0 1 Knapsack Problem in DSA C

Mental Model

Choose items one by one to maximize value without exceeding weight limit, but you can only take each item once.

Analogy: Imagine packing a backpack for a trip with limited space. You pick items to get the most useful stuff without overloading the bag.

Knapsack capacity: 5
Items: [weight, value]
[1, 6], [2, 10], [3, 12]
Backpack: capacity 5 -> can hold items up to total weight 5

Dry Run Walkthrough

Input: items: [(weight=1, value=6), (weight=2, value=10), (weight=3, value=12)], capacity=5

Goal: Find the maximum value we can carry without exceeding capacity 5

Step 1: Consider first item (weight=1, value=6), update dp for capacities 1 to 5

dp array after step 1: [0, 6, 6, 6, 6, 6]

Why: We can take the first item if capacity allows, so value 6 is possible from capacity 1 upwards

Step 2: Consider second item (weight=2, value=10), update dp for capacities 2 to 5

dp array after step 2: [0, 6, 10, 16, 16, 16]

Why: At capacity 2, taking second item gives value 10 better than previous 6; at capacity 3, combine first and second items for value 16

Step 3: Consider third item (weight=3, value=12), update dp for capacities 3 to 5

dp array after step 3: [0, 6, 10, 16, 18, 22]

Why: At capacity 4, taking third item plus first item gives 18; at capacity 5, taking third and second items gives max value 22

Result:

dp array final: [0, 6, 10, 16, 18, 22]
Maximum value: 22

Annotated Code

DSA C

#include <stdio.h>

// Function to find max of two integers
int max(int a, int b) {
    return (a > b) ? a : b;
}

// 0 1 Knapsack function
int knapsack(int capacity, int weights[], int values[], int n) {
    int dp[capacity + 1];
    for (int i = 0; i <= capacity; i++) {
        dp[i] = 0; // initialize dp array with 0
    }

    // For each item
    for (int i = 0; i < n; i++) {
        // Traverse capacities from high to low to avoid reuse of same item
        for (int w = capacity; w >= weights[i]; w--) {
            // Update dp[w] if including item i gives better value
            dp[w] = max(dp[w], dp[w - weights[i]] + values[i]);
        }
    }
    return dp[capacity];
}

int main() {
    int weights[] = {1, 2, 3};
    int values[] = {6, 10, 12};
    int capacity = 5;
    int n = sizeof(weights) / sizeof(weights[0]);

    int max_value = knapsack(capacity, weights, values, n);
    printf("Maximum value: %d\n", max_value);
    return 0;
}

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

iterate over each item to consider including it

for (int w = capacity; w >= weights[i]; w--) {

iterate backwards over capacities to avoid reusing the same item multiple times

dp[w] = max(dp[w], dp[w - weights[i]] + values[i]);

update dp[w] to max of excluding or including current item

Output

Maximum value: 22

Complexity Analysis

Time: O(n * capacity) because we process each item for each capacity up to the limit

Space: O(capacity) because we use a one-dimensional dp array of size capacity+1

vs Alternative: Compared to a 2D dp approach with O(n * capacity) time and space, this optimized 1D dp reduces space to O(capacity)

Edge Cases

capacity is zero

maximum value is zero since no items can be taken

DSA C

for (int i = 0; i <= capacity; i++) { dp[i] = 0; }

no items (n=0)

maximum value is zero because no items to choose

DSA C

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

items heavier than capacity

those items are never included as weights[i] > capacity prevents dp update

DSA C

for (int w = capacity; w >= weights[i]; w--) { ... }

When to Use This Pattern

When you need to pick items with weights and values to maximize value without exceeding capacity, and each item can be chosen once, use the 0 1 Knapsack pattern with dynamic programming.

Common Mistakes

Mistake: Iterating capacities from low to high causing reuse of the same item multiple times
Fix: Iterate capacities from high to low to ensure each item is only counted once

Mistake: Using 2D dp without space optimization when 1D dp suffices
Fix: Use 1D dp array and iterate capacities backwards to optimize space

Summary

Find the maximum total value of items that fit in a knapsack without exceeding its capacity, choosing each item at most once.

Use when you have a set of items with weights and values and want to maximize value under a weight limit with no repeats.

The key is to build up solutions for smaller capacities and items, updating from high to low capacity to avoid reusing items.