0 1 Knapsack Problem in DSA Typescript - 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 without exceeding the backpack's weight limit.

📋 What You'll Learn

Create an array of items with exact weights and values

Create a variable for the 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, resource allocation, budget planning, and many optimization problems use the 0 1 Knapsack approach.

💼 Career

Understanding dynamic programming and optimization algorithms is valuable for software engineering roles, especially in fields like logistics, finance, and game development.

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 Typescript

// Your code here

Hint

Use an array of objects with weight and value properties.

2

Set the maximum weight capacity

Create a constant called maxWeight and set it to 5

DSA Typescript

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 algorithm

Create a function called knapsack that takes items and maxWeight as parameters and returns the maximum value. Use a 2D array dp where dp[i][w] stores the max value using first i items with weight limit w. Fill dp using nested for loops over items and weights. Return dp[items.length][maxWeight].

DSA Typescript

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 dynamic programming with a 2D array to store max values for subproblems.

4

Print the maximum value

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

DSA Typescript

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

const maxWeight = 5;

function knapsack(items: { weight: number; value: number }[], maxWeight: number): number {
  const n = items.length;
  const dp: number[][] = Array.from({ length: n + 1 }, () => Array(maxWeight + 1).fill(0));

  for (let i = 1; i <= n; i++) {
    const { weight, value } = items[i - 1];
    for (let w = 0; w <= maxWeight; w++) {
      if (weight <= w) {
        dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - weight] + value);
      } else {
        dp[i][w] = dp[i - 1][w];
      }
    }
  }

  return dp[n][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 minimum weight of items selected

B. The total weight of items without exceeding the value limit

C. The number of items selected regardless of weight

D. The total value of items without exceeding the weight limit

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 constraint is the maximum weight the knapsack can carry, so total weight must not exceed this limit.

3. Final Answer:

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

4. Quick Check:

   Maximize value under weight limit = The total value of items without exceeding the weight limit [OK]

Hint: Maximize value, not weight, under capacity limit [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Trying to maximize weight instead of value
• Ignoring the weight constraint

2. Which of the following TypeScript function signatures correctly represents the 0 1 Knapsack function?

easy

A. function knapsack(weights: number[], values: number[], capacity: number): number

B. function knapsack(weights: number, values: number, capacity: number[]): number[]

C. function knapsack(weights: string[], values: string[], capacity: string): string

D. function knapsack(weights: number[], values: number[], capacity: string): void

Solution

1. Step 1: Check parameter types

   Weights and values should be arrays of numbers, capacity a number, and return type a number representing max value.

2. Step 2: Match correct signature

   function knapsack(weights: number[], values: number[], capacity: number): number matches this: arrays for weights and values, number for capacity, returns number.

3. Final Answer:

   function knapsack(weights: number[], values: number[], capacity: number): number -> Option A

4. Quick Check:

   Correct types and return = function knapsack(weights: number[], values: number[], capacity: number): number [OK]

Hint: Weights and values are number arrays; capacity is a number [OK]

Common Mistakes:

• Using string types instead of number
• Swapping array and single number types
• Incorrect return type

3. What is the output of the following TypeScript code snippet for 0 1 Knapsack?

const weights = [1, 3, 4];
const values = [15, 50, 60];
const capacity = 5;

function knapsack(w: number[], v: number[], c: number): number {
  const n = w.length;
  const dp = Array(n + 1).fill(null).map(() => Array(c + 1).fill(0));
  for (let i = 1; i <= n; i++) {
    for (let j = 1; j <= c; j++) {
      if (w[i - 1] <= j) {
        dp[i][j] = Math.max(dp[i - 1][j], v[i - 1] + dp[i - 1][j - w[i - 1]]);
      } else {
        dp[i][j] = dp[i - 1][j];
      }
    }
  }
  return dp[n][c];
}

console.log(knapsack(weights, values, capacity));

medium

A. 65

B. 60

C. 75

D. 50

Solution

1. Step 1: Understand DP table filling

   For capacity 5, items: (1,15), (3,50), (4,60). Check combinations: - Item1 + Item2: weight 4, value 65 - Item3 alone: weight 4, value 60 - Item2 alone: weight 3, value 50 - Item1 + Item3: weight 5, value 75

2. Step 2: Find max value under capacity

   Item1 + Item3 total weight 5, value 75 is max possible under capacity 5.

3. Final Answer:

   75 -> Option C

4. Quick Check:

   Max value with capacity 5 = 75 [OK]

Hint: Check all item combos fitting capacity, pick max value [OK]

Common Mistakes:

• Ignoring combined items
• Choosing single highest value item only
• Miscomputing weights or values

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

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

medium

A. Change Array(n + 1).fill([]) to Array(n + 1).fill(null)

B. Change Array(n + 1).fill([]) to Array(n + 1).fill([]).map(() => Array(capacity + 1).fill(0))

C. No bug, code is correct

D. Change Array(n + 1).fill([]) to Array(n + 1).fill(0)

Solution

1. Step 1: Examine the dp array initialization

   The code uses Array(n + 1).fill([]).map(() => Array(capacity + 1).fill(0)), but fill([]) fills all elements with the same array reference, which is then overridden by map creating new arrays. However, the initial fill([]) is unnecessary and can cause confusion or bugs if map is removed.

2. Step 2: Correct initialization

   Better to initialize dp as Array(n + 1).fill(null).map(() => Array(capacity + 1).fill(0)) or directly with map to ensure independent arrays without shared references.

3. Final Answer:

   Change Array(n + 1).fill([]) to Array(n + 1).fill([]).map(() => Array(capacity + 1).fill(0)) -> Option B

4. Quick Check:

   Ensure dp rows are independent arrays to avoid shared reference bugs [OK]

Hint: Use map to create independent arrays for dp rows [OK]

Common Mistakes:

• Assuming fill([]) creates independent arrays
• Ignoring shared reference issues
• Claiming no bug when shared references exist

5. Given items with weights [2, 3, 5, 7] and values [10, 5, 15, 7], and a knapsack capacity of 10, which set of items maximizes value without exceeding capacity?

hard

A. Items with weights 2, 3, and 5 (total value 30)

B. Items with weights 3 and 7 (total value 12)

C. Items with weights 5 and 7 (total value 22)

D. Items with weights 2 and 7 (total value 17)

Solution

1. Step 1: Calculate total weights and values for each option

   Items with weights 2, 3, and 5 (total value 30): weights 2+3+5=10, values 10+5+15=30 Items with weights 3 and 7 (total value 12): weights 3+7=10, values 5+7=12 Items with weights 5 and 7 (total value 22): weights 5+7=12 (exceeds capacity) Items with weights 2 and 7 (total value 17): weights 2+7=9, values 10+7=17

2. Step 2: Choose max value under capacity

   Items with weights 2, 3, and 5 (total value 30) fits capacity exactly and has highest value 30.

3. Final Answer:

   Items with weights 2, 3, and 5 (total value 30) -> Option A

4. Quick Check:

   Max value under capacity 10 = 30 [OK]

Hint: Sum weights ≤ capacity; pick max value combo [OK]

Common Mistakes:

• Choosing items exceeding capacity
• Ignoring combined item values
• Selecting fewer items with lower total value