0 1 Knapsack Problem in DSA Python - Build from Scratch

0 1 Knapsack Problem

📖 Scenario: You are packing a backpack for a hiking trip. You have a list of items, each with a weight and a value. Your backpack can only hold a limited weight. You want to choose items to maximize the total value without exceeding the weight limit.

🎯 Goal: Build a program that uses the 0 1 Knapsack algorithm to find the maximum value of items you can carry without exceeding the backpack's weight limit.

📋 What You'll Learn

Create a list of items with exact weights and values

Set the backpack's maximum weight capacity

Implement the 0 1 Knapsack dynamic programming solution

Print the maximum value that fits in the backpack

💡 Why This Matters

🌍 Real World

Packing efficiently for trips, budgeting resources, or selecting projects with limited resources.

💼 Career

Understanding dynamic programming and optimization problems is useful in software engineering, data science, and operations research.

1

Create the list of items

Create a list called items where each item is a tuple with weight and value. Use these exact items: (2, 3), (3, 4), (4, 5), (5, 8).

DSA Python

# Your code here

Hint

Use a list of tuples where each tuple has two numbers: weight and value.

2

Set the backpack capacity

Create a variable called capacity and set it to 5 to represent the maximum weight the backpack can hold.

DSA Python

items = [(2, 3), (3, 4), (4, 5), (5, 8)]
# Your code here

Hint

Just create a variable named capacity and assign the number 5.

3

Implement the 0 1 Knapsack solution

Create a list called dp with capacity + 1 zeros. Then, use a for loop with variable weight, value to iterate over items. Inside it, use a for loop with variable w going from capacity down to weight (inclusive). Update dp[w] to be the maximum of dp[w] and dp[w - weight] + value.

DSA Python

items = [(2, 3), (3, 4), (4, 5), (5, 8)]
capacity = 5
# Your code here

Hint

Use dynamic programming with a list dp to store max values for weights up to capacity.

4

Print the maximum value

Print the last element of the dp list using print(dp[capacity]) to show the maximum value that fits in the backpack.

DSA Python

items = [(2, 3), (3, 4), (4, 5), (5, 8)]
capacity = 5
dp = [0] * (capacity + 1)
for weight, value in items:
    for w in range(capacity, weight - 1, -1):
        dp[w] = max(dp[w], dp[w - weight] + value)
# Your code here

Hint

Print the maximum value stored at dp[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 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