of intended bounds causing wrong results or IndexError in some environments.

🚀 Application

advanced

Find selected items in 0 1 Knapsack

After computing the DP table for weights = [1, 3, 4], values = [15, 50, 60], capacity = 5, which items are selected to achieve max value?

A Items with weights 1 and 3

B Items with weights 3 and 4

C Items with weights 1 and 4

D Only item with weight 4

💡 Hint

Trace back from dp[n][capacity] to find which items were included.

Max value 75 comes from including item with weight 4 (value 60) and item with weight 1 (value 15).

🧠 Conceptual

expert

Time Complexity of 0 1 Knapsack DP

What is the time complexity of the classic 0 1 Knapsack dynamic programming solution with n items and capacity W?

A O(n * W)

B O(n + W)

C O(n^2 * W)

D O(2^n)

💡 Hint

Consider the nested loops over items and capacity.

The DP solution uses a table of size n by W and fills each cell once, so time complexity is O(n * W).

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