0 1 Knapsack Problem in DSA Javascript - 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 carry 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 in your backpack without exceeding the weight limit.

📋 What You'll Learn

Create an array of items with exact weights and values

Create a variable for the maximum weight the backpack can carry

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, resource allocation, budget management, and many optimization problems in real life use the 0 1 Knapsack approach.

💼 Career

Understanding dynamic programming and optimization problems like 0 1 Knapsack is important for software engineers, data scientists, and algorithm developers.

1

Create the items array

Create a constant array called items with these exact objects: { weight: 2, value: 3 }, { weight: 3, value: 4 }, { weight: 4, value: 5 }, { weight: 5, value: 8 }

DSA Javascript

# Your code here

Hint

Use an array of objects with weight and value properties.

2

Set the maximum backpack weight

Create a constant called maxWeight and set it to 5

DSA Javascript

const items = [
  { weight: 2, value: 3 },
  { weight: 3, value: 4 },
  { weight: 4, value: 5 },
  { weight: 5, value: 8 }
];
# Your code here

Hint

This is the weight limit your backpack can carry.

3

Implement the 0 1 Knapsack solution

Create a function called knapsack that takes items and maxWeight as parameters. Inside, create a 2D array dp with dimensions items.length + 1 by maxWeight + 1 filled with zeros. Use nested for loops with variables i from 1 to items.length and w from 1 to maxWeight. For each item, if its weight is less than or equal to w, set dp[i][w] to the maximum of dp[i-1][w] and dp[i-1][w - items[i-1].weight] + items[i-1].value. Otherwise, set dp[i][w] to dp[i-1][w]. Return dp[items.length][maxWeight].

DSA Javascript

const items = [
  { weight: 2, value: 3 },
  { weight: 3, value: 4 },
  { weight: 4, value: 5 },
  { weight: 5, value: 8 }
];

const maxWeight = 5;

# Your code here

Hint

Use a 2D array to store max values for subproblems. Compare including or excluding each item.

4

Print the maximum value

Call the knapsack function with items and maxWeight and print the result using console.log.

DSA Javascript

const items = [
  { weight: 2, value: 3 },
  { weight: 3, value: 4 },
  { weight: 4, value: 5 },
  { weight: 5, value: 8 }
];

const maxWeight = 5;

function knapsack(items, maxWeight) {
  const dp = Array(items.length + 1).fill(null).map(() => Array(maxWeight + 1).fill(0));

  for (let i = 1; i <= items.length; i++) {
    for (let w = 1; w <= maxWeight; w++) {
      if (items[i - 1].weight <= w) {
        dp[i][w] = Math.max(
          dp[i - 1][w],
          dp[i - 1][w - items[i - 1].weight] + items[i - 1].value
        );
      } else {
        dp[i][w] = dp[i - 1][w];
      }
    }
  }

  return dp[items.length][maxWeight];
}

# Your code here

Hint

The maximum value that fits in the backpack with weight limit 5 is 8.

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 without exceeding the value limit

C. The number of items selected regardless of weight

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: Identify the constraint

   The total weight of selected items must not exceed the knapsack's weight limit.

3. Final Answer:

   The total value of items without exceeding the weight limit -> Option A

4. Quick Check:

   Maximize value with weight limit = B [OK]

Hint: Remember: maximize value, not weight or count [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Thinking the problem maximizes number of items
• Ignoring the weight constraint

2. Which of the following is the correct way to initialize the DP array for the 0 1 Knapsack Problem in JavaScript?

easy

A. const dp = new Array(n + 1).fill(new Array(W + 1).fill(0));

B. const dp = Array(n + 1).fill(Array(W + 1).fill(0));

C. const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));

D. const dp = Array(n).fill(Array(W).fill(0));

Solution

1. Step 1: Understand DP array initialization

   We need a 2D array where each row is a separate array to avoid reference issues.

2. Step 2: Choose correct initialization

   Using Array.from with a function creates independent inner arrays, preventing shared references.

