0 1 Knapsack Problem in DSA Typescript - Time & Space Complexity

Time Complexity: 0 1 Knapsack Problem

O(n * W)

Understanding Time Complexity

We want to understand how the time needed to solve the 0 1 Knapsack problem grows as the number of items and the knapsack capacity increase.

How does the number of steps change when we add more items or increase the capacity?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function knapsack(weights: number[], values: number[], capacity: number): number {
  const n = weights.length;
  const dp: number[][] = Array.from({ length: n + 1 }, () => 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], dp[i - 1][w - weights[i - 1]] + values[i - 1]);
      } else {
        dp[i][w] = dp[i - 1][w];
      }
    }
  }
  return dp[n][capacity];
}
    

This code solves the 0 1 Knapsack problem using dynamic programming by filling a table of size (number of items + 1) by (capacity + 1).

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over items and capacity filling the dp table.
• How many times: Outer loop runs n times (number of items), inner loop runs capacity times.

How Execution Grows With Input

As we add more items or increase capacity, the number of steps grows by multiplying these two factors.

Input Size (n items, capacity W)	Approx. Operations
10 items, capacity 50	About 10 * 50 = 500 steps
100 items, capacity 1000	About 100 * 1000 = 100,000 steps
1000 items, capacity 10,000	About 1000 * 10,000 = 10,000,000 steps

Pattern observation: The steps grow roughly by multiplying the number of items by the capacity, so doubling either doubles the work.

Final Time Complexity

Time Complexity: O(n * W)

This means the time needed grows proportionally with both the number of items and the knapsack capacity.

Common Mistake

[X] Wrong: "The time complexity depends only on the number of items, not the capacity."

[OK] Correct: The inner loop runs for every capacity value up to W, so capacity directly affects the total steps.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over brute force and why capacity matters in performance.

Self-Check

"What if we used a recursive solution without memoization? How would the time complexity change?"

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