Fractional Knapsack Problem in DSA Go - 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 does sorting and selecting fractions affect the total steps?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import (
	"fmt"
	"sort"
)

type Item struct {
	value, weight float64
}

func fractionalKnapsack(items []Item, capacity float64) float64 {
	sort.Slice(items, func(i, j int) bool {
		return items[i].value/items[i].weight > items[j].value/items[j].weight
	})
	totalValue := 0.0
	for _, item := range items {
		if capacity == 0 {
			break
		}
		if item.weight <= capacity {
			totalValue += item.value
			capacity -= item.weight
		} else {
			totalValue += item.value * (capacity / item.weight)
			capacity = 0
		}
	}
	return totalValue
}

func main() {
	items := []Item{{60, 10}, {100, 20}, {120, 30}}
	capacity := 50.0
	fmt.Println(fractionalKnapsack(items, capacity))
}

This code sorts items by value per weight and then picks whole or fractional parts until capacity is full.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the items by value-to-weight ratio.
• How many times: Sorting takes about n log n steps for n items.
• Secondary operation: One loop through all items to pick fractions or whole items.
• How many times: This loop runs n times.

How Execution Grows With Input

As the number of items (n) grows, sorting takes more steps, roughly n log n, while the picking loop grows linearly with n.

Input Size (n)	Approx. Operations
10	About 10 * log 10 + 10 ≈ 43 steps
100	About 100 * log 100 + 100 ≈ 700 steps
1000	About 1000 * log 1000 + 1000 ≈ 10,000 steps

Pattern observation: Sorting dominates and grows faster than the linear picking loop.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than just the number of items because sorting takes extra steps.

Common Mistake

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

[OK] Correct: Sorting the items first takes more time than just picking, so sorting controls the overall speed.

Interview Connect

Understanding this time complexity helps you explain why sorting is important before selection and shows you can analyze combined steps clearly.

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. Find the item with the highest value only

B. Minimize the total weight of items selected

C. Select items without exceeding the number of items limit

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

Solution

1. Step 1: Understand the problem goal

   The fractional knapsack problem allows taking parts of items to maximize total value without exceeding weight capacity.

2. Step 2: Identify the correct objective

   The goal is to maximize value by possibly taking fractions, not just whole items or minimizing weight.

3. Final Answer:

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

4. Quick Check:

   Goal = Maximize value with fractions [OK]

Hint: Remember: fractional means partial items allowed [OK]

Common Mistakes:

• Thinking only whole items can be taken
• Confusing weight limit with item count limit
• Assuming minimizing weight is the goal

2. Which of the following Go code snippets correctly sorts items by their value-to-weight ratio in descending order for the fractional knapsack problem?

easy

A. sort.Slice(items, func(i, j int) bool { return items[i].value < items[j].value })

B. sort.Slice(items, func(i, j int) bool { return items[i].weight > items[j].weight })

C. sort.Slice(items, func(i, j int) bool { return items[i].value/items[i].weight > items[j].value/items[j].weight })

D. sort.Slice(items, func(i, j int) bool { return items[i].weight/items[i].value > items[j].weight/items[j].value })

Solution

1. Step 1: Identify sorting criteria for fractional knapsack

   Items must be sorted by value-to-weight ratio in descending order to pick the best value per weight first.

2. Step 2: Check each option's sorting logic

   sort.Slice(items, func(i, j int) bool { return items[i].value/items[i].weight > items[j].value/items[j].weight }) sorts by value/weight descending, which is correct. Others sort by weight or value incorrectly.

