0 1 Knapsack Problem in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP array for the knapsack problem.

DSA Typescript

const dp: number[][] = Array(n + 1).fill(null).map(() => Array([1] + 1).fill(0));

Drag options to blanks, or click blank then click option'

AW

Bn

Cweights.length

Dvalues.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using the number of items instead of the weight capacity for dp array size.

Confusing the length of values or weights arrays with the weight capacity.

The DP array has dimensions (n+1) x (W+1), where W is the maximum weight capacity.

2fill in blank

medium

Complete the code to check if the current item's weight fits in the current capacity.

DSA Typescript

if (weights[i - 1] [1] w) {

Drag options to blanks, or click blank then click option'

A<

B>

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which checks if the item is too heavy instead of fits.

Using '==' which excludes items lighter than capacity.

We check if the item's weight is less than or equal to the current capacity to decide if we can include it.

3fill in blank

hard

Fix the error in the code to correctly update the DP value when including the current item.

DSA Typescript

dp[i][w] = Math.max(dp[i - 1][w], dp[i - 1][w - [1]] + values[i - 1]);

Drag options to blanks, or click blank then click option'

Avalues[i - 1]

Bweights[i]

Cweights[i - 1]

Dvalues[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using weights[i] causes index out of range error.

Using values instead of weights for capacity subtraction.

We subtract the weight of the current item (weights[i - 1]) from the capacity to find the remaining capacity.

4fill in blank

hard

Fill both blanks to complete the loop headers for the DP table construction.

DSA Typescript

for (let i = 1; i [1] n + 1; i++) {
  for (let w = 0; w [2] W + 1; w++) {

Drag options to blanks, or click blank then click option'

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of range errors.

Using '>' or '>=' causes infinite loops or no iteration.

Loops run from 1 to n (exclusive of n+1) and 0 to W (exclusive of W+1) using '<' operator.

5fill in blank

hard

Fill all three blanks to complete the function signature and return statement for the knapsack solution.

DSA Typescript

function knapsack([1]: number[], [2]: number[], [3]: number): number {
  // function body
  return dp[n][W];
}

Drag options to blanks, or click blank then click option'

Aweights

BW

Cn

Dvalues

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up weights and values in the parameter list.

Incorrect order of parameters causing confusion.

The function takes values array, weights array, and weight capacity W as parameters.

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