Fractional Knapsack Problem in DSA C

Mental Model

We want to fill a bag with items to get the most value, and we can take parts of items if needed.

Analogy: Imagine filling a jar with different types of honey jars. You want the sweetest honey first, and if the jar is almost full, you take only part of the last honey jar.

Knapsack capacity: [__________] (empty)
Items sorted by value/weight ratio:
Item1(value=60, weight=10) -> ratio=6
Item2(value=100, weight=20) -> ratio=5
Item3(value=120, weight=30) -> ratio=4

Dry Run Walkthrough

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

Goal: Maximize total value in knapsack by taking whole or fractional parts of items

Step 1: Sort items by value/weight ratio descending

Items order: (60,10) ratio=6 -> (100,20) ratio=5 -> (120,30) ratio=4

Why: We want to pick items with highest value per weight first to maximize value

Step 2: Take whole item 1 (weight 10) into knapsack

Knapsack: 10/50 filled, total value=60

Why: Item 1 fits fully, so take it all

Step 3: Take whole item 2 (weight 20) into knapsack

Knapsack: 30/50 filled, total value=160

Why: Item 2 fits fully, so take it all

Step 4: Take fraction of item 3 to fill remaining capacity (20/30 fraction)

Knapsack: 50/50 filled, total value=160 + (120 * 20/30) = 240

Why: Only part of item 3 fits, so take fraction to fill knapsack

Result:

Knapsack full: total value=240

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

// Structure to store item value and weight
typedef struct {
    int value;
    int weight;
    double ratio;
} Item;

// Comparator to sort items by value/weight ratio descending
int cmp(const void *a, const void *b) {
    Item *item1 = (Item *)a;
    Item *item2 = (Item *)b;
    if (item2->ratio > item1->ratio) return 1;
    else if (item2->ratio < item1->ratio) return -1;
    else return 0;
}

// Function to solve fractional knapsack
double fractionalKnapsack(Item items[], int n, int capacity) {
    // Calculate ratio for each item
    for (int i = 0; i < n; i++) {
        items[i].ratio = (double)items[i].value / items[i].weight;
    }

    // Sort items by ratio descending
    qsort(items, n, sizeof(Item), cmp);

    double totalValue = 0.0;
    int remaining = capacity;

    for (int i = 0; i < n; i++) {
        if (items[i].weight <= remaining) {
            // Take whole item
            totalValue += items[i].value;
            remaining -= items[i].weight;
        } else {
            // Take fraction of item
            totalValue += items[i].ratio * remaining;
            break; // Knapsack full
        }
    }
    return totalValue;
}

int main() {
    Item items[] = {{60, 10, 0}, {100, 20, 0}, {120, 30, 0}};
    int n = sizeof(items) / sizeof(items[0]);
    int capacity = 50;

    double maxValue = fractionalKnapsack(items, n, capacity);
    printf("Maximum value in knapsack = %.2f\n", maxValue);
    return 0;
}

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

Calculate value per weight ratio for sorting priority

qsort(items, n, sizeof(Item), cmp);

Sort items by descending ratio to pick best value first

if (items[i].weight <= remaining) {

Check if whole item fits in remaining capacity

totalValue += items[i].value;

Add full item value to total

remaining -= items[i].weight;

Reduce remaining capacity by item weight

totalValue += items[i].ratio * remaining;

Add fractional value for partial item to fill knapsack

break;

Stop after filling knapsack with fraction

Output

Maximum value in knapsack = 240.00

Complexity Analysis

Time: O(n log n) because we sort n items by ratio, then iterate once over them

Space: O(n) for storing items and their ratios

vs Alternative: Better than trying all subsets (exponential), greedy sorting gives optimal fractional solution efficiently

Edge Cases

Empty items list

Returns 0 value since no items to take

DSA C

for (int i = 0; i < n; i++) {

Knapsack capacity zero

Returns 0 value since no capacity to take items

DSA C

int remaining = capacity;

Item with zero weight (invalid)

Division by zero avoided by input validation (not shown here), else program may crash

DSA C

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

When to Use This Pattern

When you need to maximize value with divisible items and limited capacity, use the fractional knapsack greedy approach sorting by value/weight ratio.

Common Mistakes

Mistake: Sorting items by value or weight alone instead of value/weight ratio
Fix: Sort items by value divided by weight to prioritize best value per unit weight

Mistake: Not taking fractional part when item doesn't fully fit
Fix: Calculate fraction of item to fill remaining capacity and add proportional value

Summary

Find maximum value by taking whole or fractional parts of items to fill knapsack capacity.

Use when items can be divided and you want the best value for limited capacity.

Sort items by value/weight ratio and take as much as possible from highest ratio first.