0 1 Knapsack Problem in DSA C++ - Build from Scratch

0 1 Knapsack Problem

📖 Scenario: You are packing a backpack for a hiking trip. You have a list of items, each with a weight and a value. Your backpack can only carry a limited weight. You want to choose items to maximize the total value without exceeding the weight limit.

🎯 Goal: Build a program that uses the 0 1 Knapsack algorithm to find the maximum value of items you can carry without exceeding the backpack's weight limit.

📋 What You'll Learn

Create arrays for item weights and values with exact given values

Create an integer variable for the backpack weight limit

Implement the 0 1 Knapsack dynamic programming solution using a 2D array

Print the maximum value that fits in the backpack

💡 Why This Matters

🌍 Real World

Packing efficiently for trips, resource allocation, budget management, and selecting projects with limited resources.

💼 Career

Understanding dynamic programming and optimization problems is essential for software engineers, data scientists, and algorithm developers.

1

Create item weights and values arrays

Create an integer array called weights with values {2, 3, 4, 5} and an integer array called values with values {3, 4, 5, 6}.

DSA C++

// Your code here

Hint

Use C++ array syntax to create weights and values with the exact numbers.

2

Set the backpack weight limit

Create an integer variable called maxWeight and set it to 5.

DSA C++

int weights[] = {2, 3, 4, 5};
int values[] = {3, 4, 5, 6};
// Your code here

Hint

Use a simple integer variable to store the backpack's weight limit.

3

Implement the 0 1 Knapsack dynamic programming solution

Create a 2D integer array called dp with dimensions 5 by maxWeight + 1. Use a nested for loop with variables i from 1 to 4 and w from 0 to maxWeight. Inside the loops, if weights[i-1] <= w, set dp[i][w] to the maximum of values[i-1] + dp[i-1][w - weights[i-1]] and dp[i-1][w]. Otherwise, set dp[i][w] to dp[i-1][w].

DSA C++

int weights[] = {2, 3, 4, 5};
int values[] = {3, 4, 5, 6};
int maxWeight = 5;
// Your code here

Hint

Use a 2D array dp to store max values. Loop through items and weights. Use std::max to choose the best value.

4

Print the maximum value that fits in the backpack

Print the value of dp[4][maxWeight] using std::cout.

DSA C++

int weights[] = {2, 3, 4, 5};
int values[] = {3, 4, 5, 6};
int maxWeight = 5;

int n = 4;
int dp[5][6] = {0};

for (int i = 1; i <= n; i++) {
    for (int w = 0; w <= maxWeight; w++) {
        if (weights[i - 1] <= w) {
            dp[i][w] = std::max(values[i - 1] + dp[i - 1][w - weights[i - 1]], dp[i - 1][w]);
        } else {
            dp[i][w] = dp[i - 1][w];
        }
    }
}
// Your code here

Hint

Print the last cell of the dp table which holds the maximum value.

Practice

1. What does the 0 1 Knapsack Problem aim to maximize given a set of items with weights and values?

easy

A. The total weight of items without exceeding the value limit

B. The total value of items without exceeding the weight limit

C. The number of items selected regardless of weight

D. The minimum weight of items selected

Solution

1. Step 1: Understand the problem goal

   The 0 1 Knapsack Problem tries to select items to maximize total value without exceeding the weight capacity.

2. Step 2: Identify the constraint

   The total weight of selected items must not exceed the given weight limit.

3. Final Answer:

   The total value of items without exceeding the weight limit -> Option B

4. Quick Check:

   Maximize value with weight limit = A [OK]

Hint: Maximize value, keep total weight ≤ capacity [OK]

Common Mistakes:

• Confusing weight limit with value limit
• Trying to maximize weight instead of value
• Ignoring the weight constraint

2. Which of the following is the correct way to declare the DP table for the 0 1 Knapsack Problem in C++ if n is number of items and W is max weight?

easy

A. int dp[n+1][W+1];

B. int dp[W+1][n+1];

C. int dp[n][W];

D. int dp[W][n];

Solution

1. Step 1: Understand DP table dimensions

   The DP table stores solutions for items 0 to n and weights 0 to W, so size is (n+1) x (W+1).

2. Step 2: Match correct declaration

   int dp[n+1][W+1]; declares dp with dimensions [n+1][W+1], matching the problem requirements.

