Unbounded Knapsack Problem in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Unbounded Knapsack Problem?

It is a problem where you have a knapsack with a weight limit and unlimited copies of each item. The goal is to maximize the total value without exceeding the weight limit.

beginner

How does the Unbounded Knapsack Problem differ from the 0/1 Knapsack Problem?

In the Unbounded Knapsack, you can use unlimited copies of each item, while in 0/1 Knapsack, each item can be used only once.

intermediate

What is the main idea behind the dynamic programming solution for the Unbounded Knapsack Problem?

Build a table where each entry represents the maximum value for a given weight. For each weight, try all items and update the maximum value by including the item multiple times if possible.

beginner

In the Unbounded Knapsack Problem, why do we consider the same item multiple times during the solution?

Because there is no limit on the number of copies of each item, so we can add the same item repeatedly as long as the total weight does not exceed the limit.

intermediate

What is the time complexity of the dynamic programming solution for the Unbounded Knapsack Problem?

The time complexity is O(n * W), where n is the number of items and W is the maximum weight capacity of the knapsack.

In the Unbounded Knapsack Problem, what can you do with each item?

A Use unlimited copies of each item

B Use each item only once

C Use each item at most twice

D Use no items

The Unbounded Knapsack allows unlimited copies of each item.

What does the dynamic programming table store in the Unbounded Knapsack solution?

A Number of items used

B List of items chosen

C Minimum weight for each value

D Maximum value for each weight limit

The table stores the maximum value achievable for each weight limit.

What is the main difference between Unbounded and 0/1 Knapsack problems?

A 0/1 allows multiple copies; Unbounded does not

B Unbounded allows multiple copies of items; 0/1 does not

C Both allow unlimited copies

D Neither allows multiple copies

Unbounded Knapsack allows unlimited copies of each item, unlike 0/1 Knapsack.

What is the time complexity of the Unbounded Knapsack dynamic programming solution?

A O(n * W)

B O(n^2)

C O(W^2)

D O(n + W)

The solution runs in O(n * W), where n is items count and W is weight capacity.

Why can the same item be chosen multiple times in Unbounded Knapsack?

A Because weight limit is ignored

B Because items are free

C Because there is no limit on item copies

D Because items have zero weight

The problem allows unlimited copies of each item, so the same item can be chosen multiple times.

Explain the Unbounded Knapsack Problem and how dynamic programming solves it.

Think about how you can fill the knapsack by repeatedly adding items.

Describe the difference between Unbounded Knapsack and 0/1 Knapsack problems.

Focus on how many times you can use each item.

Practice

1. What is the key difference between Unbounded Knapsack and 0/1 Knapsack?

easy

A. Unbounded Knapsack uses recursion while 0/1 Knapsack uses iteration

B. In Unbounded Knapsack each item can be chosen any number of times; in 0/1 each item is chosen at most once

C. Unbounded Knapsack has no weight constraint

D. 0/1 Knapsack allows fractional items while Unbounded does not

Solution

1. Step 1: Understand 0/1 constraint

   In 0/1 Knapsack each item has a binary choice - include once or exclude. After including, it cannot be reused.

2. Step 2: Understand Unbounded constraint

   In Unbounded Knapsack there is no limit on how many times an item can be selected. The same item can fill the bag.

3. Step 3: Practical impact

   This changes the recurrence: 0/1 uses dp[i-1][w-wt] (previous row); Unbounded uses dp[i][w-wt] (same row, enabling reuse).

4. Final Answer:

   Unbounded allows unlimited item repetition. -> Option B

5. Quick Check:

   Unbounded = unlimited copies. 0/1 = at most one per item. [OK]

Hint: Unbounded = unlimited copies. 0/1 = at most one copy per item. [OK]

Common Mistakes:

• Confusing Unbounded with Fractional Knapsack
• Thinking 0/1 allows unlimited items
• Believing recursion vs iteration distinguishes them

2. In the 1D bottom-up Unbounded Knapsack solution, why does the inner loop go left-to-right (from weight[i] to W)?

medium

A. Left-to-right is faster in JavaScript due to array caching

B. Left-to-right allows the same item to be reused in one pass - exactly what Unbounded Knapsack requires

C. Right-to-left would cause negative indices

D. Left-to-right matches the order of the items array

Solution

1. Step 1: Left-to-right reuse

   Computing dp[w] using dp[w - weight[i]] left-to-right means dp[w - weight[i]] has already been updated - it reflects using item i again. This is intentional reuse.

