Unbounded Knapsack Problem in DSA Typescript - Build from Scratch

Unbounded Knapsack Problem

📖 Scenario: You are helping a shop owner decide how to fill a knapsack with items to maximize value. The knapsack can hold a limited weight, and the owner can use unlimited quantities of each item.

🎯 Goal: Build a program that calculates the maximum value achievable with unlimited items given their weights and values and the knapsack's capacity.

📋 What You'll Learn

Create arrays for item weights and values with exact values

Create a variable for knapsack capacity with exact value

Implement the unbounded knapsack dynamic programming logic

Print the maximum value achievable

💡 Why This Matters

🌍 Real World

This problem helps in resource allocation where you can use unlimited quantities of items, like packing goods or budgeting.

💼 Career

Understanding dynamic programming and knapsack problems is important for software engineers working on optimization, logistics, and algorithm design.

1

Create item weights and values arrays

Create an array called weights with values [2, 3, 4, 5] and an array called values with values [3, 4, 5, 6].

DSA Typescript

# Your code here

Hint

Use const weights = [2, 3, 4, 5]; and const values = [3, 4, 5, 6];

2

Create knapsack capacity variable

Create a variable called capacity and set it to 8.

DSA Typescript

const weights = [2, 3, 4, 5];
const values = [3, 4, 5, 6];
# Your code here

Hint

Use const capacity = 8; to set the knapsack capacity.

3

Implement unbounded knapsack logic

Create an array called dp of length capacity + 1 filled with zeros. Use a for loop with variable w from 1 to capacity. Inside it, use another for loop with variable i from 0 to weights.length - 1. If weights[i] is less than or equal to w, update dp[w] to the maximum of dp[w] and dp[w - weights[i]] + values[i].

DSA Typescript

const weights = [2, 3, 4, 5];
const values = [3, 4, 5, 6];
const capacity = 8;
# Your code here

Hint

Use two nested loops and update dp[w] with the maximum value possible.

4

Print the maximum value achievable

Print the value of dp[capacity] using console.log(dp[capacity]).

DSA Typescript

const weights = [2, 3, 4, 5];
const values = [3, 4, 5, 6];
const capacity = 8;

const dp = new Array(capacity + 1).fill(0);
for (let w = 1; w <= capacity; w++) {
    for (let i = 0; i < weights.length; i++) {
        if (weights[i] <= w) {
            dp[w] = Math.max(dp[w], dp[w - weights[i]] + values[i]);
        }
    }
}
# Your code here

Hint

Use console.log(dp[capacity]) to print the maximum value.

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