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

Recall & Review

beginner

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

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

beginner

Explain the meaning of the 'capacity' in the Unbounded Knapsack Problem.

Capacity is the maximum weight or size the knapsack can hold. The goal is to maximize value without exceeding this capacity.

intermediate

In the Unbounded Knapsack Problem, why do we consider all items for every capacity from 1 to max capacity?

Because we can use unlimited copies of each item, we check all items for every capacity to find the best combination that maximizes value.

intermediate

What is the time complexity of the typical 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 capacity of the knapsack.

beginner

Show the base case initialization for the dynamic programming array in the Unbounded Knapsack Problem.

The base case is dp[0] = 0, meaning when the capacity is zero, the maximum value is zero because no items can be added.

In the Unbounded Knapsack Problem, what does 'unbounded' mean?

A You can use unlimited copies of each item

B You cannot use any item

C You can use each item only once

D You must use all items

Unbounded means you can use unlimited copies of each item to maximize the total value.

What is the main goal of the Unbounded Knapsack Problem?

A Maximize the total value without exceeding capacity

B Use all items exactly once

C Minimize the total weight

D Find the lightest item

The goal is to maximize the total value of items in the knapsack without exceeding its capacity.

Which approach is commonly used to solve the Unbounded Knapsack Problem efficiently?

A Greedy algorithm

B Sorting items by weight

C Dynamic programming

D Brute force without memoization

Dynamic programming is used to efficiently find the maximum value by building solutions for smaller capacities.

If the knapsack capacity is 0, what is the maximum value achievable?

A Sum of all item values

B Depends on items

C Infinity

D 0

With zero capacity, no items can be added, so the maximum value is zero.

What does dp[w] represent in the dynamic programming solution for Unbounded Knapsack?

A Weight of the heaviest item

B Maximum value for capacity w

C Number of items used

D Minimum value for capacity w

dp[w] stores the maximum value achievable with knapsack capacity w.

Describe the steps to solve the Unbounded Knapsack Problem using dynamic programming.

Think about building solutions from smaller capacities to larger.

Explain why the Unbounded Knapsack Problem allows multiple copies of the same item and how this affects the solution approach.

Focus on the difference in item usage and its impact on dp updates.

Practice

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

easy

A. In Unbounded Knapsack, you can use unlimited copies of each item.

B. In Unbounded Knapsack, items cannot be repeated.

C. Unbounded Knapsack only works with fractional items.

D. Unbounded Knapsack has a fixed number of items to choose from.

Solution

1. Step 1: Understand the item usage rules

   In 0/1 Knapsack, each item can be chosen at most once. In Unbounded Knapsack, items can be chosen unlimited times.

2. Step 2: Compare the problem constraints

   Unbounded Knapsack allows repeated use of the same item, unlike 0/1 Knapsack.

3. Final Answer:

   In Unbounded Knapsack, you can use unlimited copies of each item. -> Option A

4. Quick Check:

   Unlimited item usage = In Unbounded Knapsack, you can use unlimited copies of each item. [OK]

Hint: Unbounded means unlimited items allowed [OK]

Common Mistakes:

• Confusing unbounded with fractional knapsack
• Thinking items cannot repeat
• Assuming fixed item count

2. Which of the following TypeScript function signatures correctly represents the Unbounded Knapsack function?

easy

A. function unboundedKnapsack(W: number, wt: string[], val: number[], n: number): number

B. function unboundedKnapsack(W: string, wt: number[], val: number[], n: number): number

C. function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number

D. function unboundedKnapsack(W: number, wt: number[], val: number[], n: string): number

Solution

1. Step 1: Check parameter types for correctness

   Weight capacity W should be a number, weights wt and values val arrays of numbers, and n the number of items as a number.

2. Step 2: Identify the correct signature

   function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number matches all types correctly. Others have wrong types like string instead of number.

3. Final Answer:

   function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number -> Option C

4. Quick Check:

   Correct types for all parameters = function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number [OK]

Hint: Check parameter types carefully [OK]

Common Mistakes:

• Using string instead of number for weights or capacity
• Mixing array types
• Incorrect return type

3. What will be the output of the following TypeScript code snippet for Unbounded Knapsack?

const val = [10, 40, 50];
const wt = [1, 3, 4];
const W = 6;

function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number {
  const dp = Array(W + 1).fill(0);
  for (let i = 1; i <= W; i++) {
    for (let j = 0; j < n; j++) {
      if (wt[j] <= i) {
        dp[i] = Math.max(dp[i], dp[i - wt[j]] + val[j]);
      }
    }
  }
  return dp[W];
}

console.log(unboundedKnapsack(W, wt, val, val.length));

medium

A. 90

B. 100

C. 120

D. 80

Solution

1. Step 1: Understand the DP array construction

   We build dp array from 0 to W=6. For each capacity, we check all items and update dp[i] with max value including unlimited items.

