Fractional Knapsack Problem in DSA Python - 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.

items = [(60, 10), (100, 20), (120, 30)]  # (value, weight)
capacity = 50

# Sort items by value per weight
items.sort(key=lambda x: x[0]/x[1], reverse=True)

total_value = 0
for value, weight in items:
    if capacity == 0:
        break
    take = min(weight, capacity)
    total_value += take * (value / weight)
    capacity -= take

print(total_value)
    

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

Identify Repeating Operations

• Primary operation: Sorting the list of items by value per weight.
• How many times: Sorting compares items roughly n log n times for n items.
• Secondary operation: Looping through the sorted items once to pick fractions.
• How many times: Loop runs at most n times.

How Execution Grows With Input

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

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

Pattern observation: Sorting dominates and grows a bit faster than the number of items, while picking grows linearly.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a bit faster than the number of items because of sorting, which is the main cost.

Common Mistake

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

[OK] Correct: Sorting takes more time than just looping because it compares items many times, making it the dominant step.

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 per weight? 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. Minimize total weight by selecting whole items only

C. Maximize number of items regardless of weight

D. Select items randomly without considering value or weight

Solution

1. Step 1: Understand problem objective

   The fractional knapsack problem aims to maximize the total value of items taken, allowing fractions of items to fit within the weight limit.

2. Step 2: Compare options

   Only Maximize total value by taking fractions of items within weight limit correctly describes maximizing value with fractional items. Others describe incorrect goals.

3. Final Answer:

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

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 value maximization with weight minimization
• Ignoring the fractional part of items

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

easy

A. items.sort(key=lambda x: x[0]/x[1], reverse=True)

B. items.sort(key=lambda x: x[1]/x[0], reverse=True)

C. items.sort(key=lambda x: x[1]+x[0], reverse=True)

D. items.sort(key=lambda x: x[0]-x[1], reverse=True)

Solution

1. Step 1: Identify ratio for sorting

   Value-to-weight ratio is value divided by weight, so ratio = x[1]/x[0].

2. Step 2: Check sorting code

   items.sort(key=lambda x: x[1]/x[0], reverse=True) sorts items by this ratio descending, which is correct for fractional knapsack.

3. Final Answer:

   items.sort(key=lambda x: x[1]/x[0], reverse=True) -> Option B

4. Quick Check:

   Sort by value/weight descending [OK]

Hint: Sort by value divided by weight, descending [OK]

Common Mistakes:

• Reversing numerator and denominator
• Sorting by sum or difference instead of ratio
• Not using reverse=True for descending order

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

items = [(10, 60), (20, 100), (30, 120)]
capacity = 50
items.sort(key=lambda x: x[1]/x[0], reverse=True)
total_value = 0
for w, v in items:
    if capacity >= w:
        capacity -= w
        total_value += v
    else:
        total_value += v * (capacity / w)
        break
print(round(total_value, 2))

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: (10,60), (20,100), (30,120).

2. Step 2: Add items until capacity is full

   Take full 10 weight: value=60, capacity=40 left.
   Take full 20 weight: value=60+100=160, capacity=20 left.
   Take fraction of last item: fraction=20/30=2/3, value=160 + 120*(2/3)=160 + 80=240.
   But code prints rounded total_value after break, so total_value=240.

3. Final Answer:

   240.00 -> Option A

4. Quick Check:

   Sum values with fractions = 240.00 [OK]

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

Common Mistakes:

• Not sorting items by ratio
• Adding fractions incorrectly
• Rounding errors

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

def fractional_knapsack(items, capacity):
    items.sort(key=lambda x: x[0]/x[1], reverse=True)
    total_value = 0
    for w, v in items:
        if capacity >= w:
            capacity -= w
            total_value += v
        else:
            total_value += v * (capacity / w)
            break
    return total_value

print(fractional_knapsack([(10, 60), (20, 100), (30, 120)], 50))

medium

A. Function returns total_value before loop

B. Loop variables w and v are swapped

C. Capacity is not updated after taking fraction

D. Sorting ratio is reversed; should be value/weight, not weight/value

Solution

1. Step 1: Check sorting key

   Sorting uses x[0]/x[1] which is weight/value, but ratio should be value/weight (x[1]/x[0]).

2. Step 2: Verify other parts

   Loop variables and capacity update are correct; function returns after loop.

3. Final Answer:

   Sorting ratio is reversed; should be value/weight, not weight/value -> Option D

4. Quick Check:

   Sort by value/weight, not weight/value [OK]

Hint: Sort by value divided by weight, not the other way [OK]

Common Mistakes:

• Using weight/value instead of value/weight for sorting
• Swapping weight and value in loop
• Not updating capacity after fraction

5. Given items with weights and values: [(5, 30), (10, 40), (15, 45)], and capacity 20, what is the maximum value achievable using fractional knapsack?

hard

A. 70.0

B. 75.0

C. 85.0

D. 80.0

Solution

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

   Ratios: (30/5)=6, (40/10)=4, (45/15)=3.
   Sorted order: (5,30), (10,40), (15,45).

2. Step 2: Add items until capacity full

   Take full 5 weight: value=30, capacity=15 left.
   Take full 10 weight: value=30+40=70, capacity=5 left.
   Take fraction of last item: fraction=5/15=1/3, value=70 + 45*(1/3)=70 + 15=85.

3. Final Answer:

   85.0 -> Option C

4. Quick Check:

   Sum values with fractions = 85.0 [OK]

Hint: Take full items by ratio, then fraction last to fill capacity [OK]

Common Mistakes:

• Not sorting by ratio
• Taking items in wrong order
• Calculating fraction incorrectly