Unbounded Knapsack Problem in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Unbounded Knapsack Problem?

It is a problem where you can take unlimited quantities of each item to maximize the total value without exceeding the knapsack's capacity.

beginner

How does the Unbounded Knapsack Problem differ from the 0/1 Knapsack Problem?

In Unbounded Knapsack, you can use an item multiple times, while in 0/1 Knapsack, each item can be used only once.

intermediate

What is the main approach to solve the Unbounded Knapsack Problem efficiently?

Use dynamic programming with a one-dimensional array where dp[w] stores the maximum value for weight w, updating it by considering all items multiple times.

beginner

In the Unbounded Knapsack DP solution, what does dp[w] represent?

dp[w] represents the maximum value achievable with a knapsack capacity of w using unlimited items.

intermediate

Why do we iterate over weights from the item's weight up to the capacity in Unbounded Knapsack?

Because we can use the same item multiple times, we update dp[w] by checking if adding the item improves the value at weight w.

In the Unbounded Knapsack Problem, can an item be chosen more than once?

A Depends on the item

B No, only once

C Only twice

D Yes, unlimited times

Unbounded Knapsack allows unlimited use of each item.

What does dp[w] represent in the Unbounded Knapsack dynamic programming solution?

A Total items available

B Number of items used for capacity w

C Maximum value for capacity w

D Minimum weight for value w

dp[w] stores the maximum value achievable with capacity w.

Which loop order is correct for Unbounded Knapsack DP?

A For each item, then for each weight

B Only one loop over items

C For each weight, then for each item

D Only one loop over weights

We loop over items first, then weights to allow multiple uses of items.

If an item has weight 3 and value 5, how do we update dp[6] in Unbounded Knapsack?

A dp[6] = dp[6] + 5

B dp[6] = max(dp[6], dp[3] + 5)

C dp[6] = dp[3]

D dp[6] = max(dp[6], dp[6] + 5)

We check if adding the item again improves dp[6] by dp[3] + value.

What is the time complexity of the Unbounded Knapsack DP solution with n items and capacity W?

A O(n * W)

B O(n + W)

C O(n^2 * W)

D O(W^2)

The solution uses nested loops over items and weights, resulting in O(n * W).

Explain how dynamic programming solves the Unbounded Knapsack Problem.

Think about how you build solutions for smaller capacities to solve bigger ones.

Describe the difference between 0/1 Knapsack and Unbounded Knapsack problems.

Focus on how many times you can pick each item.

Practice

1. What is the main difference between the Unbounded Knapsack Problem and the 0/1 Knapsack Problem?

easy

A. In Unbounded Knapsack, you can use unlimited copies of each item.

B. In Unbounded Knapsack, items cannot be repeated.

C. Unbounded Knapsack only works with fractional items.

D. Unbounded Knapsack has a fixed number of items to choose from.

Solution

1. Step 1: Understand item usage rules

   In 0/1 Knapsack, each item can be chosen at most once. In Unbounded Knapsack, items can be chosen unlimited times.

2. Step 2: Compare problem constraints

   Unbounded Knapsack allows multiple copies of the same item, unlike 0/1 Knapsack.

3. Final Answer:

   In Unbounded Knapsack, you can use unlimited copies of each item. -> Option A

4. Quick Check:

   Unlimited item usage = In Unbounded Knapsack, you can use unlimited copies of each item. [OK]

Hint: Remember: Unbounded means unlimited item copies allowed [OK]

Common Mistakes:

• Confusing Unbounded with fractional knapsack
• Thinking items can only be chosen once
• Assuming fixed number of items to pick

2. Which of the following C# code snippets correctly initializes the DP array for the Unbounded Knapsack problem of capacity W?

easy

A. int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0;

B. int[] dp = new int[W]; for (int i = 0; i < W; i++) dp[i] = -1;

C. int[] dp = new int[W + 1]; for (int i = 1; i <= W; i++) dp[i] = int.MaxValue;

D. int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = int.MinValue;

Solution

1. Step 1: Check array size and initialization

   The DP array size should be W+1 to cover capacities from 0 to W. Initializing with 0 is correct for max value problems.

2. Step 2: Validate initialization values

   Setting dp[i] = 0 means no value at start, which is correct. Negative or max int values are incorrect here.

3. Final Answer:

   int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0; -> Option A

4. Quick Check:

   DP array size W+1 and zero initialization = int[] dp = new int[W + 1]; for (int i = 0; i <= W; i++) dp[i] = 0; [OK]

Hint: DP array size = capacity + 1, initialize with zero for max value [OK]

Common Mistakes:

• Using wrong array size (W instead of W+1)
• Initializing with int.MaxValue or int.MinValue incorrectly
• Starting loop from 1 instead of 0

3. What is the output of the following C# code snippet for Unbounded Knapsack?

int W = 5;
int[] wt = {1, 3, 4};
int[] val = {10, 40, 50};
int[] dp = new int[W + 1];
for (int i = 0; i <= W; i++) {
    for (int j = 0; j < wt.Length; j++) {
        if (wt[j] <= i) {
            dp[i] = Math.Max(dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}
Console.WriteLine(dp[W]);

medium

A. 120

B. 100

C. 60

D. 90

Solution

1. Step 1: Trace dp array updates for each capacity

   Calculate dp values: - dp[0]=0 - dp[1]=max(dp[1], dp[0]+10)=10 - dp[2]=max(dp[2], dp[1]+10)=20 - dp[3]=max(dp[3], dp[0]+40, dp[2]+10)=max(0,40,30)=40 - dp[4]=max(dp[4], dp[1]+40, dp[3]+10, dp[0]+50)=max(0,50,50,50)=50 - dp[5]=max(dp[5], dp[2]+40, dp[4]+10, dp[1]+50)=max(0,60,60,60)=60 But re-check dp[5] carefully: dp[5] = max(dp[5], dp[5-1]+10=dp[4]+10=50+10=60, dp[5-3]+40=dp[2]+40=20+40=60, dp[5-4]+50=dp[1]+50=10+50=60) = 60

