Why Fractional Knapsack Problem in DSA C?

The Big Idea

What if you could carry the most valuable mix by taking just parts of items instead of whole ones?

The Scenario

Imagine you have a backpack that can carry a limited weight, and you want to fill it with items to get the most value. You try to pick whole items one by one, but some items are too heavy or too valuable to skip. How do you decide what to take?

The Problem

Picking items one by one without breaking them can leave a lot of unused space in your backpack. You might miss out on getting more value because you can't take part of an item. Doing this by hand is slow and often leads to poor choices.

The Solution

The Fractional Knapsack Problem lets you take parts of items instead of whole items. This way, you can fill your backpack perfectly to get the highest total value. It uses a simple rule: take items with the best value per weight first, even if only partially.

Before vs After

✗ Before

int totalValue = 0;
for (int i = 0; i < n; i++) {
    if (weight[i] <= capacity) {
        totalValue += value[i];
        capacity -= weight[i];
    }
}

✓ After

sortItemsByValuePerWeight();
for (int i = 0; i < n && capacity > 0; i++) {
    int take = (weight[i] < capacity) ? weight[i] : capacity;
    totalValue += take * valuePerWeight[i];
    capacity -= take;
}

What It Enables

This concept allows you to maximize the total value you carry by smartly choosing whole or parts of items based on their value-to-weight ratio.

Real Life Example

Think about packing a lunchbox with snacks. Some snacks can be split (like nuts or candies), so you take just enough to fill the box and get the tastiest mix possible.

Key Takeaways

Manual picking wastes space and value.

Fractional Knapsack lets you take parts of items for best value.

It sorts items by value per weight and fills the capacity greedily.

Practice

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

easy

A. Select items only if their weight is less than a fixed number

B. Minimize total weight of items selected

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

D. Find the number of items that fit exactly in the knapsack

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

   Only Maximize total value by taking fractions of items within weight limit correctly describes maximizing value by taking fractions within weight limit.

3. Final Answer:

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

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 with minimizing weight
• Ignoring value maximization

2. Which of the following is the correct way to sort items by value-to-weight ratio in C for the fractional knapsack?

easy

A. qsort(items, n, sizeof(Item), compare); where compare sorts by value/weight descending

B. sort(items, items+n); without a comparator

C. qsort(items, n, sizeof(Item), compare); where compare sorts by weight ascending

D. Using bubble sort on values only

Solution

1. Step 1: Identify sorting requirement

   Items must be sorted by value-to-weight ratio descending for greedy selection.

2. Step 2: Match sorting method

   Using qsort with a comparator sorting by value/weight descending is correct.

3. Final Answer:

   qsort(items, n, sizeof(Item), compare); where compare sorts by value/weight descending -> Option A

4. Quick Check:

   Sort by value/weight ratio descending [OK]

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

Common Mistakes:

• Sorting by weight only
• Sorting without comparator
• Using inefficient sorting for large data

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

typedef struct { int value; int weight; } Item;

Item items[] = {{60, 10}, {100, 20}, {120, 30}};
int n = 3, W = 50;

// Assume items sorted by value/weight descending
float fractionalKnapsack(Item items[], int n, int W) {
    int curWeight = 0;
    float finalValue = 0.0;
    for (int i = 0; i < n; i++) {
        if (curWeight + items[i].weight <= W) {
            curWeight += items[i].weight;
            finalValue += items[i].value;
        } else {
            int remain = W - curWeight;
            finalValue += items[i].value * ((float) remain / items[i].weight);
            break;
        }
    }
    return finalValue;
}

printf("%.2f", fractionalKnapsack(items, n, W));

medium

A. 240.00

B. 220.00

C. 180.00

D. 200.00

Solution

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

   Ratios: 60/10=6, 100/20=5, 120/30=4. Sorted order is as given.

2. Step 2: Add items until weight limit 50

   Add first item fully: weight=10, value=60; second fully: weight=30, value=160; third partially: remain=20, value=120*(20/30)=80; total value=60+100+80=240.

3. Final Answer:

   240.00 -> Option A

4. Quick Check:

   Sum values with fractions = 240.00 [OK]

Hint: Add full items then fraction of next to fill knapsack [OK]

Common Mistakes:

• Not sorting by ratio
• Adding partial value incorrectly
• Ignoring remaining capacity

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

float fractionalKnapsack(Item items[], int n, int W) {
    int curWeight = 0;
    float finalValue = 0.0;
    for (int i = 0; i < n; i++) {
        if (curWeight + items[i].weight < W) {
            curWeight += items[i].weight;
            finalValue += items[i].value;
        } else {
            int remain = W - curWeight;
            finalValue += items[i].value * ((float) remain / items[i].weight);
            break;
        }
    }
    return finalValue;
}

medium

A. The remain calculation is incorrect

B. The finalValue should be int, not float

C. The loop should run from i=1 to n

D. The condition should be <= instead of < to include full weight equal to W

Solution

1. Step 1: Analyze the weight check condition

   The condition curWeight + items[i].weight < W excludes items that exactly fill the knapsack.

2. Step 2: Correct the condition

   Changing to <= allows taking items that fit exactly, avoiding missing full items.

3. Final Answer:

   The condition should be <= instead of < to include full weight equal to W -> Option D

4. Quick Check:

   Use <= to include exact weight [OK]

Hint: Use <= to allow full item if it fits exactly [OK]

Common Mistakes:

• Using < excludes exact fit items
• Changing variable types unnecessarily
• Incorrect loop indices

5. Given items with values and weights: {value: 40, weight: 2}, {value: 100, weight: 20}, {value: 50, weight: 10}, and a knapsack capacity of 26, what is the maximum value achievable using fractional knapsack?

hard

A. 190

B. 160

C. 180

D. 140

Solution

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

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

2. Step 2: Greedy selection with capacity 26

   Take {40,2} full: w=2, v=40, remain=24.
   Take {100,20} full: w=22, v=140, remain=4.
   Fraction of {50,10}: (4/10)*50=20, total v=160.

3. Final Answer:

   160 -> Option B

4. Quick Check:

   Max value by greedy fractions = 160 [OK]

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

Common Mistakes:

• Ignoring sorting by ratio
• Taking items in wrong order
• Miscomputing fractional value