Fractional Knapsack Problem in DSA Javascript - Time & Space Complexity

Time Complexity: Fractional Knapsack Problem

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to solve the Fractional Knapsack Problem changes as the number of items grows.

Specifically, how the sorting and selection steps affect the total work done.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function fractionalKnapsack(items, capacity) {
  items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight));
  let totalValue = 0;
  for (let item of items) {
    if (capacity === 0) break;
    let take = Math.min(item.weight, capacity);
    totalValue += take * (item.value / item.weight);
    capacity -= take;
  }
  return totalValue;
}
    

This code sorts items by value per weight, then takes as much as possible from the best items until capacity is full.

Identify Repeating Operations

• Primary operation: Sorting the items array by value-to-weight ratio.
• How many times: Sorting typically takes about n log n steps for n items.
• Secondary operation: One loop through all items to pick fractions.
• How many times: Loop runs at most n times, once per item.

How Execution Grows With Input

As the number of items increases, sorting takes more time, growing faster than just going through the list.

Input Size (n)	Approx. Operations
10	About 10 * log 10 ≈ 33 steps for sorting + 10 steps for picking
100	About 100 * log 100 ≈ 664 steps for sorting + 100 steps for picking
1000	About 1000 * log 1000 ≈ 9966 steps for sorting + 1000 steps for picking

Pattern observation: Sorting dominates and grows a bit faster than just the number of items.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the number of items because of sorting, but the picking step grows linearly.

Common Mistake

[X] Wrong: "The loop to pick items is the slowest part, so time complexity is O(n)."

[OK] Correct: Sorting the items first takes more time than just looping, so it controls the total time.

Interview Connect

Understanding how sorting affects the overall time helps you explain your solution clearly and shows you know where the main work happens.

Self-Check

"What if the items were already sorted by value-to-weight ratio? How would the time complexity change?"

Practice

1. What is the main goal of the Fractional Knapsack Problem?

easy

A. Maximize total value by taking fractions of items within weight limit

B. Maximize total weight without considering value

C. Select only whole items to maximize value

D. Minimize the number of items selected

Solution

1. Step 1: Understand the problem goal

   The fractional knapsack problem allows taking parts of items to maximize total value within a weight limit.

2. Step 2: Compare options

   Maximize total value by taking fractions of items within weight limit correctly states maximizing value by taking fractions. Others ignore fractions or value maximization.

3. Final Answer:

   Maximize total value by taking fractions of items within weight limit -> Option A

4. Quick Check:

   Fractional knapsack = maximize value with fractions [OK]

Hint: Remember: fractions allowed to maximize value [OK]

Common Mistakes:

• Thinking only whole items can be taken
• Ignoring value and focusing on weight
• Confusing with 0/1 knapsack problem

2. Which of the following JavaScript code snippets correctly sorts items by value-to-weight ratio for the fractional knapsack?

easy

A. items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight));

B. items.sort((a, b) => (a.weight / a.value) - (b.weight / b.value));

C. items.sort((a, b) => a.value - b.value);

D. items.sort((a, b) => a.weight - b.weight);

Solution

1. Step 1: Identify sorting criteria

   We must sort items by descending value-to-weight ratio (value/weight) to pick best items first.

2. Step 2: Check each code snippet

   items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight)); sorts descending by (value/weight). Others sort incorrectly by inverse or by value or weight alone.

3. Final Answer:

   items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight)); -> Option A

4. Quick Check:

   Sort by descending value/weight ratio [OK]

Hint: Sort items by (value/weight) descending [OK]

Common Mistakes:

• Sorting by weight or value alone
• Sorting ascending instead of descending
• Using inverse ratio (weight/value)

3. What is the output of the following JavaScript code?

const items = [
  {value: 60, weight: 10},
  {value: 100, weight: 20},
  {value: 120, weight: 30}
];
const capacity = 50;

function fractionalKnapsack(items, capacity) {
  items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight));
  let totalValue = 0;
  let remaining = capacity;
  for (const item of items) {
    if (item.weight <= remaining) {
      totalValue += item.value;
      remaining -= item.weight;
    } else {
      totalValue += item.value * (remaining / item.weight);
      break;
    }
  }
  return totalValue;
}

console.log(fractionalKnapsack(items, capacity));

medium

A. 220

B. 240

C. 180

D. 200

Solution

1. Step 1: Sort items by value/weight ratio

   Ratios: 60/10=6, 100/20=5, 120/30=4. Sorted order: item1, item2, item3.

2. Step 2: Add items until capacity 50

   Take item1 fully (weight 10, value 60), remaining 40.
   Take item2 fully (weight 20, value 100), remaining 20.
   Take fraction of item3: (20/30)*120 = 80.
   Total value = 60 + 100 + 80 = 240.

3. Final Answer:

   240 -> Option B

4. Quick Check:

   Sum values with fractions = 240 [OK]

Hint: Sort by ratio, take full items, then fraction [OK]

Common Mistakes:

• Not sorting by ratio correctly
• Taking items without fractions
• Miscomputing fraction value

4. Find the error in this fractional knapsack code snippet:

function fractionalKnapsack(items, capacity) {
  items.sort((a, b) => (a.value / a.weight) - (b.value / b.weight));
  let totalValue = 0;
  let remaining = capacity;
  for (const item of items) {
    if (item.weight <= remaining) {
      totalValue += item.value;
      remaining -= item.weight;
    } else {
      totalValue += item.value * (remaining / item.weight);
      break;
    }
  }
  return totalValue;
}

medium

A. Incorrect fraction calculation formula

B. Using 'let' instead of 'const' for totalValue

C. Missing return statement

D. Sorting is ascending; should be descending by value/weight ratio

Solution

1. Step 1: Analyze sorting order

   The code sorts items ascending by value/weight ratio, but we need descending to pick best items first.

2. Step 2: Check other parts

   Return statement exists, fraction formula is correct, 'let' is appropriate for totalValue.

3. Final Answer:

   Sorting is ascending; should be descending by value/weight ratio -> Option D

4. Quick Check:

   Sort descending by ratio to maximize value [OK]

Hint: Sort descending by value/weight ratio [OK]

Common Mistakes:

• Sorting ascending instead of descending
• Ignoring fraction calculation correctness
• Confusing variable declarations

5. Given items with values and weights:

items = [
  {value: 40, weight: 2},
  {value: 100, weight: 20},
  {value: 50, weight: 10},
  {value: 60, weight: 12}
];
capacity = 30;

What is the maximum total value achievable using the fractional knapsack approach?

hard

A. 190

B. 160

C. 180

D. 200

Solution

1. Step 1: Calculate value/weight ratios and sort descending

   Ratios:
   Item1: 40/2=20
   Item2: 100/20=5
   Item3: 50/10=5
   Item4: 60/12=5
   Sorted order: Item1, then Items 2,3,4 (equal ratio 5).

2. Step 2: Pick items until capacity 30

   Take Item1 fully: weight 2, value 40, remaining 28.
   Next, take Item2 fully: weight 20, value 100, remaining 8.
   Then take fraction of Item3: fraction = 8/10 = 0.8, value = 50 * 0.8 = 40.
   Total value = 40 + 100 + 40 = 180.

3. Step 3: Check if taking Item4 fraction is better

   Item4 has same ratio as Item3, but we already used remaining capacity on Item3 fraction.
   So total value remains 180.

4. Final Answer:

   180 -> Option C

5. Quick Check:

   Sum values with fractions = 180 [OK]

Hint: Pick highest ratio items first, then fractions [OK]

Common Mistakes:

• Ignoring highest ratio item first
• Not taking fractions correctly
• Adding values without considering capacity