3. Final Answer:

   int dp[n+1][W+1]; -> Option A

4. Quick Check:

   DP table size = (n+1) x (W+1) [OK]

Hint: DP table size = items+1 by weight+1 [OK]

Common Mistakes:

• Using dp[n][W] misses base case row/column
• Swapping dimensions incorrectly
• Declaring smaller arrays causing out-of-bound errors

3. What is the output of the following C++ code snippet for 0 1 Knapsack Problem?

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

int dp[4][51] = {0};

for (int i = 1; i <= n; i++) {
    for (int w = 1; w <= W; w++) {
        if (weights[i-1] <= w) {
            dp[i][w] = std::max(dp[i-1][w], dp[i-1][w - weights[i-1]] + values[i-1]);
        } else {
            dp[i][w] = dp[i-1][w];
        }
    }
}

std::cout << dp[n][W] << std::endl;

medium

A. 160

B. 180

C. 220

D. 250

Solution

1. Step 1: Understand the DP logic

   The code fills dp table to find max value for weight limit 50 with items of given weights and values.

2. Step 2: Calculate max value manually

   Items: (10,60), (20,100), (30,120). Max value for weight 50 is choosing items 2 and 3: 100 + 120 = 220.

3. Final Answer:

   220 -> Option C

4. Quick Check:

   Max value for W=50 = 220 [OK]

Hint: Pick items with max value under weight limit [OK]

Common Mistakes:

• Adding weights instead of values
• Choosing items exceeding weight limit
• Misindexing arrays causing wrong dp values

4. Identify the bug in this 0 1 Knapsack DP code snippet:

for (int i = 1; i <= n; i++) {
    for (int w = 1; w <= W; w++) {
        if (weights[i] <= w) {
            dp[i][w] = std::max(dp[i-1][w], dp[i-1][w - weights[i]] + values[i]);
        } else {
            dp[i][w] = dp[i-1][w];
        }
    }
}

medium

A. Indexing weights and values with i instead of i-1

B. Loop ranges should start from 0 instead of 1

C. Using std::max instead of a custom max function

D. Not initializing dp array to zero

Solution

1. Step 1: Check array indexing

   Arrays weights and values are zero-indexed, but code uses i which starts at 1, causing out-of-bound or wrong access.

2. Step 2: Correct indexing

   Use weights[i-1] and values[i-1] to access correct elements.

3. Final Answer:

   Indexing weights and values with i instead of i-1 -> Option A

4. Quick Check:

   Zero-based arrays need i-1 index [OK]

Hint: Use i-1 for zero-based arrays in loops starting at 1 [OK]

Common Mistakes:

• Accessing arrays with wrong index causing runtime errors
• Starting loops at 0 but using i-1 indexing
• Ignoring dp initialization

5. Given items with values {20, 5, 10, 40, 15, 25} and weights {1, 2, 3, 8, 7, 4}, and max weight 10, what is the maximum value achievable using 0 1 Knapsack?

hard

A. 75

B. 65

C. 70

D. 60

Solution

1. Step 1: Analyze items and weight limit

   Items: values = {20,5,10,40,15,25}, weights = {1,2,3,8,7,4}, max weight = 10.

2. Step 2: Find best combination under weight 10

   Choose items with weights 1,3,4 (values 20,10,25) total weight 8, value 55; or items 1,2,6 (weights 1,2,4) value 20+5+25=50; best is items 1,4 (weights 1+8=9) value 20+40=60; or items 4 and 6 (8+4=12 >10) invalid; items 1,5,6 (1+7+4=12 >10) invalid; items 1,3,5 (1+3+7=11 >10) invalid; items 2,3,6 (2+3+4=9) value 5+10+25=40; items 1,2,3,6 (1+2+3+4=10) value 20+5+10+25=60; items 1,2,3,5 (1+2+3+7=13 >10) invalid; items 1,2,3,4 (1+2+3+8=14 >10) invalid; items 1,2,6,5 (1+2+4+7=14 >10) invalid; items 1,3,6 (1+3+4=8) value 20+10+25=55; items 1,2,3,6 total value 60 and weight 10 fits exactly.

3. Final Answer:

   60 -> Option D

4. Quick Check:

   Max value with weight 10 = 60 [OK]

Hint: Try combinations fitting weight limit, sum max values [OK]

Common Mistakes:

• Ignoring weight limit when summing values
• Missing best combination with multiple items
• Assuming greedy choice works always