2. Step 2: Calculate dp values step-by-step

   dp[1]=max(0, dp[0]+10)=10
   dp[2]=max(0, dp[1]+10)=20
   dp[3]=max(dp[3], dp[0]+40)=40, also dp[3]=max(40, dp[2]+10)=40
   dp[4]=max(dp[4], dp[1]+40)=50, also dp[4]=max(50, dp[0]+50)=50
   dp[5]=max(dp[5], dp[2]+40)=60, also dp[5]=max(60, dp[1]+50)=60
   dp[6]=max(dp[6], dp[3]+40)=80, also dp[6]=max(80, dp[2]+50)=90

3. Final Answer:

   90 -> Option A

4. Quick Check:

   Max value for W=6 = 90 [OK]

Hint: Build dp array incrementally for each capacity [OK]

Common Mistakes:

• Not considering multiple copies of items
• Incorrect dp array updates
• Mixing 0/1 and unbounded logic

4. Identify the bug in this Unbounded Knapsack TypeScript code snippet:

function unboundedKnapsack(W: number, wt: number[], val: number[], n: number): number {
  const dp = Array(W + 1).fill(0);
  for (let j = 0; j < n; j++) {
    for (let i = 1; i <= W; i++) {
      if (wt[j] <= i) {
        dp[i] = Math.max(dp[i], dp[i - wt[j]] + val[j]);
      }
    }
  }
  return dp[W];
}

medium

A. The condition wt[j] <= i should be wt[j] < i

B. Initialization of dp array is incorrect

C. Return statement should be dp[0] instead of dp[W]

D. Loops are reversed; inner loop should be over items, outer over capacity

Solution

1. Step 1: Analyze loop order for unbounded knapsack

   For unbounded knapsack, outer loop should iterate over capacities (i), inner loop over items (j) to allow multiple uses of items.

2. Step 2: Identify the bug in loop order

   Here, outer loop is over items and inner over capacity, which is incorrect and leads to wrong results.

3. Final Answer:

   Loops are reversed; inner loop should be over items, outer over capacity -> Option D

4. Quick Check:

   Correct loop order = capacity outer, items inner [OK]

Hint: Outer loop on capacity for unbounded knapsack [OK]

Common Mistakes:

• Swapping loop order
• Wrong dp initialization
• Incorrect return index

5. Given items with values val = [5, 10, 15] and weights wt = [1, 3, 4], and a knapsack capacity W = 7, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 25

B. 30

C. 35

D. 40

Solution

1. Step 1: Calculate dp array values for capacities 1 to 7

   dp[0]=0
   dp[1]=max(0, dp[0]+5)=5
   dp[2]=max(0, dp[1]+5)=10
   dp[3]=max(dp[3], dp[0]+10)=10, also dp[3]=max(10, dp[2]+5)=10
   dp[4]=max(dp[4], dp[1]+10)=15, also dp[4]=max(15, dp[0]+15)=15
   dp[5]=max(dp[5], dp[2]+10)=20, also dp[5]=max(20, dp[1]+15)=20
   dp[6]=max(dp[6], dp[3]+10)=20, also dp[6]=max(20, dp[2]+15)=25
   dp[7]=max(dp[7], dp[4]+10)=25, also dp[7]=max(25, dp[3]+15)=25

2. Step 2: Find maximum value at capacity 7

   The maximum value achievable is 25, but check if better combinations exist:
   Try dp[7] = dp[3] + val[2] = 10 + 15 = 25
   Try dp[7] = dp[4] + val[1] = 15 + 10 = 25
   Try dp[7] = dp[6] + val[0] = 25 + 5 = 30 (better)
   Update dp[7] = 30
   Try dp[7] = dp[3] + val[2] = 10 + 15 = 25 (less than 30)
   So final dp[7] = 30

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Maximum value for W=7 = 30 [OK]

Hint: Check all item combinations for each capacity [OK]

Common Mistakes:

• Ignoring multiple item usage
• Stopping at first max value found
• Not updating dp array correctly