3. Final Answer:

   const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0)); -> Option C

4. Quick Check:

   Independent inner arrays = A [OK]

Hint: Use Array.from with function to avoid shared references [OK]

Common Mistakes:

• Using fill with array directly causing shared references
• Initializing with wrong dimensions
• Forgetting to fill inner arrays

3. What is the output of the following code snippet for weights = [1, 3, 4], values = [15, 50, 60], and W = 5?

function knapsack(weights, values, W) {
  const n = weights.length;
  const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));
  for (let i = 1; i <= n; i++) {
    for (let w = 1; w <= W; w++) {
      if (weights[i - 1] <= w) {
        dp[i][w] = Math.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];
}

console.log(knapsack([1, 3, 4], [15, 50, 60], 5));

medium

A. 65

B. 80

C. 60

D. 75

Solution

1. Step 1: Trace DP table filling

   Check items one by one and weights up to 5:
   - Item 1 (weight 1, value 15): dp[1][1..5] = 15
   - Item 2 (weight 3, value 50): dp[2][3] = max(15, 0+50)=50, dp[2][4]=max(15,15+50)=65, dp[2][5]=max(15,15+50)=65
   - Item 3 (weight 4, value 60): dp[3][4]=max(65,0+60)=65, dp[3][5]=max(65,15+60)=75

2. Step 2: Find final answer

   The maximum value for weight 5 is dp[3][5] = 75.

3. Final Answer:

   75 -> Option D

4. Quick Check:

   Max value at W=5 = 75 [OK]

Hint: Trace DP table row by row for each item [OK]

Common Mistakes:

• Adding values without checking weight limit
• Confusing indices in dp array
• Forgetting to compare including vs excluding item

4. Identify the bug in this 0 1 Knapsack implementation and select the fix:

function knapsack(weights, values, W) {
  const n = weights.length;
  const dp = Array.from({ length: n + 1 }, () => Array(W + 1).fill(0));
  for (let i = 1; i <= n; i++) {
    for (let w = 1; w <= W; w++) {
      if (weights[i] <= w) {
        dp[i][w] = Math.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 weights[i] and values[i] to weights[i - 1] and values[i - 1]

B. Change loop to start i from 0 instead of 1

C. Initialize dp with zeros using fill(1)

D. Change inner loop to run from w = 0 to W - 1

Solution

1. Step 1: Identify index error in array access

   Arrays in JavaScript are zero-indexed, but loop starts at 1, so weights[i] accesses out-of-bound or wrong element.

2. Step 2: Correct index usage

   Use weights[i - 1] and values[i - 1] to access correct item corresponding to dp row i.

3. Final Answer:

   Change weights[i] and values[i] to weights[i - 1] and values[i - 1] -> Option A

4. Quick Check:

   Fix index offset = D [OK]

Hint: Remember dp row i corresponds to item i-1 [OK]

Common Mistakes:

• Using wrong indices causing undefined values
• Changing loop start incorrectly
• Incorrect dp initialization

5. Given items with weights = [2, 3, 5, 7], values = [10, 5, 15, 7], and knapsack capacity W = 10, which set of items maximizes value without exceeding weight? Select the correct maximum value.
Note: You can only take each item once.

hard

A. 25

B. 30

C. 27

D. 22

Solution

1. Step 1: Consider all item combinations within weight limit 10

   Possible combinations:
   - Items 1,2,3: weights 2+3+5=10, values 10+5+15=30 (exceeds capacity? No, equals 10)
   - Items 1,3,4: weights 2+5+7=14 (exceeds 10)
   - Items 2,3,4: weights 3+5+7=15 (exceeds 10)
   - Items 1,4: weights 2+7=9, values 10+7=17
   - Items 3 alone: weight 5, value 15
   - Items 1,2,4: weights 2+3+7=12 (exceeds 10)

2. Step 2: Find max value without exceeding weight

   Best valid combination is items 1,2,3 with total weight 10 and value 30. But since weight is exactly 10, it's valid.
   Check if any other combination gives more value:
   Items 1,3: weight 7, value 25
   Items 3,4: weight 12 (exceeds)
   So max value is 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Max value with W=10 = 30 [OK]

Hint: Try all combos within weight limit to find max value [OK]

Common Mistakes:

• Ignoring exact weight limit equality
• Adding values of items exceeding capacity
• Missing best combination