2. Step 2: Contrast with 0/1

   0/1 Knapsack goes right-to-left to prevent reuse - ensuring dp[w - weight[i]] holds the pre-item value.

3. Final Answer:

   Left-to-right enables intentional item reuse. -> Option B

4. Quick Check:

   Left-to-right = reuse. Right-to-left = no reuse. [OK]

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

Common Mistakes:

• Thinking direction affects performance not semantics
• Not knowing left-to-right enables reuse
• Confusing with 0/1 which requires right-to-left

3. Given items with weights [1, 3, 4] and values [10, 40, 50] and capacity W=6, what is the maximum value for Unbounded Knapsack?

medium

A. 60

B. 80

C. 90

D. 100

Solution

1. Step 1: Try repeating each item

   Item 1 (w=1, v=10) × 6 = value 60. Item 2 (w=3, v=40) × 2 = value 80, weight=6 exactly. Item 3 (w=4, v=50) + item 1 × 2 = value 70.

2. Step 2: Find maximum

   Two copies of item 2 gives value 80 using exactly capacity 6 - this is the maximum.

3. Final Answer:

   80 -> Option B

4. Quick Check:

   Item 2 × 2: 40×2=80, weight 3×2=6. [OK]

Hint: Try repeating each item to fill capacity: item 2 twice (3×2=6, 40×2=80) wins. [OK]

Common Mistakes:

• Only trying single-item combinations
• Not trying repeated use of the same item
• Confusing with 0/1 where each item used at most once

4. What is the time and space complexity of the optimized 1D Unbounded Knapsack solution?

medium

A. Time O(n×W), Space O(n×W)

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

C. Time O(n×W), Space O(W)

D. Time O(n log W), Space O(W)

Solution

1. Step 1: Time complexity

   Two nested loops - n items × W+1 capacities. Each cell O(1). Total O(n×W).

2. Step 2: Space complexity

   Only a 1D array dp[W+1] is needed, updated in place. Space O(W).

3. Step 3: Contrast with 2D

   Standard 2D DP uses O(n×W) space. 1D optimization reduces to O(W).

4. Final Answer:

   Time O(n×W), Space O(W) -> Option C

5. Quick Check:

   1D = O(n×W) time, O(W) space. [OK]

Hint: 1D Unbounded = O(n×W) time, O(W) space - one array updated in place. [OK]

Common Mistakes:

• Reporting O(n×W) space for 1D version
• Confusing with O(2^n) exponential recursion
• Forgetting 1D reduces space to O(W)

5. Which real-world problem is a direct application of Unbounded Knapsack?

hard

A. Job scheduling where each job can be done only once

B. Coin change - finding minimum coins to make an amount where each denomination can be used unlimited times

C. Activity selection where activities have deadlines

D. Matrix chain multiplication to minimize operations

Solution

1. Step 1: Map coin change to Knapsack

   Coin denominations = item weights. Target amount = capacity W. Each denomination usable unlimited times = Unbounded constraint.

2. Step 2: Rule out other options

   Job scheduling once per job = 0/1 Knapsack. Activity selection = greedy. Matrix chain = interval DP.

3. Final Answer:

   Coin change with unlimited denominations. -> Option B

4. Quick Check:

   Coin change = Unbounded. Coins used unlimited times, amount = capacity. [OK]

Hint: Coin change = Unbounded Knapsack. Each coin denomination reusable, amount is capacity. [OK]

Common Mistakes:

• Mapping job scheduling to Unbounded instead of 0/1
• Not recognizing coin change as Knapsack
• Thinking activity selection uses Unbounded DP