2. Step 2: Recalculate carefully for maximum value

   Check if multiple items can be combined: - For dp[5], try dp[5-1]+10=dp[4]+10=50+10=60 - For dp[4], dp[4-4]+50=dp[0]+50=50 - For dp[3], dp[3-3]+40=dp[0]+40=40 - For dp[2], dp[2-1]+10=dp[1]+10=10+10=20 - For dp[1], dp[1-1]+10=dp[0]+10=10 So dp[5] = 60

3. Final Answer:

   60 -> Option C

4. Quick Check:

   Maximum value for capacity 5 = 60 [OK]

Hint: Trace dp array from 0 to W, update max values [OK]

Common Mistakes:

• Not considering multiple copies of items
• Incorrect dp array indexing
• Mixing up weights and values

4. Identify the bug in this Unbounded Knapsack C# code snippet:

int W = 7;
int[] wt = {1, 3, 4};
int[] val = {10, 40, 50};
int[] dp = new int[W + 1];
for (int i = 0; i < W; i++) {
    for (int j = 0; j < wt.Length; j++) {
        if (wt[j] <= i) {
            dp[i] = Math.Max(dp[i], dp[i - wt[j]] + val[j]);
        }
    }
}
Console.WriteLine(dp[W]);

medium

A. The inner loop should run from 1 to wt.Length.

B. The condition should be wt[j] < i instead of wt[j] <= i.

C. The dp array size should be W instead of W + 1.

D. The outer loop should run from 0 to W inclusive (i <= W).

Solution

1. Step 1: Check loop ranges for dp array access

   The dp array has size W+1, so valid indices are 0 to W inclusive. The outer loop runs only to i < W, missing dp[W].

2. Step 2: Identify impact of loop range

   Because dp[W] is not computed, the final answer dp[W] will be incorrect or default zero.

3. Final Answer:

   The outer loop should run from 0 to W inclusive (i <= W). -> Option D

4. Quick Check:

   Loop must cover full dp array size = The outer loop should run from 0 to W inclusive (i <= W). [OK]

Hint: Loop dp array fully from 0 to W inclusive [OK]

Common Mistakes:

• Using wrong loop bounds causing incomplete dp calculation
• Incorrect dp array size
• Wrong comparison operators in conditions

5. Given items with weights [2, 3, 5] and values [50, 70, 100], and a knapsack capacity of 8, what is the maximum value achievable using the Unbounded Knapsack approach?

hard

A. 270

B. 200

C. 250

D. 220

Solution

1. Step 1: Calculate dp array values for capacities 0 to 8

   Initialize dp[0]=0. - dp[1]=0 (no item fits) - dp[2]=max(dp[2], dp[0]+50)=50 - dp[3]=max(dp[3], dp[0]+70)=70 - dp[4]=max(dp[4], dp[2]+50=50+50=100, dp[1]+70=0+70=70)=100 - dp[5]=max(dp[5], dp[0]+100=100, dp[3]+50=70+50=120, dp[2]+70=50+70=120)=120 - dp[6]=max(dp[6], dp[4]+50=100+50=150, dp[3]+70=70+70=140, dp[1]+100=0+100=100)=150 - dp[7]=max(dp[7], dp[5]+50=120+50=170, dp[4]+70=100+70=170, dp[2]+100=50+100=150)=170 - dp[8]=max(dp[8], dp[6]+50=150+50=200, dp[5]+70=120+70=190, dp[3]+100=70+100=170)=200

2. Step 2: Check for better combinations

   Try combining items differently: - Using four 2-weight items: 4*50=200 - Using two 3-weight + one 2-weight: 2*70+50=190 - Using one 5-weight + one 3-weight: 100+70=170 - Using one 5-weight + one 3-weight + one 2-weight (over capacity) No better than 200 found, but check dp[8] carefully: - dp[8] = max(200, 190, 170) = 200

3. Step 3: Re-examine for missed combinations

   Try dp[8] = dp[3] + dp[5] = 70 + 120 = 190 (less than 200) Try dp[8] = dp[2] + dp[6] = 50 + 150 = 200 Try dp[8] = dp[0] + 100 (weight 5) + dp[3] (weight 3) = 100 + 70 = 170 Try dp[8] = dp[4] + dp[4] = 100 + 100 = 200 Try dp[8] = dp[7] + dp[1] = 170 + 0 = 170 Try dp[8] = dp[6] + dp[2] = 150 + 50 = 200 Try dp[8] = dp[5] + dp[3] = 120 + 70 = 190 Try dp[8] = dp[8-5] + 100 = dp[3] + 100 = 70 + 100 = 170 Try dp[8] = dp[8-3] + 70 = dp[5] + 70 = 120 + 70 = 190 Try dp[8] = dp[8-2] + 50 = dp[6] + 50 = 150 + 50 = 200 No combination exceeds 200, so 200 is max.

4. Final Answer:

   200 -> Option B

5. Quick Check:

   Max value for capacity 8 = 200 [OK]

Hint: Build dp array stepwise, combine items for max value [OK]

Common Mistakes:

• Ignoring multiple copies of items
• Not checking all item combinations
• Incorrect dp array updates