Fractional Knapsack Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the structure for an item with weight and value.

DSA C

typedef struct {
    int value;
    int [1];
} Item;

Drag options to blanks, or click blank then click option'

Aweight

Bsize

Ccost

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' or 'cost' instead of 'weight'.

The structure must have a field named 'weight' to store the item's weight.

2fill in blank

medium

Complete the code to calculate the value per weight ratio for an item.

DSA C

double ratio = (double)items[i].value / items[i].[1];

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Acost

Bweight

Csize

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Dividing by 'cost' or 'size' which are not defined.

The ratio is value divided by weight, so 'weight' is the correct field.

3fill in blank

hard

Fix the error in the comparison function used for sorting items by ratio.

DSA C

int cmp(const void *a, const void *b) {
    Item *itemA = (Item *)a;
    Item *itemB = (Item *)b;
    double rA = (double)itemA->value / itemA->weight;
    double rB = (double)itemB->value / itemB->weight;
    if (rA < rB) return [1];
    else if (rA > rB) return -1;
    else return 0;
}

Drag options to blanks, or click blank then click option'

A1

B-1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning -1 instead of 1 causes ascending order.

For sorting in descending order, return 1 when rA < rB so itemB comes first.

4fill in blank

hard

Fill both blanks to update the total value and remaining capacity when taking a fraction of an item.

DSA C

if (items[i].weight <= capacity) {
    totalValue += items[i].value;
    capacity -= items[i].weight;
} else {
    double fraction = (double)capacity / items[i].weight;
    totalValue += items[i].value [1] fraction;
    capacity[2]=capacity;
    break;
}

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A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Adding fraction instead of multiplying, or subtracting capacity incorrectly.

Multiply value by fraction to add partial value, then reduce capacity to zero.

5fill in blank

hard

Fill all three blanks to complete the main loop that processes items for the fractional knapsack.

DSA C

for (int i = 0; i < n; i++) {
    if (items[i].weight <= [1]) {
        totalValue += items[i].value;
        [2] -= items[i].weight;
    } else {
        double fraction = (double)[3] / items[i].weight;
        totalValue += items[i].value * fraction;
        break;
    }
}

Drag options to blanks, or click blank then click option'

Acapacity

BtotalValue

Cweight

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variables like 'totalValue' or 'weight' in place of capacity.

Use 'capacity' to check and update remaining space and calculate fraction.

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