3. Final Answer:

   sort.Slice(items, func(i, j int) bool { return items[i].value/items[i].weight > items[j].value/items[j].weight }) -> Option C

4. Quick Check:

   Sort by value/weight descending = sort.Slice(items, func(i, j int) bool { return items[i].value/items[i].weight > items[j].value/items[j].weight }) [OK]

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

Common Mistakes:

• Sorting by weight only
• Sorting by value only
• Using ascending order instead of descending

3. What is the output of the following Go code snippet for fractional knapsack?

type Item struct {
    value, weight float64
}

items := []Item{{60, 10}, {100, 20}, {120, 30}}
capacity := 50.0

// Sort by value/weight descending
sort.Slice(items, func(i, j int) bool {
    return items[i].value/items[i].weight > items[j].value/items[j].weight
})

var totalValue float64
for _, item := range items {
    if capacity == 0 {
        break
    }
    if item.weight <= capacity {
        capacity -= item.weight
        totalValue += item.value
    } else {
        fraction := capacity / item.weight
        totalValue += item.value * fraction
        capacity = 0
    }
}
fmt.Printf("%.2f", totalValue)

medium

A. 240.00

B. 220.00

C. 180.00

D. 200.00

Solution

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

   Ratios: 60/10=6, 100/20=5, 120/30=4. Sorted order: {60,10}, {100,20}, {120,30}.

2. Step 2: Pick items until capacity 50

   Take full {60,10} (capacity=40, value=60), full {100,20} (capacity=20, value=160), then fraction of {120,30} with 20 weight (fraction=20/30=2/3), value added=120*(2/3)=80, total=240.

3. Final Answer:

   240.00 -> Option A

4. Quick Check:

   Total value = 60 + 100 + 80 = 240.00 [OK]

Hint: Pick full items first, then fraction of last item [OK]

Common Mistakes:

• Not sorting by value/weight ratio
• Taking items without checking capacity
• Incorrect fraction calculation

4. Identify the bug in this Go code snippet for fractional knapsack and select the fix:

type Item struct {
    value, weight float64
}

items := []Item{{60, 10}, {100, 20}, {120, 30}}
capacity := 50.0

sort.Slice(items, func(i, j int) bool {
    return items[i].value > items[j].value
})

var totalValue float64
for _, item := range items {
    if capacity == 0 {
        break
    }
    if item.weight <= capacity {
        capacity -= item.weight
        totalValue += item.value
    } else {
        fraction := capacity / item.weight
        totalValue += item.value * fraction
        capacity = 0
    }
}
fmt.Printf("%.2f", totalValue)

medium

A. Change capacity to int type

B. Sort items by value/weight ratio descending instead of value only

C. Remove the fraction calculation

D. Use a for loop with index instead of range

Solution

1. Step 1: Analyze sorting criteria

   The code sorts by value only, which is incorrect for fractional knapsack; it should sort by value/weight ratio.

2. Step 2: Identify correct fix

   Sorting by value/weight ratio descending ensures maximum value per weight is chosen first, fixing the bug.

3. Final Answer:

   Sort items by value/weight ratio descending instead of value only -> Option B

4. Quick Check:

   Sort by value/weight ratio fixes selection [OK]

Hint: Sort by value/weight ratio, not just value [OK]

Common Mistakes:

• Sorting by value only
• Ignoring fraction calculation
• Changing variable types unnecessarily

5. Given items with values and weights: {value: 40, weight: 2}, {value: 100, weight: 20}, {value: 50, weight: 10}, and a knapsack capacity of 15, what is the maximum total value achievable using the fractional knapsack approach?

hard

A. 140.00

B. 120.00

C. 150.00

D. 160.00

Solution

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

   Ratios: 40/2=20, 100/20=5, 50/10=5. Sorted order: {40,2}, {100,20}, {50,10}.

2. Step 2: Pick items until capacity 15

   Take full {40,2} (capacity=13, value=40), then full {50,10} (capacity=3, value=90), then fraction of {100,20} with 3 weight (fraction=3/20=0.15), value added=100*0.15=15, total=40 + 50 + 15=105.

3. Final Answer:

   105.00 -> Option A

4. Quick Check:

   Total value = 40 + 50 + 15 = 105.00 [OK]

Hint: Pick highest ratio items fully, then fraction last [OK]

Common Mistakes:

• Not sorting by value/weight ratio
• Taking heavier items first
• Incorrect fraction calculation